Using polar coordinates evaluate the integral

Question

Can you help me to solve these questions? I'm totally lost here.

a) $\iint_D e^{-(x^2+y^2)}~dx~dy, D=\{(x,y):x^2+y^2\leq 2\}$

b) $\iint_D \frac{1}{x^2+y^2-1}~dx~dy, D=\{(x,y):x^2+y^2\geq 9, x^2+y^2\leq 25\}$

Answer 1

Hint

For $a,b\ge 0$ $$ \int_{a^2<x^2+y^2<b^2} f(x,y)dxdy=\int_{a<r<b} rf(r\cos\theta,r\sin\theta)drd\theta $$

Answer 2

When you go to polar coordinates, you have $r^2=x^2+y^2$. So for your first integral, the region $D$ consists of those points where $r^2\leq 2$; that is, $r\leq\sqrt2$. Then $$ \iint_D e^{-(x^2+y^2)}~dx~dy=\int_0^{2\pi}\int_0^{\sqrt2}e^{-r^2}\,r\,dr\,d\theta=2\pi\,\int_0^{\sqrt2}e^{-r^2}\,r\,dr=\pi(1-e^{-2}) $$ For your second integral you now have $9\leq r^2\leq 25$, so your integral becomes $$ \iint_D \frac{1}{x^2+y^2-1}~dx~dy=\int_0^{2\pi}\int_3^5\frac{r}{r^2-1}\,dr\,d\theta. $$