Finding the volume of 3 dimensional region under the graph of a function.

Question

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.

Nicolas · Answer 1 · 2015-08-09T20:02:03.323

0

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

edited Aug 09 '15 at 20:02

answered Aug 09 '15 at 19:20

Nicolas

3,326

Your integrals are in reverse order. $\mathrm{dz}$ should come first, shouldn't it? – Jack's wasted life Aug 09 '15 at 19:25
@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! – Nicolas Aug 09 '15 at 19:27
You're welcome :D – Jack's wasted life Aug 09 '15 at 19:28
Can you give me a bit more explanation as I don't understand still. – user3472448 Aug 09 '15 at 19:32
@dk1 I've edited my post, please check if you can understand it better. – Nicolas Aug 09 '15 at 19:37
So do you integrate ex2+y2? – user3472448 Aug 09 '15 at 19:51
@dk1 No, you never integrate the function : it is just a part of the boundary of the integration domain. – Nicolas Aug 09 '15 at 19:53
So how do you get the answer? – user3472448 Aug 09 '15 at 19:57
@dk1 Edited anew : there is the final answer (hope with no mistake!). Remember that the general formula is the first expression for $V$ (with $x$ and $y$ in the domain given by the problem). – Nicolas Aug 09 '15 at 20:03

score 0 · Answer 2 · answered Aug 09 '15 at 20:20

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

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

Finding the volume of 3 dimensional region under the graph of a function.

2 Answers2