# Thread: Triple integral in spherical coordinates

1. ## Triple integral in spherical coordinates

Um is the answer wrong for this question? I can't seem to get the right answer... it is off by a factor of 2 grrr anyways:

Evaluate $\displaystyle \int \int \int_E x^2 dV$ where E is bounded by the xz-plane and the hemispheres $\displaystyle y = \sqrt{9-x^2-z^2}$ and $\displaystyle y = \sqrt{16-x^2-z^2}$.

So my solution is this:

Using spherical coordinates, we have $\displaystyle 3 \le \rho \le 4, 0 \le \theta \le \pi, 0 \le \phi \le \frac{\pi}{2}$

Thus $\displaystyle \int \int \int_E x^2 dV = \int_{0}^{\frac{\pi}{2}} \int_0^{\pi} \int_3^4 (\rho \sin(\phi)\cos(\theta))^2\rho^2\sin(\phi) d\rho d\theta d\phi = \frac{781\pi}{15}$
But the answer is $\displaystyle \frac{1562\pi}{15}$ hmmm how did they get this?

Thanks!

2. The azimuthal angle $\displaystyle \theta$ will range from $\displaystyle 0$ to $\displaystyle \pi.$ That you have correct. What's not correct is that the polar angle $\displaystyle \phi$ will also range from $\displaystyle 0$ to $\displaystyle \pi.$ That may change things a bit for you. I'm not seeing any other obvious problems. See what you get with this.

3. ohhhhh right, lol omg I sketched the graph wrong, it's y = f(x,z) so y is the axis of symmetry!

EDIT: thanks Ackbeet!!

4. Right. Did you get the correct answer, then?

5. Originally Posted by Ackbeet
The azimuthal angle $\displaystyle \theta$ will range from $\displaystyle 0$ to $\displaystyle \pi.$ That you have correct. What's not correct is that the polar angle $\displaystyle \phi$ will also range from $\displaystyle 0$ to $\displaystyle \pi.$ That may change things a bit for you. I'm not seeing any other obvious problems. See what you get with this.
Edit- I am incorrect. I did not see that we were bounded by the xz plane. OOPS!

The azimuthal will range from 0 to $\displaystyle \frac{ \pi }{2}$ and theta will range from 0 to $\displaystyle 2 \pi$....we want to cover the entire top half the hemisphere so there is no reason why theta would not be 360 degrees. Likewise pi over two corresponds with the top half (measuring from the Z-axis to the XY plane).

6. Reply to AllanCuz at Post # 5:

I think you're a bit confused. The polar angle in this problem is $\displaystyle \phi,$ and inhabits the natural range $\displaystyle [0,\pi]$ in any spherical coordinates problem. The azimuthal angle in this problem is $\displaystyle \theta,$ and inhabits the natural range $\displaystyle [0,2\pi]$ in any spherical coordinates problem. However, in this problem, the shape described, which is that shape bounded by the $\displaystyle xz$ plane, and the two hemispheres $\displaystyle \rho=3$ and $\displaystyle \rho=4$, you have the artificial restriction $\displaystyle \theta\in[0,\pi].$ There is no restriction on the polar angle.

Hence your limits, while producing the same answer for the integral as my limits, do not describe the shape in the original problem.

[EDIT] Ah, I see you saw your mistake.

7. Originally Posted by Ackbeet
Reply to AllanCuz at Post # 5:

I think you're a bit confused. The polar angle in this problem is $\displaystyle \phi,$ and inhabits the natural range $\displaystyle [0,\pi]$ in any spherical coordinates problem. The azimuthal angle in this problem is $\displaystyle \theta,$ and inhabits the natural range $\displaystyle [0,2\pi]$ in any spherical coordinates problem. However, in this problem, the shape described, which is that shape bounded by the $\displaystyle xz$ plane, and the two hemispheres $\displaystyle \rho=3$ and $\displaystyle \rho=4$, you have the artificial restriction $\displaystyle \theta\in[0,\pi].$ There is no restriction on the polar angle.

Hence your limits, while producing the same answer for the integral as my limits, do not describe the shape in the original problem.

8. Your limits, actually, are a legitimate way of solving the problem if you think of the whole shape as rotated about the $\displaystyle x$ axis through $\displaystyle \pi/2$ radians such that $\displaystyle \hat{j}$ becomes $\displaystyle \hat{k}$. So, if you believe that rotations preserve volumes and such, and since the integrand is insensitive to that rotation, your limits produce the same result.