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Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 1 PART 3 INTEGRAL CALCULUS IN SPACE REVIEW OF RIEMANN INTEGRAL FOR y = f(x) Recall that we started with a continuous function defined on a closed interval [a,b] and above the x – axis. And we asked what is the area underneath that function. We did: a) Partition interval [a,b] into n – subintervals: a = x0 < x1 < ....xi -1 < xi < ..... < b = xn (For simplicity, we will use uniform partition so that each subinterval has equal length) b) Chose a value on each subinterval: ci Î [ xi -1 , xi ] c) Calculated the area of elementary rectangle: A( Ri ) = f (ci ) × ( xi - xi -1 ) = f (ci )Dx n d) Add it all up and we got an approximate answer: A » å A( Ri ) this is called the i =1 Riemann sum e) The answer gets better and better the more rectangles we use (refine the partitioning). n With uniform partition we can now take the limit A = lim n®¥ å A( Ri ) and if this limit i =1 existed we called it the definite integral. b n f) In general, we have ò f ( x)dx = lim n®¥ å f (ci )Dx which doesn’t have to represent a i =1 area anymore. If the limit exists, we call the function integrable over the interval [a,b] and the integral is called the Riemann integral. The major result was that continuous functions are integrable. Notes and caveats: a) We don’t have to use uniform partitioning, we can use any partitioning but then the limit would need to be specified in different terms. b) If a function is continuous on [a,b] then it is integrable, but you can have a function with finitely many jump discontinuities and still integrable. (Advanced Calculus) c) Monotone functions are integrable d) The Riemann integral is not “synonymous with area” Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 2 DEFINITION OF DOUBLE INTEGRALS OVER RECTANGLES We will motivate the definition by volumes underneath surfaces (defined by non-negative functions z = f ( x, y ) ³ 0 ) just the same way as we did for the 2 dimensional case with areas underneath non-negative functions y = f ( x) ³ 0 . Suppose that the function z = f ( x, y ) ³ 0 is defined over a rectangle R = {( x, y ) : a £ x £ b, c £ y £ d } = [a, b] ´ [c, d ] a) Consider a two-dimensional partition P of the rectangle into subrectangles: a = x0 £ x1 £ .... £ xi -1 £ xi £ ... £ xn = b c = y0 £ y1 £ .... £ y j -1 £ y j £ ... £ ym = d b) Select a subrectangle Rij = [ xi -1 , xi ] ´ [ y j -1 , y j ] and select a point inside of it: ( xi* , y* ) Î Rij . The area of this rectangle is DAij = ( xi - xi -1 ) × ( y j - y j -1 ) = Dxi Dy j j The elementary volume of the function defined over this elementary rectangle is DVij = f ( xi* , y* ) × DAij j c) The approximation to the volume underneath the function over the entire rectangle would be, approximately, the sum of all these elementary volumes and this sum is going to be precise in the limit as both n and m (the number of subrectangles) goes to infinity. In order to express such a limit as a single limit, we define the “norm” P as the maximal of the diagonals or each subrectangle: P = max i , j {diagonalRij } . If this norm goes to zero (over all possible partitionings), we are getting more and more subrectangles and the approximation is getting better. d) The double sum (we need double sum since we have two indices over which we are n m n m finding): åå DV = åå f ( x , y )DA i =0 j =0 ij i =0 j =0 * i * j ij is called the double Riemann sum. We thus have the following formula for double Riemann integral over the rectangle R: m n V = lim P ®0 åå DVij @ òò f ( x, y )dA . If this limit exists, then the function z = f ( x, y ) is i =1 j =1 R called integrable and if z = f ( x, y ) ³ 0 then the limit V is called the volume. Note that, Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 3 in general, the double integral is not the same thing as volume (since we think of volumes as non-negative numbers). THE MEANING OF INTEGRABILITY It could be shown that the above limit does not matter on how we partition the rectangle R. The meaning of the above limit is that we can make the distance between the sum and the limit arbitrarily small by choosing an appropriate partitioning. (for every possible partitioning), This means that there exists a partitioning P, such that åå f ( x , y ) - V i j i j is arbitrarily small. This means that if someone gives me an arbitrarily small positive number e > 0 I can always find a correspondingly small d > 0 and a partition P such that P < ¶ Þ åå f ( x , y ) - V i j i j <e It should be noticed that as before, if the function f(x,y) is continuous, then it is integrable. (So, continuity is stronger then integrability). In fact, even if the function has finitely many jump discontinuities, then it still remains integrable. PROPERTIES OF DOUBLE INTEGRALS OVER RECTANGLE We also have the following properties, which could be proven directly from the definition. These properties hold for general regions, not just for rectangles: a) Linearity: òò [af ( x, y) ± bg ( x, y)]dA = a òò f ( x, y)dA ± b òò g ( x, y)dA R R R b) Monotonicity: f ( x, y ) £ g ( x, y ) Þ òò f ( x, y )dA £ òò g ( x, y )dA R R c) Additivity: òò R1 È R2 f ( x, y )dA = òò f ( x, y )dA + òò f ( x, y )dA (for disjoint regions) R1 R2 d) Generalized triangle inequality: òò f ( x, y)dA £ òò R R f ( x, y ) dA Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 4 ITERATED INTEGRALS AND THE FUBINI THEOREM Now, after the definition, we need some way of calculating the limit. For a simple integral, we have the Fundamental Theorem of Calculus (the Newton-Leibniz formula). For the double integral we have the FUBINI’S THEOREM: This theorem transforms double integrals into iterated (nested) integrals. Theorem: (Fubini) If f(x,y) is integrable of a rectangle R = {( x, y ) : a £ x £ b, c £ y £ d } = [a, b] ´ [c, d ] é b d ù é d b ù Then òò R f ( x, y )dA = ò ê ò f ( x, y )dy údx = ò ê ò f ( x, y )dx údy a ëc û c ëa û The “partial integrals” inside the parenthesis are evaluated by treating the other variable as constants (as with partial derivatives). Usually one does not write the parenthesis, for ébd ù d b example: ò ê ò f ( x, y )dx údy = ò ò f ( x, y )dxdy . Notice that if the region is a rectangle, the c ëa û c a order of integration is not important. Problem: òò ( x - 3 y )dA where R is the rectangle [0, 2] ´ [1, 2] described by equations: 2 Evaluate R 0£ x£2 1£ y £ 2 Solution: We obtain: 2 2 2 y 4 òò ( x y)dA = ò ò x ydydx = ò 9 dy = 18 . We should get the same answer if we changed the 2 2 R 0 1 0 2 2 òò x ydA = ò ò x ydxdy 2 2 order of integration: R 1 0 Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 5 Problem: Let A(R) denote the area of the region R. Show that òò kdA = k × A( R) R Solution: Using linearity property òò kdA = k òò dA = k × A( R) R R Problem: (Using the additivity property for piecewise continuous function) ì1 [0,3] ´ [0,1] Evaluate òò f ( x, y)dA where R f ( x, y ) = í î2 [0,3] ´ (1, 2] and the rectangle R is R = {( x, y ) : 0 £ x £ 3, 0 £ y £ 1} = [0,3] ´ [0, 2] Solution: The Region R splits into two subregions: R1 = {( x, y ) : 0 £ x £ 3, 0 £ x £ 1} = [0,3] ´ [0,1] and the value is f(x,y) = 1 and R2 = {( x, y ) : 0 £ x £ 3, 0 £ x £ 1} = [0,3] ´ [1, 2] and the value is f(x,y) =2 We calculate: òò R = R1 È R2 f ( x, y )dA = òò f ( x, y )dA + òò f ( x, y )dA = 1A( R1 ) + 2 A( R2 ) = 9 = R1 R2 CAVALIERI’S PRINCIPLE (Volumes of solids with known cross section) Gives a partial justification of the Fubini theorem for a non negative function. Proposition: (Cavalieri’s Principle) If the cross sectional area is known to be A(x), then the volume of the solid from x = a to b x = b is given by the integral V = ò A( x)dx a Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 6 Proof (Sketch): Suppose that f(x,y) is non-negative. Partition the interval [a,b] and fix a point inside, say ck Î [ xk -1 , xk ] . Consider the vertical plane given by the equation x = ck . This is a vertical plane, perpendicular to the x – axis. It intersects the surface in a curve h( y ) = f (ck , y ) . d d The area underneath this curve is: A(ck ) = ò h( y )dy = ò f (ck , y )dy . We can consider the c c area to be approximately constant on this interval. Now, consider the totality of these vertical areas. The volume is approximately the sum of the vertical cylinders: n n d V » å A(ck )Dxk = å ò f (ck , y )dy which is a Riemann sum. The limit of this sum (if it k =0 k =0 c n d é b d ù exists) is the Riemann integral: V = lim n®¥ å ò f (ck , y )dy = ò ê ò f ( x, y )dy ú dx which k =0 c a ëc û agrees (for non-negative functions over rectangles with the Fubini Theorem.) Of course, one would have to show as a matter of proof that the volume from the Fubini is the same as the volume from the Cavaliery. Problem (Volume of a pyramid) Calculate the volume of a square base pyramid with height h and base b Solution: Place the pyramid in such such a way that the height is on the x – axis and the base square is perpendicular to it. And the vertex is at the origin. The interval of integration will then be [0, h]. The cross section is a square perpendicular to the x axis and we need a formula for A(x). Imagine that the intersecting plane is x – units away from the center. Also suppose that the square base (x units away from the center) has sides = s (which depends on x of course). The pyramid with height x is geometrically similar to the s b b pyramid with height h. This implies that = and from here we get s = x and the x h h Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 7 b2 2 area of the cross sectional square is A( x) = s 2 = x and using Cavaliery we obtain h2 h h b2 2 1 V = ò A( x)dx = ò 2 x dx = b 2 h 0 0 h 3 DOUBLE INTEGRALS OVER (MORE GENERAL) SIMPLE REGIONS Fubini theorem extends to regions which are either x-simple or y-simple. (Also called horizontally or vertically simple). Also called type1 and type 2 elementary regions. We say that an elementary region R is x-simple, if it could be described as: ì a£ x£b ü R = í( x, y ) : ý In this case the Fubini Theorem has the form: î f ( x) £ y £ g ( x) þ b g ( x) òò f ( x, y)dA = ò ò R a f ( x) f ( x, y )dydx Notice that x-simple region is characterized by constant limits for x and variable limits for y. The constant limits always go to the outer integral We say that the region R is y-simple, if it could be described as: ì c£ y£d ü R = í ( x, y ) : ý In this case the Fubini Theorem has the form: î h( y ) £ x £ k ( y ) þ d k ( y) òò f ( x, y)dA = ò ò R c h( y ) f ( x, y )dxdy An elementary region R is simple if it is either x-simple or y-simple. Every simple region could be described as either x or y simple. For example, we can describe the x-simple Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 8 ì a£ x£b ü region R = í( x, y ) : ý as a y-simple region (provided that the functions f î f ( x) £ y £ g ( x) þ ì f -1 ( y ) £ x £ g -1 ( y ) ü and g are invertible as R = í( x, y ) : ý î f (a ) £ y £ g (b) þ Problem: Calculate the integral òò xydA where R is the triangular region in the first quadrant with R vertices (0,0), (1,0) and (0,1) Solution: First, draw the region. Notice that you get a right triangle and the hypotenuse goes from (1,0) to (0,1). The equation of this hypotenuse is y = -x + 1 And now it is easy to describe it: {( x, y ) : 0 £ x £ 1, 0 £ y £ - x + 1} and you have: 1 - x +1 1 é y2 ù 1 x 1 òò xydA = ò R 0 ò 0 xydydx = ò x ê údx = ò dx = 0 ë2û 0 2 4 Problem: Calculate the integral òò xydA where R is the triangular region in the first R quadrant with vertices (0,0), (1,0) and (1,1) regarding the region as y – simple Solution: We can describe the region as a x – simple region: R2 = {( x, y ) : 0 £ x £ 1, 0 £ y £ 1 - x} 1 1- x 1- x 1 é y2 ù 1 1 And we have: òò xydA = ò ò R 0 0 xydydx = ò x ê ú dx = ò x(1 - x) 2 dx...... 0 ë 2 û0 20 CHANGE OF THE ORDER OF INTEGRATION Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 9 Problem: Calculate the integral òò xydA where R is the triangular region in the first R quadrant with vertices (0,0), (1,0) and (0,1) using x– simple region and then as a y – simple region. Solution: (Sketch the triangle) a) As a x-simple region we can describe the triangle as: ì 0 £ x £1 ü R = í ( x, y ) : ý and the integral becomes: î 0 £ y £ - x + 1þ 1 1- x 1- x é y2 ù 1 1 1 1 òò 0 0 xydydx = ò x ê ú dx = ò x(1 - x) 2 dx = 0 ë 2 û0 20 24 (Notice that once the limits for x are fixed…to find the variable limits for y, you move horizontally, enter the region at y = 0 and leave the region at the line y = -x + 1) ì 0 £ y £1 ü b) As a y – simple region we describe it as R = í( x, y ) : ý and the integral î 0 £ x £ 1- yþ 1 1- y becomes ò ò xydxdy 0 0 Problem: 1 x2 Calculate the integral ò ò xydydx directly and then change the order of integration 0 0 Solution: 2 1 x2 éx ù x 2 1 1 é y2 ù 1 5 x 1 ò ò xydydx = ò ê ò xydy údx = ò x ê 2 ú dx = ò 2 dx = 12 . 0 ê0 ú ë û0 0 0 ë û 0 0 Now, let us evaluate the same integral by changing the order of integration. To do that, draw the picture of the x-simple region (underneath the parabola) described by the ì 0 £ x £1 ü ì 0 £ y £1 ü ï ï equations í 2ý and rewrite it as a y-simple region, we get: í ý . The î0 £ y £ x þ ï y £ x £ 1ï î þ Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 10 1 1 1 1 1 é x2 ù 1 æ1 yö 1 é y 4 y3 ù 1 integral now becomes: ò 0 ò xydxdy = ò y ê 2 ú dy = ò y ç 2 - 2 ÷ dy = ò ê 8 - 6 ú dy = 12 0 ë û y 0 è ø 0 ë û0 y which is the same result. Problem: 1 1 ò ò sin y dydx 2 Show that the only way you can evaluate is by changing the order of 0 x integration. Answer: The reason for the order change is the impossibility to integrate ò sin y 2 dy . The 1 final answer, after you change the order is: . 2 Problem (Region splits into two different parts) òò sin xydA over the region in the xy-plane bounded by x = y 2 Evaluate and the line R y+x=2. Solution: If we view the region as x-simple, it splits into the following two regions: ì 0 £ x £1 ü ï ï R1 = í ý ï- x £ y £ x ï î þ ì 1£ x £ 4 ï ü ï R2 = í ý ï- x £ y £ 2 - x ï î þ The resulting integral will be evaluated using the additivity property: òò R1 È R2 f ( x, y )dA = òò f ( x, y )dA + òò f ( x, y )dA (which you can perform yourself). R1 R2 Problem: Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 11 Evaluate the above integral using the change of order Solution: Viewing the region as a y-simple region, we can describe it as: ì -2 £ y £ 1 ü R=í 2 ý which is easy to evaluate. î y £ x £ 2 - yþ Answer: 99 In both cases we get 2 AREA OF A BOUNDED REGION: R is given clearly by the integral: A( R) = òò 1dA = òò dA . R R Problem: (Formula for the area between two curves from Calculus2) If the region R is bounded by the curves f(x) and g(x), assuming that g ( x) ³ f ( x) for all values in the interval [a,b] intersecting at the points x = a and x = b then the area is given b by the integral A = ò [ f ( x) - g ( x)]dx a Solution: ì a£ x£b ü The region could be described as R = í ý and we have î f ( x) £ y £ g ( x) þ b g ( x) b A=ò ò dydx = ò [ g ( x) - f ( x)]dx which is the formula from Calculus2 course. (Area a f ( x) a between two curves). Problem: Use double integrals to calculate the area of the triangular region with vertices (0,0), (1,0) and (0,1) Solution: Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 12 1 - x +1 1 1 1 dydx = ò x [ y ]0 dx = ò x(- x + 1)dx = - x +1 òò dA = ò ò R 0 0 0 0 6 VOLUMES OF SOLIDS AND MASSES OF PLANAR LAMINAS If z = f ( x, y ) is a “regular” non-negative function, then the volume under it and over a region R is given by: Volume = òò f ( x, y )dA R If the function represents density (which now can vary in both directions, not just ones like in Calculus2), we have the mass (of a planar lamina): Mass ( R) = òò f ( x, y )dA R Problem: Calculate the mass of the planar region R = {( x, y ) : 0 £ x £ 1, 0 £ y £ 2} with variable density given by function f ( x, y ) = 4 - x 2 - y Solution: 2 1 16 M = ò ò (4 - x 2 - y )dxdy = 0 0 3 Problem: Calculate the volume under f ( x, y ) = 4 - x 2 - y over the region R = {( x, y ) : 0 £ x £ 1, 0 £ y £ 2} Solution: Notice that this is exactly the same as the problem above. Problem: Calculate the mass of the region bounded by y = x 2 and y = 2 - x 2 the density is r ( x, y ) = xy Answer: 1 2- x2 7 M = òò r ( x, y )dA = ò ò xydydx = R 0 x2 24 Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 13 Problem: (Volume of a sphere) Find the volume of a sphere with radius R, with equation: x2 + y 2 + z 2 £ R2 . Solution: The projection onto the plane is x 2 + y 2 £ R 2 The projection onto the x – axis is: x 2 £ R 2 . The resulting region R could be described by these equations: -R £ x £ R - R2 - x2 £ y £ R2 - x2 Solving for z explicitly gives: z = ± R 2 - x 2 - y 2 . We will take the positive value and multiply the result by 2. Or, even better, restrict ourselves to the first octant and multiply the result by 8. (And also by 2, so altogether by 16). R R2 - x2 R R2 - x2 We have: V = 2 × ò ò R - y - x dydx = 16 × ò ò R 2 - y 2 - x 2 dzdydx 2 2 2 - R - R2 - x2 0 0 Problem: (Volume under a paraboloid) Find the volume of the solid that lies under the paraboloid z = x 2 + y 2 and above the region in the xy-plane bounded by y = x 2 and y = x . Solution: Drawing the region, we could describe it is a x-simple region: 0 £ x £1 x2 £ y £ x 1 x We get: V = òò ( x + y )dA = ò ò ( x 2 + y 2 )dydx 2 2 R 0 x2 Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 14 Problem: (Volume of a Tetrahedron ...given by equation) Find the volume of a tetrahedron bounded by the coordinate planes and the plane 3x + 6y + 4z -12 = 0 Solution (Draw the picture) The projection of the tetrahedron onto the xy plane we get by setting z = 0, which gives the equation 3x + 6y = 12 or x + 2y = 4 which is a line. Since the tetrahedron is bounded by the coordinate planes, we can describe this triangular region as: {( x, y ) : 0 £ y £ 2, 0 £ x £ 4 - 2 y} as a y – simple region. 2 4- 2 y 1 And we have V = ò 0 ò 4 [12 - 3x - 6 y ]dxdy ….and the rest is easy 0 Problem (Volume of a Tetrahedron… given by vertices) Find the volume of a tetrahedron given by vertices (0,0,0), (1,1,0), (0,1,0), (0,1,1) Solution: (Draw all the points to see how the tetrahedron is formed) The projection of this tetrahedron into the xy – plane is the triangular region given by vertices (0,0), (1,1) and (0,1)….sketch the xy- region. This will give us the limits for x and y. Next, notice that if you enter the tetrahedron from the bottom (z = 0), you leave it from the plane given by vertices (0,0,0), (1,1,0), (0,1,1). We need the equation of plane. Recall that if a plane is given by points A, B,C, then the vector equation is X = A + t(B – A) +s (C – A) and we get the parametric equations by rewriting into coordinates. In our case, we have: x=1–t–s y=1–t z=s And eliminating the parameter t yields equation z = y – x. and thus, we can describe the region as: Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 15 ì 0 £ x £ 1ü 1 1 R = í( x, y ) : ý and we have V = ò ò ( y - x)dydx î x £ y £ 1þ 0 x Problem (Volume between two surfaces) Calculate the volume of the solid bounded by the paraboloids z = 2 - x 2 - y 2 and z = x2 + y 2 Solution: Notice that we have a paraboloid which opens down (with z – intercept (0,0,2) and paraboloid that opens up with vertex at the origin. We need to calculate the intersection so that we know where the x and y enter and leave the solid. And then we use the formula òò z R top - òò zbottom = òò ( ztop - zbottom ) using the additivity of integrals. R R The intersection: z = 2 - x2 - y 2 = 2 - ( x2 + y 2 ) = 2 - z Þ 2z = 2 Þ z = 1 So, in the horizontal plane z = 1 the two surfaces intersect in a circle given by x 2 + y 2 = 1 and now it is easy to set up the integral: 1 1- x 2 1 1- x 2 V= ò ò é(2 - x - y ) - ( x + y ) ùdydx = ò ò [2 - 2 x 2 - 2 y 2 ] 2 2 2 2 ë û -1 - 1- x 2 -1 - 1- x 2 Problem: (Above paraboloid and inside a sphere) Calculate the volume which lies above the paraboloid z = x 2 + y 2 and inside the sphere x2 + y 2 + z 2 = 2 Solution: Try to imagine what it looks like. The rotating paraboloid goes up from the origin and meets the sphere. The z coordinate changes from the paraboloid below to the sphere. We again need to subtract the two integrals (corresponding to the z): Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 16 òò R 2 - x 2 - y 2 dA - òò ( x 2 + y 2 )dA . What is the region of integration for the x and y R coordinate? We find it by looking at the intersection between the paraboloid and the sphere, or, solving the equations z = x 2 + y 2 and x 2 + y 2 + z 2 = 2 simultaneously, we get z = 1. (The other solution z = -2 does not apply because the region is above the xy-plane). Thus, the region R could be described by the equations: -1 £ x £ 1 - 1 - x2 £ y £ 1 - x2 1 1- x 2 é 2 - x 2 - y 2 - ( x 2 + y 2 ùdydx And the resulting integral is: V = ò ò -1 - 1- x 2 ë û Problem: (Volume of a solid bounded by two parabolic cylinders and planes) find the volume of a solid which is bounded by: z =1 z=y y = x2 y = 2 - x2 Solution: The first two equations are planes and the second two equations are parabolic cylinders. First, notice that: òò 1dA gives the volume underneath the plane z = 1 R òò ydA gives the volume underneath the plane z = y R The resulting volume is: V = òò 1dA - òò ydA = òò (1 - y )dA R R R Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 17 The region R is the intersection of the two parabolas in the plane (the projected cylinders). Setting x 2 = 2 - x 2 Þ x = ±1 . Now, we can describe the region R using the equations: -1 £ x £ 1 x2 £ y £ 2 - x2 1 2- x2 And, we get: V = ò ò -1 x 2 (1 - 2 y )dydx Problem (Same as above, different numbers) Find the volume of the solid bounded by planes z = 6, z = 2y and the parabolic cylinders y = x2 , y = 2 - x2 Solution: 1 2- x2 32 òò (6 - 2 y)dA = ò ò R -1 x 2 (6 - 2 y )dydx = ....... - 3 Problem (Volume of the region common to the intersection of two cylinders) Calculate the volume of the region common to the circular cylinder x 2 + y 2 = a 2 and x2 + z 2 = a2 Solution: (Once cylinder is around the x-axis and the other cylinder is around the z axis) The intersection is: x 2 + y 2 = a 2 = x 2 + z 2 Þ y 2 = z 2 Û y = ± z = ± a 2 - x 2 And we can describe the region as: -a £ x £ a - a2 - x2 £ y £ a2 - x2 And we have: Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 18 ( ) a a2 - x2 a a2 - x2 é a -z - a - z ùdydx = 2 ò ò ò ò a 2 - z 2 dydx 2 2 2 2 ê ë ú û - a - a2 - x 2 -a - a2 - x2 a a -x 2 2 a 16a 3 = 8ò ò a - z dydx = 8ò (a - x )dx = 2 2 2 2 0 0 0 3 FURTHER APPLICATIONS OF DOUBLE INTEGRALS AVERAGE VALUE OF A FUNCTION z = f(x,y) Recall, that in the one-dimensional case the Mean Value Theorem for Integrals said that there exists a point in the interval [a,b], say c Î [a, b] such that the area underneath the curve y = f(x) over this interval is the same as the area of the rectangle with the interval b as its base and f(c) as its height. In math symbols: f (c) × (b - a ) = ò f ( x)dx . The a b 1 b-a ò value f (c) = f ( x)dx was called the average value of the function on the interval a [a,b]. The average value exists whenever the function is continuous. We have a simple generalization for double integrals. The Mean Value Theorem for Double Integrals says that there exists a point ( x0 , y0 ) Î R such that the volume of underneath the function f(x,y) is the same as the volume of the box with base R and height f ( x0 , y0 ) . If A( R) denotes the area of this (bounded) region, the Intermediate Value Theorem for double integrals has the form: f ( x0 , y0 ) × A( R) = òò f ( x, y )dA . The R 1 A( R) òò value f ( x0 , y0 ) = f ( x, y )dA is called the Average Value of the Function f(x,y) R over the region R. Proof: (it is simple and instructive, but optional) Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 19 a) Since f is continuous on the (compact) region R, it attains its maximum (M) and minimum (m) on or inside the region R We have m £ f ( x, y ) £ M and now we integrate the inequality and use the monotonicity of integral: òò mdA £ òò f ( x, y)dA £ òò MdA R R R m òò dA £ òò f ( x, y )dA £ M òò dA R R r m × A( R) £ òò f ( x, y )dA £ M × A( R) R Now, by the Intermediate Value Theorem, which says that a continuous function f(x,y) attains all the values between m and M, we conclude that there exists a value for which the inequality will result. Problem: Calculate the average value of a function z = xy over the region bounded by the parabola y = x 2 from x = 0 to x =1. Solution: 1 1 First, we calculate the area: A( R ) = ò x 2 dx = 0 3 2 1 x2 éx ù x 2 1 1 é y2 ù 1 5 x 1 Next, we calculate òò f ( x, y )dA = ò ò xydydx = ò ê ò xydy údx = ò x ê ú dx = ò dx = 0 ê0 ú ë 2 û0 2 12 R 0 0 ë û 0 0 1 1 1 The average value is then: f ( x0 , y0 ) = òò f ( x, y)dA = 3 × 12 = 4 A( R ) R Notice that we did not have to calculate at which point the average value occurs. SURFACE AREA We know how to integrate the surface area for surfaces of revolution. Recall, that if we rotate the curve y = f ( x) on the interval [a,b]around, say, the x-axis, then the surface Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 20 2 æ dy ö b area is given by S = 2p ò f ( x)ds where ds = 1 + ç ÷ dx is the elementary arc length. a è dx ø Now, suppose that you have a surface parametrized by: r (u , v) = [ x(u, v), y (u, v), z (u, v) ] . The domain is a rectangle (u , v) = [a, b] ´ [c, d ] . Now, the usual partitioning of this domain will give a corresponding partitioning of the surface into elementary patches. Consider such an elementary patch. Starting with the left corner, construct the two ¶r ¶r tangent vectors Tu = and Tv = . Now, from the definition of partial derivative we ¶u ¶v ¶r r (u + Du.v) - r (u , v) have: Tu = » Þ Tu Du » r (u + Du , v) - r (u , v) . Similarly we have ¶u Du Tv × Dv » r (u , v + Dv) - r (u , v) and now look at the parallelogram spanned by Tu × Du and Tv × Dv where Du and Dv are the length and with of the domain elementary rectangle. We know that the cross product Tu Du ´ Tv Dv = Tu ´ Tv Du Dv = N Du Dv where N is the normal vector, will give the area of a parallelogram spanned by the two tangent vectors Tu × Du and Tv × Dv . We will use this as the approximation to the surface area of the elementary patch. Summing (integrating), we get the formula for the surface area of the ¶r ¶r surface: S = òò Tu ´ Tv dudv = òò ´ dudv = òò N (u , v) dudv where N(u,v) is the R R ¶u ¶v R normal vector. which works well for surfaces given parametrically. Problem: Show that i j k éy zu xu zu xu yu ù N = xu yu zu = ê u , , ú y zv xv zv xv yv û xv yv zv ë v é ¶ ( y , z ) ¶ ( x, z ) ¶ ( x, y ) ù ê ¶ (u , v) , ¶ (u , v) , ¶ (u, v) ú ë û The components of the normal vector are the Jacobians. Problem (Surface Area formula for explicitly defined function) Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 21 Derive a formula for the surface area of a function z = f(x,y) Solution: æ ¶f ¶f ö We know that the unit normal vector is N = ç , , -1÷ and its length is è ¶x ¶y ø 2 æ ¶f ö æ ¶f ö 2 N = ç ÷ + ç ÷ +1 è ¶x ø è ¶y ø 2 æ ¶f ö æ ¶f ö 2 The formula for such surfaces becomes then: S = òò ç ÷ + ç ÷ + 1dA R è ¶x ø è ¶y ø Problem: Find the surface area of a surface given parametrically as: x=u y=v z = uv 0 £ u £1 0 £ v £1 Solution: ¶r ¶r Here we use the formula: S = òò Tu ´ Tv dudv = òò ´ dudv R R ¶u ¶v We have: r (u , v) = (u , v, uv) ¶r = (1, 0, v) ¶u ¶r = (0,1, u ) ¶v i j k ¶r ¶r ¶r ¶r ´ = 1 0 v = (-v, -u ,1) Þ ´ = 1 + u 2 + v2 ¶u ¶v ¶u ¶v 0 1 u 1 1 S = ò ò 1 + u 2 + v 2 dudv 0 0 Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 22 Problem: (Surface Area of a Sphere) Find the surface area of a sphere of radius R = a using the spherical coordinates parametrization. Solution: (Only a hint, the details you should fill in) We have: using u = q and v = f r (u , v) = [a sin v cos u, a sin v sin u , a cos v] 0 £ u £ 2p 0£v £p 2p p S After all the calculation, we end up with: = ò òa sin vdvdu = 2p a 2 2 2 0 0 Problem: Set up the surface area integral of the paraboloid z = 1 - x 2 - y 2 above the xy-plane Solution: This surface is given explicitly and therefore, we use the formula: 2 1 1- x 2 æ ¶f ö æ ¶f ö 2 S = òò ç ÷ + ç ÷ + 1dA = ò ò 1 + 4 x 2 + 4 y 2 dydx R è ¶x ø è ¶y ø -1 - 1- x 2 Problem: (Formula for the surface area of solids of revolution) Derive the formula for the surface area of surfaces of revolution using double integrals. Solution hint: Suppose that we generate a surface by revolving the graph of y = f ( x) and [a,b] around the x-axis. The equation of the resulting surface is: y 2 + z 2 = ( f ( x) ) 2 We can parametrize this equation as: Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 23 x=u y = f (u ) cos v z = f (u ) sin v a£u£b 0 £ v £ 2p From here we calculate: i j k ¶r ¶r ´ = 1 f '(u ) cos v f '(u ) sin v ¶u ¶v 0 - f '(u ) sin v f '(u ) cos v ¶r ¶r b S = òò ´ dA = 2p ò f (u ) 1 + ( f '(u ) ) du 2 D ¶u ¶v a FIRST MOMENTS, CENTROIDS OF LAMINAS (with variable densities) In Calculus 2 we learned how to calculate moments and centroids of planar regions (laminas) with either constant densities or with densities that are proportional to the coordinates. We were not able to calculate moments or centroids for densities which were functions of both x, y. (Like for example, if the density is proportional to the distance from the origin). With double integrals, we can accomplish this task. (But with triple integrals, we can even consider densities which are function of all three coordinates. Recall the equation mass = density X area (We are in a 2D case). If the density is constant, then we have mass = r × òò dA . In general, we have mass = m = òò r ( x, y )dA . R R (Notice, that you have another physical interpretation of a double integral. It is the mass of a planar object with variable density). Now, subdivide the region R into subrectangles in the usual way and look at a typical subrectangle Rij = [ xi -1 , xi ] ´ [ y j , y j -1 ] and choose a point inside of this elementary rectangle: ( xi* , y* ) Î Rij . If this point is chosen to be the j center of this rectangle, then it is the centroid and it represents the rectangle. The mass is concentrated there. Thus, the mass of the rectangle is the density times the area, or: m( Rij ) = r ( xi* , y* ) × DAij . Recall that moment = mass X distance j Thus, we calculate the moments for the elementary rectangle: a) About x-axis: DM x = é r ( xi* , y* )DAij ù × y* = y* × r ( xi* , y* )DAij ë j û j j j Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 24 b) About y-axis: DM y = é r ( xi* , y* )DAij ù × xi* = xi* × r ( xi* , y* )DAij ë j û j Now, the total moment of the lamina will be approximately the sum of the moments of the rectangles. For example, the moment with respect to the x – axis is: n m M x = åå y* × r ( xi* , y* )DAij which we recognize to be a Riemann sum. Now, assuming j j i =1 j =1 that the density is continuous, the sum converges and its limit is the integral. Thus we get the moments of the entire lamina: M x = òò y × r ( x, y )dA R M y = òò x × r ( x, y )dA Finally, recall that centroid = moment / mass We are getting the following formulas for the coordinates of the centroid: My 1 m òò X= = x r ( x, y )dA m R Mx 1 m m òò Y = = y r ( x, y )dA R m = òò r ( x, y )dA R Problem (Formulas for centroid from Calculus 2) Using the above formulas, derive the formulas for Calculus1 Now, notice that in the case that the density is constant, and the region is bounded by a function y = f(x) over the interval [a,b], the formulas agree with the formulas we obtained in Calculus2: Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 25 m = òò r ( x, y )dA = r òò dA = r A( R ) R R b f ( x) b 1 M x = r òò ydA = r ò ò ydydx = r ò ( f ( x) ) dx 2 R a 0 a 2 b M y = r òò ydA = ... = r ò xf ( x)dx R a b 1 Aò X= xf ( x)dx a b 1 1 Y = ò 2 ( f ( x) ) dx 2 Aa which is precisely what we obtained in Calculus 2. Notice that the Calc3 derivation was much “cleaner”. Adding a dimension many times results in simplification of equations. Unlike the volumes, calculating the moments and centroids is easy because the region R is (of course) given. Problem: (Formulas for centroid when the density is constant) If the density is constant, we have r ( x, y ) = K mass = K × òò dA = K × A( R) R M y = K òò xdA R M x = K òò ydA R My 1 A( R) òò X= = xdA m R Mx 1 A( R) òò Y = = ydA m R Problem: Calculate the moments and centroid of the triangular region with vertices (0,0), (1,0), (0,1), and the density r ( x, y ) = x + y Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 26 Hint for the solution: Note first that this problem would be impossible to calculate in Calculus2. Draw the triangle and notice that the region could be described as: ì 0 £ x £1 ü R=í ý î0 £ y £ - x + 1þ 1 -2 x + 2 The mass is: m = ò ò ( x + y )dydx = .......( finish) 0 0 The moments are: 1 - x +1 Mx = ò ò y ( x + 1)dydx = ... 0 0 1 - x +1 My = ò ò x( x + 1)dydx = ... 0 0 The centroid is then: My X= = ... m M Y = x = ... m ROTATIONAL MOMENTS OF INERTIA (SECOND MOMENTS) 1) Single particle rotating: Suppose that a particle rotates around a circle with radius. (Rotational motion). We define the following: dq q s a) Angular velocity: w = = where q = dt t r ds s b) Linear velocity: v = = dt t Problem: (Relation between angular and linear velocity) Show that w = rv Solution: Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 27 s qr v= = = wr t t 1 2 1 1 c) Rotational kinetic energy: K .E = mv = m(rw) 2 = mr 2 w2 2 2 2 2) Discrete system of particles rotating about a line Now, suppose that you have a system of particles rotating with the same angular velocity (but they have different different radii and masses). The rotational kinetic energy for 1 particle (i) is K .E (i ) = mi ri 2 w2 and for the entire system it is: 2 n n 1 1 1 1 K .E = å mi ri 2 w2 = w2 å mi ri 2 = w2 I = Iw2 and comparing this with the formula i =1 2 2 i =1 2 2 for kinetic energy in linear motion (with v replaced by w) we see that the quantity n I = å mi ri 2 is analogous to the mass. It represents the resistance of the body (discrete i =1 system of particles) to rotational motion and it is called the moment of inertia. Mass represents the resistance to linear motion caused by a force F. The force which causes the rotational motion is called the torque and in analogy to the second Newton’s Law dw d 2q (which says that F = ma) we have t = Ia where a = = 2 is called the angular dt dt acceleration. (We talked about torque in connection with vector products, the two definitions are the same.) For example, if we rotate with respect to the x – axis, we get: I x = å yk mk . 2 k 3) Continuous particle distribution rotating: Suppose that the body is continuous (solid, as opposed to a system of discrete particles). Then, instead of summing, we integrate. Replace masses by the product of density and area. The idea is again to partition the body and select points inside each cell and calculate the discrete sum and take the limit. Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 28 We get the following formulas: I l = òò r 2 r ( x, y )dA A I x = òò y 2 r ( x, y )dA A I y = òò x 2 r ( x, y )dA A I 0 = òò ( x 2 + y 2 ) r ( x, y )dA = I x + I y A The last moment is called the polar moment (or moment about the origin). You can think of it as the moment around the z-axis. The identity I 0 = I x + I y is called the Perpendicular Axis Theorem. Problem: Imagine a rectangle with base b and height h. Calculate the second moment with respect to the x-axis assuming constant density Solution: h b h é y3 ù 1 I x = òò y r dA = r × ò ò y dydx = b × r × ê ú = b r h3 . Similarly, we calculate: 2 2 R 0 0 ë 3 û0 3 1 I y = b3 r h . 3 Problem: As an exercise, show that the inertial moments with respect to the coordinate axes of a right triangle with vertices (0,b) and (0,h) [base is b and height is h) is 1 3 Iy = b rh 12 1 I x = b r h3 12 and verify the formula on a particular triangle, with density r = 1 and vertices (1,0) and (0,1). Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 29 Answer: 1 1- x 1 Ix = ò ò y 2 dydx = 0 0 12 DOUBLE INTEGRALS IN POLAR COORDINATES Sometimes it is convenient to express the integral in polar coordinates. Recall that the ì x = r cos q ü transformation equations are given by: í ý î y = r sin q þ We have to transform both the function and the region of integration. It is always a good idea to set up the integral in Cartesian coordinates first. ì a £ r £ bü To derive the formula, take the polar region described by R = í(r , q ) : ý which is î c £q £ dþ (by the way) called polar rectangle. Notice the similarity with the Cartesian derivation. It is important that you understand all these derivations!! Partition the region in the usual way: a = r0 £ r1 £ ... £ ri -1 £ ri £ ... = b c = q 0 £ q1 £ ...q j £ q j -1 £ ... = d Now, consider the polar patch: Rij = [ri -1 , ri ] ´ [q j -1 , q j ] Recall, from Precalculus2 that the area of a circular sector with central angle q (in 1 2 radians) and radius r is A = rq 2 The patch is the region between two circular sectors with central angle Dq j = q j - q j -1 and the radii ri and ri -1 respectively. Thus, the elementary area of this patch is: ri Dq j - ri 21Dq j = ( ri 2 - ri 21 ) Dq j 1 2 1 1 DAij = - - 2 2 2 1 = [(ri + ri -1 )(ri - ri -1 )] Dq j = ri*Dri Dq j 2 Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 30 1 where ri* = (ri + ri -1 ) is the average radius. Now, as usual, select a point inside of the 2 patch: (ri # , q j# ) Î Rij . Now, we are trying to evaluate the function f(x, y) in polar coordinates. Now, suppose that: xi# = ri # cos q j# y # = ri # sin q j# j n m and consider the Riemann Sum åå f (r i =0 j =0 i # cos q j# , ri # sin q j# )DAij which will converge (in the usual sense) to the integral òò f (r cos q , r sin q )rdrdq where R* is the transformed R* polar region (under the transformation equations). We have just derived the following: ì x = r cos q ü òò f ( x, y)dA = òò f (r cos q , r sin q ) × rdA * Let í ý Then î y = r sin q þ R R* Notice that three things are involved in this: a) Transforming the function into polar coordinates b) Transforming the region of integration c) The factor r in the integrand (called the local change of area factor) Problem: 6 y Change the integral ò ò xdxdy 0 0 into polar coordinates. Solution: Draw the triangular region. Notice that you can describe as: ì pü p p ï 0 £q £ ï 4 6 4 6 R = í( r , q ) : ò ò r cos q × rdrdq = ò ò r cos q drdq * 2 4 ý and the integral becomes ï î 0£r £6ï þ 0 0 0 0 Problem (Area of a polar region formula from Calculus2) Let r = r (q ) : a £ q £ b be a closed polar region. Develop the formula for the area. Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 31 Solution: We can describe the polar region as follows: ì a £q £ b ü R * = í( r , q ) : ý î 0 £ r £ r (q ) þ b r (q ) r (q ) b é r2 ù b 1 2 And we have: A = ò ò rdrdq = ò ê ú dq = ò r (q )dq …which is the formula. a ë a 0 2 û0 2a Problem: Find the volume under the paraboloid z = 1 - x 2 - y 2 (above the xy-plane and bounded by this vertical, opening down paraboloid). Solution: In Cartesian coordinates the integral would be V = òò (1 - x 2 - y 2 )dA where the region R R of integration is the circle x 2 + y 2 = 1 . We will follow the three steps: a) f (r cos q , r sin q ) = 1 - (r cos) 2 - (r sin q ) 2 = 1 - r 2 b) The region R could be expressed in polar coordinates as a rectangular region: ì0 £ q £ 2p ü R* = í ý î 0 £ r £1 þ 2p 1 p òò f ( x, y )dA = òò f (r cos q , r sin q ) × rdA* = ò ò (1 - r )rdrdq = ... = 2 c) R R* 0 0 2 Problem: Find the volume of the solid bounded by paraboloid z = x 2 + y 2 and sphere x2 + y 2 + z 2 = 6 Solution Lets set it up in Cartesian coordinates first. The intersection is: z + z 2 = 6 Þ z = 2 and in this horizontal plane the intersection curve is a circle x 2 + y 2 = 2 . This is the region R. Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 32 R = {( x, y ) : - 2 £ x £ 2, - 2 - x 2 £ y £ 2 - x 2 } and we have: 2 2- x2 V = òò é 6 - x 2 - y 2 - ( x 2 + y 2 ) ù = ò ò é 6 - x 2 - y 2 - ( x 2 + y 2 ) ùdydx ë û ë û R - 2 - 2- x 2 The region R gets transformed in polar coordinates into: 2p 2 R* = {(r , q ) : 0 £ r £ 2;0 £ q £ 2p } and the integral is: V = ò òr 0 0 16 - 4r 2 drdq and the inside integral is independent of the outside and a basic substitution works: u = 4 - r2 2 2 du = -2rdr -1 1 0 16 ò 0 r 16 - 4r 2 dr = 2 ò r 4 - r 2 dr = 0 r =0Þu =4 = 2 × ò u 2 du = 2 4 3 r =2Þu =0 2p 16 64p And then we have: V = ò0 3 dq = 3 Problem: Set up the integral (in polar coordinates) for the volume of the solid bounded by the upper hemisphere x 2 + y 2 + z 2 = 16 and below by the cylindrical surface x 2 + y 2 £ 4 Solution: In Cartesian coordinates, the integral is: V = òò 16 - x 2 - y 2 dA where R is the region of R intersection: x 2 + y 2 + z 2 = 16 and x 2 + y 2 = 4 . Solving simultaneously gives z = 12 which is the plane of intersection. Plugging this into the first equation gives the circle x 2 + y 2 = 4 which is the region of integration. Again, in polar coordinates we have a ì0 £ q £ 2p ü rectangular region: R* = í ý and the integral becomes (after you transform the î 0£r £2 þ 2p 2 integrand): V = òò 0 0 16 - r 2 × rdrdq = .... Problem: Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 33 Set-up the surface area under the paraboloid z = 1 - x 2 - y 2 above the xy-plane using polar conversion: Solution: The integral (we derived this before) was: 2 1 1- x 2 æ ¶f ö æ ¶f ö 2 S = òò ç ÷ + ç ÷ + 1dA = ò ò 1 + 4 x 2 + 4 y 2 dydx R è ¶x ø è ¶y ø -1 - 1- x 2 Converting into the polar coordinates, we have the region: 0 £ r £1 0 £ q £ 2p and 1 + 4 x 2 + 4 y 2 = 1 + 4r 2 and the integral becomes: 2p 2 S= ò òr× 0 0 1 + 4r 2 drdq = ......etc Now, let us develop the Calc2 formula for the area of a polar region, bounded by the polar function r = r (q ) where a £ q £ b . The integration region could be described as: ì a £q £ b ü R=í ý î0 £ r £ r (q ) þ b r b 1 1 A = òò 1dxdy = ò ò rdrdq = ò ( r (q ) ) dq which again agrees with our well known 2 R 2a0 2a formula. GENERAL TRANSFORMATIONS IN DOUBLE INTEGRALS, JACOBIANS The polar transformation given by x = r cos q y = r sin q are special cases of many other transformations that could be useful. We need to find out how the integrals change under a general transformation: Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 34 x = x(u , v) y = y (u , v)) The above transformation maps regions in uv plane to the corresponding regions in the xy-plane. For example, a rectangle in the uv-plane gets mapped into a curved patch in the xy-plane. Like we did when we talked about the surface area integral, partitioning the region in the uv plane into subrectangles will induce corresponding partitioning in the xy plane. The left corner of one of these elementary rectangles corresponds to the left corner of the corresponding elementary patch. As before, consider the surface area approximation by the elementary parallelogram spanned by the two tangent vectors Tu Du and Tv × Dv æ ¶x ¶y ö ¶x ¶y æ ¶x ¶y ö ¶x ¶y Now, Tu = ç ; ÷ = i + j + 0k / x0 and Tv = ç ; ÷ = i+ j + 0k / x0 è ¶u ¶u ø x0 ¶u ¶u y0 è ¶v ¶v ø x = x0 ¶v ¶v y0 y0 y = y0 Crossing these gives i j k ¶y 0 ¶x 0 ¶x ¶y ¶y ¶u ¶u ¶u ¶u Tu ´ Tv = ¶x 0 =i -j +k ¶u ¶u ¶y ¶x ¶x ¶x 0 0 ¶x ¶y ¶v ¶v ¶v ¶v 0 ¶v ¶v ¶x ¶y ¶u ¶u ¶ ( x, y ) Þ Tu ´ Tv = = ¶x ¶y ¶ (u, v) ¶v ¶v The last determinant is called the Jacobi determinant, or Jacobian. With all this, the elementary area could be expressed as: ¶ ( x, y ) DA » Du Dv and from here we can view the Jacobian as the local change in ¶ (u, v) area factor. Now, we will skip many steps and proceed right to the formula: Suppose that we have a transformation x = x(u, v) y = y (u , v)) which transforms the region R into the region R*. Then we have: Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 35 ¶ ( x, y ) òò f ( x, y)dA = òò f [ x(u, v), y(u, v)] × ¶(u, v) dA * R R* Problem: Calculate the Jacobian for the polar coordinates Solution: we have the transformation: x = r cos q y = r sin q ¶ ( x, y ) cos q -r sin q = =r ¶ (u , v) sin q r cos q And the general formula become our well known formula: òò f ( x, y)dA = òò f [r cos q , r sin q ] × rdA * from before R R* Problem 1 1- x Evaluate òò 0 0 x + y ( y - 2 x) 2 dydx by using the substitution: u = x+ y v = y - 2x Solution First, let us convert the above equation into a form we are familiar with: u v x= - 3 3 2u v y= + 3 3 Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 36 Next, let us how the integration region changes: In the xy-plane it was the right triangle with vectices (0,1) and (1,0) and the hypotenuse was the line y = 1-x æ u v ö 2u v a) This line gets transformed as: 1 - ç - ÷ = + è 3 3ø 3 3 u v b) The line segment: x = 0 as: - =0Þu =v 3 3 2u v v c)The line segment: y =0 as: + =0Þu =- 3 3 2 1 -1 1 3 3 The Jacobian is: = 2 1 3 3 3 Now draw the region in the uv- plane so that you can describe it clearly: 0 £ u £1 -2u £ v £ u And now we are ready to integrate: We get the following integral: æ1ö 1 u òòu 1 v × ç ÷ dvdu = ....etc. It would be impossible to do the original integral without 2 3 0 -2 u è3ø this change of the variables, so stop fretting over this! TRIPLE INTEGRALS We will define triple integral basically by including one more variable. So, we will consider a function of three variables: h = f ( x, y, z ) defined over a box B = {( x, y ) : a £ x £ b, c £ y £ d , r £ z £ s} = [a, b] ´ [c, d ] ´ [r , s ] a) Consider a three-dimensional partition of the box into subboxes: a = x0 £ x1 £ .... £ xi -1 £ xi £ ... £ xn = b c = y0 £ y1 £ .... £ y j -1 £ y j £ ... £ ym = d r = z0 £ z1 £ ...... £ zk -1 £ zk £ .... £ zl = s Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 37 b) Select a subbox Bijk = [ xi -1 , xi ] ´ [ y j -1 , y j ] ´ [ zk -1 , zk ] and select a point inside of it: ( xi* , y* , zk ) Î Rij . The volume of this box is j * DVijk = ( xi - xi -1 ) × ( y j - y j -1 ) × ( zk - zk -1 ) = Dxi Dy j Dzk n m l And form a triple Riemann sum: DM ijk = ååå f ( xi* , y* , zk ) × DVijk . You can actually j * i =0 j =0 k =0 interpret this as an approximation of the mass of the box if the function h = f(x, y, z) represents density. c) Let again P be a partition with norm P = max{diagonalBijk } . We again take the limit of the Riemann sum (as the norm gets closer and closer to zero). And we obtain the definition of triple integral over the rectangle R: V = lim P ®0 DM ijk = òòò f ( x, y, z )dV . B If this limit exists, then the function z = f ( x, y, z ) is called integrable and if z = f ( x, y, z ) ³ 0 then the limit V is called the mass. It should be noticed that as before, if the function f(x,y) is continuous, then it is integrable. The properties (linearity, additivity, monotonicity) are easily extended to triple integrals. The Fubini Theorem for triple integrals defined over box is an extension again: b d s s d b òòò f ( x, y, z )dA = ò ò ò f ( x, y, z)dzdydx = ò ò ò f ( x, y, z )dxdydz = ....... B a c r r c a There are 3! = 6 possible ways we can change the order of integration. The calculation of the nested (iterated) integrals is the same as that for double integrals. We also have simple regions and Fubini Theorem for each: x-simple region: ì a£ x£b ü ï ï R = í f1 ( x) £ y £ f 2 ( x) ý ï ï î g1 ( x, y ) £ z £ g 2 ( x, y ) þ b f2 ( x ) g2 ( x , y ) òòò f ( x, y, z )dV = ò ò ò R a f1 ( x ) g1 ( x , y ) f ( x, y, z )dzdydx Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 38 y-simple region: ì c£ y£d ü ï ï R = í f1 ( y ) £ x £ f 2 ( y ) ý ï g ( x, y ) £ z £ g ( x, y ) ï î 1 2 þ b f2 ( y ) g2 ( x , y ) òòò f ( x, y, z )dV = ò ò ò R a f1 ( y ) g1 ( x , y ) f ( x, y, z )dzdxdy z-simple region: (Do this as an exercise) In fact, for each of the above simple region, we have two possibilities for a variable change in the projection. For example, for the z-simple region we can write: f é ù òòò f ( x, y, z )dV = ò ê òò f ( x, y)dE ú D e ë E û where E is the projection of the region D onto the xy- plane and the inside integral could be evaluated by two ways by changing the order of integration. Problem: Suppose that we want to calculate òòò ydV E where E is the region which lies under the plane z = x + 2y and above the region in the xy- plane bounded by y = x 2 , y = 0, x = 1 Solution: We have a region inside of the parabola in the xy-plane and under the given plane. Draw a picture in the xy-plane first. Notice that we could describe the region by the following equation (as a x-simple region): 0 £ x £1 0 £ y £ x2 0 £ z £ x + 2y Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 39 Correspondingly, we have: ì x é x+2 y ù ü 1 ìx ü 2 2 ï ï ï ï 1 òòò ydV = ò í ò ê ò ydz ú dy ýdx = ò í ò ( yx + 2 y 2 )dy ýdx 0ï0 ë 0 E î û ï þ 0ï0 î ï þ x2 1 1 é y2 y3 ù 1 é x4 x6 ù é x5 x7 ù 1 2 = ò ê x + 2 ú dx = ò ê + 2 ú dx = ê + 2 ú = + 0 ë 2 3 û0 0 ë 2 3û ë10 21 û 0 10 21 APPLICATION OF TRIPLE INTEGRALS: a) Volumes of bounded solids are calculated by the obvious formula: V = òòò dV E Problem Calculate the volume of the tetrahedron with vertices (1,0,0), (0,1,0) and (0,0,1) Solution: Notice that the equation of the tetrahedron plane (above the xy-plane) is: x+y+z=1 How would you find it (if the vertices had “worse” numbers)? The problem is to find the equation of plane given by three points. Recall that if you have a plane given by three points A, B, C, you can write parametric equations like this: X = (x, y, z) = A + t(C-A) + s (B-A). Here are the calculations: A = (1, 0, 0) B = (0,1, 0) C = (0, 0,1) C - A = (-1, 0,1) B - A = (-1,1, 0) X = A + t (C - A) + s ( B - A) ( x, y, z ) = (1, 0, 0) + t (-1, 0,1) + s(-1,1, 0) x = 1- t - s y=s z =t Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 40 The last three equations are the parametric equations for the plane. Now, eliminating the parameters t and s gives x = 1 – z – y which is our equation of the plane x + y + z =1. Problem Calculate the volume underneath the plane x + y + z =1. Notice that the z coordinate changes from 0 to 1 – x – y and the projection of the tetrahedron on the xy-plane is a triangle with vertices (1,0) and (0,1). The equation of the line going through these two points is y = 1 – x clearly. Thus, we have the following description of the region: 0 £ x £1 0 £ y £ 1- x 0 £ z £ 1- x - y 1 1- x 1- x - y and the integral is: V = ò ò ò dzdydx = ..... (finish up the calculations) 0 0 0 FIRST AND SECOND MOMENTS OF SOLIDS WITH DENSITIES THAT COULD BE FUNCTIONS OF x, y, z We will again, proceed heuristically, without the Riemann Sums. Students who finally got the idea should have no problems supplying the theoretical details. First,we have: mass = density X volume. This density is not allowed to be a function of three variables, namely r ( x, y, z ) (which was impossible for double integrals). Thus we have the following formulas: a) Mass and moments with respect to the coordinate planes: m = òòò r ( x, y, z )dV E M yz = òòò x r ( x, y, z )dV E M xz = òòò y r ( x, y, z )dV E M xy = òòò z r ( x, y, z )dV E Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 41 æ _ ö æM M M ö The centroid we calculate as: ç X , Y , Z ÷ = ç yz , xz , xy ÷ è ø è m m m ø The inertial moments we calculate as: I x = òòò ( y 2 + z 2 ) r ( x, y, z )dV E I y = òò ( x 2 + z 2 ) r ( x. y.z )dV E I z = òòò ( x 2 + y 2 ) r ( x, y, z )dV e Those of you, with some physics background also recall that with respect to the inertial moments we also talk about the gyration radius. These are given by: Ix Rx = m Iy Ry = m Iz Rz = m In general, the inertial moment with respect to a rotational line L we have: I L = òòò r 2 dV e IL RL = m where r is the perpendicular distance between the solid and the rotational line. Problem: Find the center of mass of the unit cube with vertices (1,0,0), (0,1,0),(0,0,1) and (0,0,0). The density is proportional to the square of the distance from the origin. Solution: Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 42 Since the density is proportional to the square of the distance from the origin, we have: r ( x, y , z ) = k ( x 2 + y 2 + z 2 ) 1 1 1 The mass is: m = òòò k ( x 2 + y 2 + z 2 )dA = ò ò ò k ( x 2 + y 2 + z 2 )dzdydx = k C 0 0 0 Next, the moment about yz-plane is 1 1 1 7k M yz = òòò x r ( x, y, z )dV = ò ò ò kx( x 2 + y 2 + z 2 )dzdydx = E 0 0 0 12 M yz 7 k /12 7 Thus, X = = = m k 12 æ7 7 7ö From symmetry,we conclude that ç , , ÷ is the centroid. è 12 12 12 ø Problem: Find the mass of a solid with constant density bounded by the parabolic cylinder x = y 2 and the planes x = z, z = 0 and x =1 Answers: (Draw pictures) Here we have to come up with the limits of integration. Careful examination of the situation leads to the following description: ì -1 £ y £ 1ü ï ï S = í( x, y, z ) : y 2 £ x £ 1ý ï 0£ z£ xï î þ 1 1 x 45 m = òòò r dV = ò ò ò r dzdxdy = ..... = E -1 y 2 0 5 TRANSFORMATIONS IN TRIPLE INTEGRALS, CYLINDRICAL AND SPHERICAL TRANSFORMATIONS Suppose that you have the following transformation equations: Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 43 x = x(u , v, w) y = y (u , v, w) z = z (u , v, w) ¶x ¶x ¶x ¶w ¶u ¶v ¶ ( x, y, z ) ¶y ¶y ¶y The Jacobian for this transformation is given by: = ¶ (u , v, w) ¶u ¶v ¶w ¶z ¶z ¶z ¶u ¶v ¶w The general statement is then: ¶ ( x, y , z ) òòò f ( x, y, z )dV = òòò f [ x(u, v, w), y(u, v, w), z(u, v, w)] × ¶((u, v, w) dV * E E* Recall that cylindrical coordinates are given by: x = r cos q y = r sin q z=z We first calculate the Jacobian: cos q -r sin q 0 ¶ ( x, y , z ) = sin q r cos q 0 =r ¶ (u , v, w) 0 0 1 And the cylindrical transformation has the form: òòò f ( x, y, z )dV = òòò [ f (r cos q , r sin q , z ) × rdV E E* If you do the partitioning and all that stuff, you get the following formula: òòò f ( x, y, z )dV = òòò f (r cosq , r sin q ) × rdV * where R* is the transformed region. R R* Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 44 Problem: For example, suppose that you want to find the mass of the surface created by intersecting the sphere x 2 + y 2 + z 2 = 1 and a cylinder r = sin q . Assume that the density is proportional to the distance from the xy-plane. Solution: We need to calculate m = òòò r ( x, y, z )dV . Since the density is proportional to the R distance from the xy- plane, we have r ( x, y, z ) = k × z = kz The cylinder r = sin q is traced (trace it!) for 0 £ q £ p . The sphere has a cylindrical equation z 2 = 1 - r 2 (which gives the bounds for z). Thus, the region can be described by ì 0 £q £p ü ï ï equations: í 0 £ r £ sin q ý . The integral becomes: ï 2ï î- 1 - r £ z £ 1 - r þ 2 p sin q 1- r 2 m = kò ò ò z × rdzdrdq = ...... (easy to finish) 0 0 - 1- r 2 Problem: Find the mass of the ellipsoidal solid E given by 4 x 2 + 4 y 2 + z 2 = 16 lying above the xy- plane. The density is proportional to the distance from the xy plane. Meaning that r ( x, y, z ) = kz Solution: (Cylindrical coordinates). 4 x 2 + 4 y 2 + z 2 = 16 4r 2 + z 2 = 16 z 2 = 16 - 4r 2 The xy- projection gives a circle with radius two and thus, we proceed as follows: 2p 2 2 4 - x 2 m= òò ò 0 0 0 kzdzdrdq = .... = 16kp Problem: Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 45 Find the moment of inertia about the axis of symmetry of the lid bounded by paraboloid z = x 2 + y 2 and the plane x = 4. Assume that the density is of the form r ( x, y , z ) = k x 2 + y 2 Solution: It is clear that the axis of symmetry is the z axis. It seems reasonable to convert into cylindrical coordinates The surface has a simple equation z = r 2 and the region of integration could be described as: 0£r £ z 0£ z£4 0 £ q £ 2p 4 2p z 512 We calculate: I z = òòò k ( x + y ) x + y dV = k ò 2 2 2 2 ò òr 2 × r × rdrdq dz = ... = kp E 0 0 0 35 Now, what happens when we transform into spherical coordinates? Recall the following transformation equations: r = r sin f x = r cos q = r sin f cos q y = r sin q = r sin f sin q z = r cos f r = x2 + y 2 + z 2 z cos f = r And recall that the angle f is measured down from the z-axis and it is on the interval [0, p ] . Cones are very easily described by f = f0 and spheres are described by r = r0 . ¶ ( x, y , z ) The Jacobian of the transformation is: = r 2 sin f and the spherical ¶ ( r , f ,q ) transformation formula is: Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 46 òòò f ( x, y, z )dV = òòò f ( r , f ,q ) r sin f dV * 2 R R* Problem: For example, suppose that you want to calculate volume of the solid which is bounded by p the cone f = and sphere r = cos f 6 Solution: Notice that the sphere is changed into Cartesian representation as: 2 z æ 1ö 1 r = cos f Þ r = Þ r = z Û x + y + z = z Û x + y + ç z - ÷ = Now we need 2 2 2 2 2 2 r è 2ø 4 to find the limits for all three parameters. Notice that: p 0 £f £ 6 0 £ q £ 2p 0 £ r £ cos q Is the spherical description of the region. The integral is: p 6 2p cos f V= òò ò 0 0 0 r 2 sin f d r dq df = .... EXAMPLE OF TEST 3 1) Set up the integral for the volume of the “icecream cone” surface created as the p intersection of the sphere r £ 1 and the cone f = 3 p 2p 3 1 Answer: V = ò ò òr sin f d r df dq 2 0 0 0 2) Set up the integral for the moment of inertia about the z-axis for the above ice cream cone, assuming that the density is proportional to the xy-plane: Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 47 p 2p 3 1 Answer: I z = ò ò ò ( r cos f ) × r × r sin f d r df dq 2 0 0 0 3) Calculate the centroid of the solid inside of a sphere with radius R , in the first octant, with constant density, using the cylindrical coordinates. - M xy òòò kydV 1 Answer: z = = R = òòò zdV . In cylindrical coordinates, the region is: ( The m òòò kdV R V R 1æ 4 ö volume we can calculate directly as: V = ç p R 3 ÷ 8è 3 ø 0£r £ R p 0 £q £ 2 0 £ z £ R2 - z 2 p _ a 2 R2 - z 2 1 3 z= V òò ò 0 0 zrdzdrdq = R 8 æ 3R 3R 3R ö The remaining coordinates are the same, so the centroid is: ç ; ; ÷ è 8 8 8 ø 1 x2 4) Evaluate ò ò sin xydxdy 0 x 5) Evaluate the above integral, but change the order of integration (from the one you did before) 6) Set up the integral for the first moment with respect to the xy-plane of the surface bounded by the cylinder x 2 + y 2 = 4 and paraboloid z = x 2 + y 2 Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 48 2p 2 r 2 32p Answer: ò ò ò rzdzdq dr = 0 0 0 3 7) identify the surface, whose volume is given by the following triple integral: p 2 2cosq 4 - r 2 ò ò ò 0 0 0 rdzdrdq Solution: From the integration variables, we recognize a cylindrical coordinate transformation. The region is described by: p 0 £q £ 2 0 £ r £ 2 cos q 0 £ z £ 4 - r2 To see it better, convert to Cartesian coordinates and you have a region bounded by: a) paraboloid z = 4 - x 2 - y 2 b) Planes z = 0 and y = 0 c) Cylinder ( x - 1) 2 + y 2 = 1 8) Evaluate òò ( x + y)dA where R is the triangular region with vertices (0,0), (2,1) and R (1,2) using the transformation: x = 2y + v y = u + 2v Dr. Tomas Kovarik: Calculus3: Part3 (Integral Calculus in Space) 49