Question

Use a double integral to find the area of the region.
The region inside the circle
(x − 5)^2 + y^2 = 25
and outside the circle
x^2 + y^2 = 25

Which one is the inside Integral and how do I find the bounds for it. Is x=r, or is x=rcos(theta)? I'm confused.

Answer 1

\[(x-5)^2+y^2=25... and...x^2+y^2=25\]

Answer 2

inside the circle, r <= 5

Answer 3

remember, x = r cos theta and y = r sin theta

Answer 4

okay. But is one of the limits of integration 0 and 5?

Answer 5

yeah, for inside the circle. 0 <= r <= 5

Answer 6

which equation is the "inside" intagral? For example, to keep dbl integrals straight I write them \[\int\limits_{a}^{b}[\int\limits_{c}^{d}f(x,y) say, dx]dy\]. Here, the inside integral is the dx.

Answer 7

Rewrite your function in terms of r and theta... and dx dy becomes r dr dtheta

\[\left( x-5\right)^2 +\left( y \right)^2= 5^2 \Rightarrow x-5 = r \cos \theta \text{, }y=r \sin \theta\]

Answer 8

I did that and found r=10cos(theta)

Answer 9

?

Answer 10

sorry, i changed x2+y2=25 into P.C and got r=+-5, rejectiing -5, so some boundary is 0,5. Not correct?

Answer 11

what is getting me, is that there are 2 equations. One that's the inside of the circle and one that's the outside. Which one is getting integrated on my "inside" integral

Answer 12

same equation... different limits of integration

Answer 13

Again, which one is inside and why?

Answer 14

:( I'm in deep doodoo.

Answer 15

if you want the area of the circle, you just use 1 as the function.

\[\int\limits_{0}^{2 \pi}\int\limits_{0}^{5}1r\,dr\,d \theta = \cdots\]

Answer 16

that's for the area inside the circle.

for the area outside the circle, the limits of integration on r would be from 5 to infinity.

Answer 17

It says find the area inside the circle AND the area outside. so do I still just use equn 1? iF SO, how do I know not to use x^2+y^2=25 as my Integrand?

Answer 18

when using a double integral merely to find an area, you integrate over the appropriate region using only dA as the integrand...