find area between 2 circles: x^2+y^2=1 and x^2+y^2=4
using (x+y) da
I got the answer of (3*pi)/2 is this a correct answer? To check I just took the difference between the two circles and then got the half of that. Does this seem right?
$\displaystyle \int\int_A(x+y)\,dA$
Use polar coordinates.
$\displaystyle x=r\cos\theta$
$\displaystyle y=r\sin\theta$
$\displaystyle dA=r\,dr\,d\theta$
$\displaystyle r=1..2$
$\displaystyle \theta=0..2\pi$
$\displaystyle \int_0^{2\pi}\int_1^2 r^2(\cos\theta+\sin\theta)\,dr\,d\theta$
When I integrate this, I got zero. It's possible I made a sign goof somewhere though.
The integral over the region is zero because
$\displaystyle \int\int_{\omega} x ~dA$ = ( x-center of gravity ) ( area of the region )
$\displaystyle \int\int_{\omega} y ~dA$ = ( y-center of gravity ) ( area of the region )
the center of gravity is $\displaystyle (0,0) $