# Area of a Circle - Download as DOC

Document Sample

```					                                                                    Area of a Circle
(1) Using horizontal strips and
integrating from y=0 to y=R:
yR
dA  x  dy  A  4                          x  dy
y 0
(x,y)
dy                                                                         multiplying by 4 to get all the area.
(2) Using: x 2  y 2  R 2 solve for x.
R
x R y                                 R 2  y 2 dy  ?
2      2
but
0
(3) We’ll need: cos 2  2cos 2   1
So: cos2   1  1 cos 2 to get…
2   2

 cos   d  1   1 sin 2  C .
2
2    4
and now for our Trig Substitution!
Using polar coordinates:
 x  R cos
              with dy = R cos d
 y  R sin 
R
(4) Now let’s get back to: A  4  R 2  y 2 dy . We’ll substitute for y and dy (see above).
0

We’ll also need to change our limits of integration: y  0    0 and y  R     .
2
  2                                                                      
2
Now: A  4                 R  R sin   R cos d  4 R                                      cos                d (Use: cos 2   sin 2   1 )
2       2         2                                   2                   2

 0                                                                       0


 2   1 sin    0  0  4
2                                                     
Using result (3), we get:                    cos  d = 1   1 sin 2
2                                                        2
 1
2     4                                      0         2              4       
0

So the area of circle is: 4R 2     R 2 AP Calculus BC, anyone?!
4

Easier is to use the “area of a sector” with radians: A                                                           Rad      R2     R2
(vs           R2 )
2 Rad                2            360
R2   d and integrate
Now consider a slice of pizza (with a very small angle, d !) dA                                                                 2
from   0 to    (90) and multiply by 4 to get all the area.
2
                              
2                              2

A4                                 d 
2                 2
4 R2                    4 R2
R
2
d      4R
2                         2
   0
2
     2
  2  0  =  R 2
                                    (Much easier!)
0                             0

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 19 posted: 9/22/2012 language: simple pages: 1
How are you planning on using Docstoc?