Im trying to do the following question but im confused.
Let W be the three dimensional region under the graph of the function $f(x,y) = \mathrm{e}^{x^2+y^2}$ and over the region in the $(x,y)$ plane defined by $1\leq x^2+y^2 \leq 2$.
I know I have to use double integrals but what will the limits be? Would be $0$ to $1$ on the outside integral and $1$ to $2$ in the inside integral.
On
Since your function is positive over the given region, the volume under the graph is given by $V=\int\int_{R} e^{x^2+y^2}dxdy$
Where $R$ is the region in the plane consisting of the points $(x,y)$ that satisfy $0\leq x^2+y^2\leq 2$. Notice that $x^2+y^2$ is simply square of the radius in polar coordinates, so switching to polar coordinates, we have
$V=\int_{0}^{2\pi}\int_1^\sqrt{2}re^{r^2}drd\theta=\int_{0}^{2\pi}d\theta\int_1^\sqrt{2}re^{r^2}dr=2\pi*\frac{1}{2}\int_0^2e^udu=\pi(e^2-e)$
By definition of the volume $V$ under the graph, we have $$V:=\iint_{1\leq x^2+y^2\leq 2}\int_{z=0}^{f(x,y)}\mathrm{d}x\mathrm{d}y\mathrm{d}z.$$
Now use polar coordinates in the plane $\{z=0\}$ to get a simpler expression $$V=\int_{\theta=0}^{2\pi}\int_{r=1}^{\sqrt{2}}\int_{z=0}^{\mathrm{e}^{r^2}}r\mathrm{d}z\mathrm{d}r\mathrm{d}\theta$$ as the jacobian is $r$ under the previous change of variables. The final result here is $$V=\int_{\theta=0}^{2\pi}\mathrm{d}\theta\int_{r=1}^{\sqrt{2}}\left(\int_{z=0}^{\mathrm{e}^{r^2}}\mathrm{d}z\right)r\mathrm{d}r=2\pi\int_{r=1}^{\sqrt{2}}r\mathrm{e}^{r^2}\mathrm{d}r=2\pi\left[\frac{\mathrm{e}^{r^2}}{2}\right]_{r=1}^{r=\sqrt{2}}=\pi\left(\mathrm{e}^{2}-\mathrm{e}\right)=\pi\mathrm{e}\left(\mathrm{e}-1\right).$$