Evaluating a Triple Integral in Polar Coordinates

Question

I'm trying to evaluate $\iiint_{E}\sqrt{3x^{2} + 3z^{2}}~dV$ where $E$ is the solid bounded by $y = 2x^{2} + 2z^{2}$ and $y=8$.

My thought was to covert this to polar coordinates using $x^{2} + z^{2} = r^{2}$. Then the solid would be a cone originating in the $xz$-plane with radius of 0 which stretches along the $y$-axis until it terminates as a circle of radius 2. So I converted $\sqrt{3x^{2} + 3z^{2}}$ to $\sqrt{3r^{2}}$ to $r\sqrt{3}$ and set the integral up this way $$\int_{\theta = 0}^{\theta = 2\pi}\int_{r=0}^{r=2}\left(\int_{y=0}^{y=2r^{2}} r\sqrt{3} dy\right) r~dr~d\theta$$ which equals $\frac{128\sqrt{3}}{5}\pi$. But the correct answer is $\frac{256\sqrt{3}\pi}{15}$, which means I'm off by a factor of $\frac{2}{3}$. Any help figuring out what I'm doing wrong would be greatly appreciated!

When you say “polar” coordinates, do you mean cylindrical coordinates with $y$ as the axis of the cylinder? — David K, Mar 25 '20 at 23:43
I don't see any reason why you are incorrect. Maybe it's a typo? — ProfOak, Mar 25 '20 at 23:45
I don't know. I pulled this off of some online notes with very credible solutions. But the solution in the notes didn't make the conversion to polar coordinates until after evaluating the inner integral in x and z. I just converted in into cylindrical coordinates, and am coming up short in my solution. — Scentless Apprentice, Mar 25 '20 at 23:48
Your evaluation of the integral you set up is correct. Either it's a typo in the answer or the set up is wrong somehow. — ProfOak, Mar 25 '20 at 23:51
The limits of integration of the posted solution were equivalent to y=2r^(2) as the lower limit, and y=8 as the upper limit. Could this be the source of my error? If so, I don't understand why this would matter. — Scentless Apprentice, Mar 25 '20 at 23:54
Oh, I see it too. Sorry, I was being careless. Your first integral should go from $y=2r^2$ to $y=8$. I was so focused on the polar conversion that I missed the $y$ integral mistake. Doh. — ProfOak, Mar 25 '20 at 23:56
Can you show how you solved the integrals? The solution with the factor $256/15$ is correct. — Fakemistake, Mar 26 '20 at 00:24

score 2 · Accepted Answer · answered Mar 25 '20 at 23:53

2

For each radius $r$, the minimum value of $y$ is $2r^2$ and the maximum value is $8$.

You should be integrating over values $y$ can take. Instead you are integrating over values of $y$ between $0$ and $2r^2$.

answered Mar 25 '20 at 23:53

David K

98,388

I know you must be right, per my comment above. But if r ranges from 0 to 2 in the middle integral, doesn't that mean y is ranging from 0 to 8? – Scentless Apprentice Mar 25 '20 at 23:56
I see now. Thanks again! You guys are great. – Scentless Apprentice Mar 26 '20 at 00:17

Evaluating a Triple Integral in Polar Coordinates

1 Answers1