Using polar coordinates, evaluate the double integral over r sin(x^2+y^2)dA where R is the region 4<x^2+y^2<36
Need to know where to start, or an example? I keep getting 8.something and it's incorrect.
I get that. But when I integrate and plug in, I get the wrong answer. So I integrate sin(r) from r=2 to 6? that give me -cos(r) from r=2 to 6. Then, integrate -cos(2)+cos(6) from theta=0 to 2pi?
This gives me 8.647 which is incorrect.
The width of an infinitessimal element expressed in polar parameters is $\displaystyle r\mathrm{d}\theta$, and its length is $\displaystyle \mathrm{d}r$. So the area of the infinitessimal element is:
$\displaystyle \mathrm{d}A=r\mathrm{d}r\mathrm{d}\theta$
That's where the 'extra' factor $\displaystyle r$ comes from.
As far integrating it, I don't have time to check it myself right now, but integration by parts might be the way to go.
I have a feeling your original integrand must have been $\displaystyle \displaystyle \begin{align*} \sin{\left(x^2 + y^2\right)} \end{align*} $, not $\displaystyle \displaystyle \begin{align*} r\sin{\left(x^2 + y^2\right)} \end{align*} $ seeing as it doesn't make sense to have mixed polars with cartesians in the first place. Also, $\displaystyle \displaystyle \begin{align*} \int{r^2\sin{\left(r^2\right)}\,dr} \end{align*} $ doesn't have a closed form answer, whereas the first will.