Results 1 to 7 of 7

Math Help - Evaluating a triple integral using spherical coordinates

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    9

    Evaluating a triple integral using spherical coordinates

    Evaluate:

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

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

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2008
    Posts
    903
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    55
    Quote Originally Posted by shawsend View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2009
    Posts
    9
    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...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    55
    Quote Originally Posted by Dwill90 View Post
    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).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Oct 2009
    Posts
    9
    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!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    55
    Quote Originally Posted by Dwill90 View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Triple integral in spherical coordinates
    Posted in the Calculus Forum
    Replies: 7
    Last Post: September 23rd 2010, 01:00 PM
  2. Triple Integral in Spherical Coordinates
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 27th 2010, 08:23 PM
  3. Triple Integral Using Spherical Coordinates
    Posted in the Calculus Forum
    Replies: 5
    Last Post: August 30th 2008, 10:50 PM
  4. spherical coordinates -triple integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 13th 2007, 10:29 AM
  5. spherical coordinates -triple integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 12th 2007, 06:41 PM

Search Tags


/mathhelpforum @mathhelpforum