Results 1 to 11 of 11

Math Help - Sphere volume derivation with the correct variable of integration

  1. #1
    Newbie
    Joined
    Sep 2010
    Posts
    20

    Sphere volume derivation with the correct variable of integration

    Hello all,

    So I'm trying to figure out why setting up integrals to find volume of a sphere with the correct variable of integration.

    For example, for volume,

     \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \pi r^2 \cos^2(\theta)d\theta=\frac{\pi r^2}{4} does not work.

    I believe my teacher explained that it had to do with the  d\theta that were in fact small wedges that, in sum, did not produce a sphere. However, the substitution using  x=r*\sin(\theta) and  dx=r*\cos(\theta)d\theta did produce the correct result, which is actually the same as the conventionally derived (single-variable) shells integral for a sphere:

     \pi \int_{-r}^{r} x \sqrt{r^2-x^2} dx

    Now, shouldn't these integrals be the same, despite the substitution? What was the first integral really solving for- I don't think I can picture it in my head.

    Thank you for taking the time to read this post and help me out.
    Last edited by progressive; December 22nd 2010 at 09:56 PM. Reason: fixed 2nd equation
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Top integral should be cos^3 not squared and you forget a r. Those two integrals aren't the same otherwise.

    And the volume of a sphere is \displaystyle \frac{4\pi r^3}{3}=\pi r^3\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\cos^3{(\theta)}d\theta
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2010
    Posts
    20
    Actually, you may be right. Do you by any chance know the proper substitution? The first integral was essentially a "washers" integral using  d\theta as the variable of integration. I'm actually not sure if is the washers or shells  dx integral that results after the substitution.

    I changed the 2nd integral in my post to the shells integral, btw.

    Thanks so much for helping out!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Shell method is of the form:

    \displaystyle 2\pi\int_{a}^{b}p(x)h(x)dx
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2010
    Posts
    20
    So I forgot another  r*\cos^(\theta) term! But how? I thought I had the general form of shells correct, just:
     \int_{\theta_1}^{\theta_2} \pi r(\theta) d\theta

    and in the 1st integral's case,  r(\theta)=r*\cos(\theta) , no?

    Thanks again.

    EDIT: I meant washers, sorry.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Sep 2010
    Posts
    20
    EDIT to post below: I meant washers for the 1st integral, not shells, sorry.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Everything was correct except for forgetting the r*\cos(\theta).

    If you would do the integration with r*\cos(\theta), you would obtain your desired results.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Shell with x=rsin substitution

    \displaystyle 4\pi r^3\int_{0}^{\frac{\pi}{2}}(\sin{\theta}*\cos^2{\t  heta})d\theta=\frac{4\pi r^3}{2}

    Washer with x=rsin substitution

    \displaystyle \pi r^3\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\cos^3{\theta}d\theta=\fra  c{4\pi r^3}{3}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Sep 2010
    Posts
    20
    Hmmm... I distinctly remember my 1st integral the one we used in class and it worked, but I think you are showing me what I should do for the shells method. I think I need the washers method actually.

    EDIT: I just saw your reply above, thanks for helping me understand shells using  d\theta so much better. But I was wondering if you could still show me washers- thats the one I initially wanted.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by progressive View Post
    Hmmm... I distinctly remember my 1st integral the one we used in class and it worked, but I think you are showing me what I should do for the shells method. I think I need the washers method actually.

    EDIT: I just saw your reply above, thanks for helping me understand shells using  d\theta so much better. But I was wondering if you could still show me washers- thats the one I initially wanted.
    I edited my post #8 with both methods and with your desired substitutions.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Sep 2010
    Posts
    20
    I think I get it now. Thanks very much for all your help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Volume of a Sphere
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 22nd 2010, 05:43 PM
  2. Replies: 1
    Last Post: August 8th 2010, 12:22 AM
  3. Volume and Sphere
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 17th 2010, 12:06 PM
  4. The volume/SA of a sphere
    Posted in the Algebra Forum
    Replies: 4
    Last Post: February 10th 2010, 05:53 AM
  5. [SOLVED] Volume cut out of a sphere
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 9th 2007, 07:48 PM

/mathhelpforum @mathhelpforum