# Setting up a Double Integral in Polar Coordinates

1. Oct 18, 2016

### Staff: Mentor

1. The problem statement, all variables and given/known data
Consider the 'ice cream cone' bounded by
z = 14 − x2 − y2 and z = x2 + y2
.(a) Find the equation of the intersection of the two surfaces in terms of x and y.
(b) Set up the integral in polar coordinates.

2. Relevant equations

3. The attempt at a solution

I got part a without any trouble. You just set each equation equal to one another and get:
$x^2+y^2=7$

I also found my limits of integration just fine: $\int_0^{2\pi} \int_0^{\sqrt{7}} rdrd\theta$

I just can't seem to set up the integrand correctly. I thought it would be: $x^2+y^2-7 = (rcos(\theta))^2+(rsin(\theta))^2 - 7 = r^2 (cos^2(\theta)+sin^2(\theta)) - 7 = r^2-7$

That would make the integral: $\int_0^{2\pi} \int_0^{\sqrt{7}} (r^3-7r) drd\theta$

However that's not correct and I can't seem to find the correct one.

2. Oct 18, 2016

### SammyS

Staff Emeritus
I suppose those equations should read

$z = \sqrt{ 14 − x^2 − y^2\ }\$ and $\ z = \sqrt{ x^2 + y^2\ }$ .

What is it that integrating is supposed to give you ?

3. Oct 18, 2016

### Staff: Mentor

Whoops, I meant to fix those before I posted...

Some sort of ice cream cone shape. Perhaps I'm supposed to integrate using those other two equations instead of where they intersect?

4. Oct 18, 2016

### LCKurtz

You need to integrate$$\iint_R z_{upper} - z_{lower}~dA$$where $R$ is the region you have figured out, and the $z$'s are expressed in terms of $r,\theta$.

5. Oct 19, 2016

### SammyS

Staff Emeritus
My question could have been more specific.

Are you integrating to find surface area, or to find volume?

6. Oct 19, 2016

### Staff: Mentor

I'm pretty sure it's volume. We haven't done anything with surface area yet.

Ah, of course. It's just like Calc 2 where you find the region between two lines, except this time it's 2 surfaces.

So that should be:
$\int_0^{2\pi}\int_0^\sqrt{7} (\sqrt{14-r^2} - r) rdrd\theta = \int_0^{2\pi}\int_0^\sqrt{7} (r\sqrt{14-r^2} - r^2) drd\theta$
Because $x^2+y^2 = r^2$, so $\sqrt{14-x^2-y^2} = \sqrt{14-(x^2+y^2)} = \sqrt{14-r^2}$
And $\sqrt{x^2+y^2} = \sqrt{r^2} = r$

Now to integrate...