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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.

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2 Answers 2

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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 $$V=\int_{\theta=0}^{2\pi}\mathrm{d}\theta\int_{r=1}^{\sqrt{2}}\left(\int_{z=0}^{\mathrm{e}^{r^2}}\mathrm{d}z\right)r\mathrm{d}r=2\pi\int_{r=1}^{\sqrt{2}}r\mathrm{e}^{r^2}\mathrm{d}r=2\pi\left[\frac{\mathrm{e}^{r^2}}{2}\right]_{r=1}^{r=\sqrt{2}}=\pi\left(\mathrm{e}^{2}-\mathrm{e}\right)=\pi\mathrm{e}\left(\mathrm{e}-1\right).$$

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  • $\begingroup$ Your integrals are in reverse order. $\mathrm{dz}$ should come first, shouldn't it? $\endgroup$ Aug 9, 2015 at 19:25
  • $\begingroup$ @Jack'swastedlife You're right, my notation is not clear. Or I write the triple integral or I write three distinct integrals with the good order. I edit my answer and I thank you! $\endgroup$
    – Nicolas
    Aug 9, 2015 at 19:27
  • $\begingroup$ You're welcome :D $\endgroup$ Aug 9, 2015 at 19:28
  • $\begingroup$ Can you give me a bit more explanation as I don't understand still. $\endgroup$ Aug 9, 2015 at 19:32
  • $\begingroup$ @dk1 I've edited my post, please check if you can understand it better. $\endgroup$
    – Nicolas
    Aug 9, 2015 at 19:37
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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

$V=\int_{0}^{2\pi}\int_1^\sqrt{2}re^{r^2}drd\theta=\int_{0}^{2\pi}d\theta\int_1^\sqrt{2}re^{r^2}dr=2\pi*\frac{1}{2}\int_0^2e^udu=\pi(e^2-e)$

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