# Thread: Evaluating a triple integral using spherical coordinates

1. ## Evaluating a triple integral using spherical coordinates

Evaluate:

Triple integral over domain H of $z^3 * sqrt(x^2 + y^2 + z^2) dv$

$
where H = { (x, y, z) | x^2 + y^2 + z^2 <= 1, y >= 0 }
$

To the point: I used spherical coordinates and during the integration with respect to phi (the vertical angle), it resulted in 0, causing the entire triple integral to evaluate to 0, which I assume to be wrong.

Elaborate: Using...
x = p cos(theta) sin(phi)
y = p sin(phi) sin(theta)
z = p cos(phi)
p = $sqrt(x^2 + y^2 + z^2)$
and H = { (p, theta, phi) | 0 <= p <= 1, 0 <= theta <= pi, 0 <= phi <= pi }

First I'll evaluate the integral with respect to phi:

integral(0, pi, p^6 * cos^3(phi) * sin(phi), dphi)

Using u = cos(phi) as the substitution, the integral becomes:

integral(1, -1, - u^3 / 7, du)

Which evalutes to zero.

Could anyone show me what I'm doing wrong, or confirm that zero is the answer?

Thank you!
David

2. Looks to me it's:

$\int_0^{\pi} \int_0^{\pi} \int_0^1 \rho^5 \cos^3(\phi) \sin(\phi) d\rho d\theta d\phi=0$

3. Originally Posted by shawsend
Looks to me it's:

$\int_0^{\pi} \int_0^{\pi} \int_0^1 \rho^5 \cos^3(\phi) \sin(\phi) d\rho d\theta d\phi=0$
Ya - I got the same.

4. Hmm. The only difference between your integral and the one I had was that I had $p^6$ as a factor, but you had $p^5$. So I tracked the ps in this manner:

$z^3$ becomes $p^3 * cos^3(phi)$;
$sqrt(x^2 + y^2 + z^2)$ becomes $p$;
and the jacobian contributes $p^2 * sin(phi)$.

$p^3 * p * p^2 = p^6$. Have I made an error?

Using the integral you've provided, I evaluated it fully out, leaving the phi integral for last, and obtained:

$pi/6 * integral(0, pi, cos^3(phi) * sin(phi), dphi)$
After using a u-substitution of u = cos(phi), I have the integral
$integral(1, -1, u^3, du)$
Which then evaluates to 0. I guess the answer really is zero...

5. Originally Posted by Dwill90
Hmm. The only difference between your integral and the one I had was that I had $p^6$ as a factor, but you had $p^5$. So I tracked the ps in this manner:

$z^3$ becomes $p^3 * cos^3(phi)$;
$sqrt(x^2 + y^2 + z^2)$ becomes $p$;
and the jacobian contributes $p^2 * sin(phi)$.

$p^3 * p * p^2 = p^6$. Have I made an error?

Using the integral you've provided, I evaluated it fully out, leaving the phi integral for last, and obtained:

$pi/6 * integral(0, pi, cos^3(phi) * sin(phi), dphi)$
After using a u-substitution of u = cos(phi), I have the integral
$integral(1, -1, u^3, du)$
Which then evaluates to 0. I guess the answer really is zero...
Your right $3 \rho$'s from $z^3$, $1 \rho$ from $\sqrt{x^2+y^2+z^2}$ and 2 $\rho$'s from $dV$. BTW - I was agreeing with the answer (zero).

6. Oh! That's awesome. Thanks a ton for your replies. My instructor is playing mind games with us; he gives us six review questions for the final exam, but doesn't give us the answers, and so far I've found the answer to three of them to be zero!

7. Originally Posted by Dwill90
Oh! That's awesome. Thanks a ton for your replies. My instructor is playing mind games with us; he gives us six review questions for the final exam, but doesn't give us the answers, and so far I've found the answer to three of them to be zero!
Post your questions and answers here on MHF and we'll check if you like.