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Math Help - [SOLVED] Find a volume

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Find a volume

    Find the volume inside the sphere x^2+y^2+z^2=4a^2 and outside the cylinder x^2+y^2=2ax.
    My attempt : I visualized the volume I must calculate. I can first calculate the volume of the sphere ( \frac{32\pi a^3}{3} ) and then calculate the volume of the cylinder inside the sphere. This is where I'm having problems.
    I think its volume is the following one, but I'm unsure :  \int_{-\pi}^{\pi} \int_{0}^{2a\cos \theta} \int_{-\sqrt{4a^2-r^2}}^{\sqrt{4a^2-r^2}} r dz dr d \theta = \int_{-\pi}^{\pi} \int_{0}^{2a\cos \theta} 2r \sqrt{4a^2-r^2} dr d\theta =...
    (I've used cylindrical coordinates for the volume of the cylinder inside the sphere, a better said, the part of the cylinder.)
    I'd like to know if what I did until now is right. If so I'd have to remember how to solve the iterated integral I got.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    For the circle r = 2acos(theta) The upper half is generated as theta varies from 0 to pi/2 and the lower half as theta varies from -pi/2 to zero.

    your outer limits should then be -pi/2 to pi /2

    more simply go 0 to pi/2 and double it.

    As for the integral let u = 4a^2 -r^2

    after solving you'll end up with an integral involving sin^3(theta)

    which can easily be solved as (1-cos^2(theta))sin(theta)
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  3. #3
    MHF Contributor arbolis's Avatar
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    I didn't reach the result yet and I probably made an error somewhere.
    Calculus of the volume of the cylinder inside the sphere : I=2\int_{0}^{\frac{\pi}{2}} \int_{0}^{2a\cos (\theta)} \int_{-\sqrt{4a^2-r^2}}^{\sqrt{4a^2-r^2}} rdzdrd\theta.

    \int_{-\sqrt{4a^2-r^2}}^{\sqrt{4a^2-r^2}} rdz=2r\sqrt{4a^2-r^2}.

    So I=2\int_{0}^{\frac{\pi}{2}} \int_{0}^{2a\cos (\theta)} 2r\sqrt{4a^2-r^2} dr d\theta.

    I call J=\int_{0}^{2a\cos(\theta)} 2r\sqrt{4a^2-r^2}dr.
    Let u=4a^2-r^2 \Rightarrow du=-2rdr, hence J=-\int_{0}^{2a\cos (theta)} \sqrt{u} du=-\frac{2}{3} (2a\cos(\theta))^{\frac{3}{2}}.
    Thus I=-\frac{4}{3} \int_{0}^{\frac{\pi}{2}} (2a\cos (\theta))^{\frac{3}{2}}d\theta.
    I don't know how to solve this integral. I didn't reach \sin ^3 (\theta) yet so I think I made an error.
    Could you tell me if I made an error? If no, I'd like to know which method to use to solve the integral because I've no idea.
    Thank you for your help.
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by arbolis View Post
    I didn't reach the result yet and I probably made an error somewhere.
    Calculus of the volume of the cylinder inside the sphere : I=2\int_{0}^{\frac{\pi}{2}} \int_{0}^{2a\cos (\theta)} \int_{-\sqrt{4a^2-r^2}}^{\sqrt{4a^2-r^2}} rdzdrd\theta.

    \int_{-\sqrt{4a^2-r^2}}^{\sqrt{4a^2-r^2}} rdz=2r\sqrt{4a^2-r^2}.

    So I=2\int_{0}^{\frac{\pi}{2}} \int_{0}^{2a\cos (\theta)} 2r\sqrt{4a^2-r^2} dr d\theta.

    I call J=\int_{0}^{2a\cos(\theta)} 2r\sqrt{4a^2-r^2}dr.
    Let u=4a^2-r^2 \Rightarrow du=-2rdr, hence {\color{red}J=-\int_{0}^{2a\cos (theta)} \sqrt{u} du=-\frac{2}{3} (2a\cos(\theta))^{\frac{3}{2}}}.
    Thus I=-\frac{4}{3} \int_{0}^{\frac{\pi}{2}} (2a\cos (\theta))^{\frac{3}{2}}d\theta.
    I don't know how to solve this integral. I didn't reach \sin ^3 (\theta) yet so I think I made an error.
    Could you tell me if I made an error? If no, I'd like to know which method to use to solve the integral because I've no idea.
    Thank you for your help.
    you just did not change the boundaries of the integral when you let
    Let u=4a^2-r^2 \Rightarrow du=-2rdr,

    u = 4a^2 - (2acos(\theta))^2 \Rightarrow u = 4a^2(1-cos^2(\theta))

    u=4a^2 - (2(0))^2 = 4a^2
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  5. #5
    MHF Contributor arbolis's Avatar
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    Oh... thanks a lot.
    So J=\int_{4a^2 \sin ^2{\theta}}^{4a^2} u^{\frac{1}{2}} du = \frac{2}{3} \big( (4a^2)^{\frac{3}{2}}-(4a^2 \sin ^2 {\theta})^{\frac{3}{2}} \big)  ?
    And I still don't see any integral involving \sin^3 (\theta). Again, I made an error.
    Last edited by arbolis; June 13th 2009 at 12:03 PM.
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  6. #6
    MHF Contributor Calculus26's Avatar
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    In general (x^2)^(3/2) = x^3

    sin^2(t)^(3/2) = sin^3(t)

    Of course you also have to take care of the constant 4a^2 part also
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  7. #7
    MHF Contributor arbolis's Avatar
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    Lol, I missed this! Ok, I'll try to finish it up.


    Ok, I've done it. My result for the volume of the cylinder inside the sphere is \frac{64-192 \pi a^3}{54}. I don't know if I made an arithmetic error but at least I know how to do it.
    So the final answer would be \frac{32\pi a^3}{3}+\frac{192\pi a^3-64 }{54}=\frac{384\pi a^3-64}{54} which is not beautiful.
    Last edited by arbolis; June 13th 2009 at 01:37 PM.
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