# Homework Help: Find the exact volume of a bounded surface. Multiple Integrals.

1. Nov 4, 2012

### notorious_lx

1. Ok, so the question is.. Find the exact volume of the solid bounded above by the surface $z=e^{-x^2-y^2}$, below by the xy-plane, and on the side by $x^2+y^2=1$.

2. Alright. So, I know that I can use a double integral to find the volume, and switching to polar coordinates would be simpler.$z=e^{-r^2}$ and $r^2 =1$, therefore $z=e^{-1}$.

3. I'm not sure how to calculate the double integral. I know volume would be the double integral of the top surface minus the bottom surface, but do i need to split the surface into two separate double integrals and add them together? This should be easy but I can't seem to figure this out.

Last edited: Nov 4, 2012
2. Nov 4, 2012

### LCKurtz

Not quite. $r=1$ only on the bounding surface. But inside the volume you need to use $e^{-r^2}$.
Not sure what is bothering you here. What do you get when you set up the integral in polar coordinates using "top - bottom" surface in the integrand as you suggest?

3. Nov 4, 2012

### HallsofIvy

You integrate $e^{-r^2}[itex] with r from 0 to 1, [itex]\theta$ from 0 to $2\pi$. The "differential of area" in polar coordinates is $r dr d\theta$.

4. Nov 4, 2012

### notorious_lx

OK here is what I have so far..

$\int_0^{2\pi} \int_{-rsinθ}^{rsinθ} (e^{-r^2}-{\frac{1}{e}})rdrdθ + \int_0^{2\pi} \int_{-rsinθ}^{rsinθ} ({\frac{1}{e}}rdrdθ)$

I'm not sure if this is right.. I may be overthinking this way too much.. haha

edit: This is just equal to $\int_0^{2\pi} \int_{-rsinθ}^{rsinθ} (e^{-r^2})rdrdθ$ right?

Last edited: Nov 4, 2012
5. Nov 4, 2012

### LCKurtz

Well, they aren't equal because you dropped the $\frac 1 e$. But where did that come from anyway?? And your inner limits, the $r$ limits, are incorrect. What is the equation of the circle $x^2+y^2=1$ in polar coordinates? Where did the $\sin\theta$ terms come from?

6. Nov 4, 2012

### notorious_lx

ok i had $\int_0^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (e^{-(x^2+y^2)})$
Then I did this and used $r^2=1$, but this is only on the boundry so this is incorrect.
\begin{align} &= \sqrt{1-x^2}\\ &= \sqrt{r^2-r^2\cos^2{\theta}}\\ &= r\sqrt{1-\cos^2{\theta}}\\ &= r\sin{\theta} \end{align}

7. Nov 4, 2012

### LCKurtz

In the integrand $r$ varies throughout the circular region. But think about how $r,\theta$ vary to cover a circle. It is the limiting values that go in the limits. You have already figured out that $\theta$ must go from $0\rightarrow 2\pi$. Now if you start at the origin and move in any particular $\theta$ direction, what are the lower and higher limits $r$ will take?

8. Nov 4, 2012

### notorious_lx

r would go from 0 to 1.
so the equation would be $\int_0^{2\pi} \int_0^1 (e^{-r^2})rdrd\theta$?

9. Nov 4, 2012

### LCKurtz

Yes. Not only is that correct, it is easier eh?