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Math Help - Find the mass of the cylinder

  1. #1
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    Find the mass of the cylinder

    Find the mass of the cylinder S:0<=z<=h, x^2+y^2<=a^2 if the density at the point (x,y,z) is \delta = 10z^4+6(x^2-y^2)^2.

    I think we should use the cylincal polar method to do this, however i have problem converting this into the polar form, which is x=rcos\theta, y =rsin\theta, z=z
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  2. #2
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    You're having trouble converting the integrand into cylindrical coordinates? What happens when you plug in for x and y?
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  3. #3
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    You might find this trig identity helpful:

    cos^2(x)- sin^2(x)= sin(2x).

    Perhaps you will remember the more basic identity
    cos(a+ b)= cos(a)cos(b)- sin(a)sin(b).

    Let a= b= x.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    You might find this trig identity helpful:

    cos^2(x)- sin^2(x)= sin(2x).

    Perhaps you will remember the more basic identity
    cos(a+ b)= cos(a)cos(b)- sin(a)sin(b).

    Let a= b= x.
    so, after converting, I got \int _\,^\! \int _0\,^h\! \int _0\,^{a^2}\! 10z^4+6rsin^2(2 \theta ) dr dz d\theta

    is this correct? I'm still not sure with the integrand of \theta however.
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  5. #5
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    No, I don't think you've quite got it yet. Just plugging in gives:

    \delta=10z^4+6(x^2-y^2)^2=10z^4+6((r\cos(\theta))^2-(r\sin(\theta))^2)^2=10z^{4}+6(r^{2}\cos^{2}(\thet  a)-r^{2}\sin^{2}(\theta))^{2}

    =10z^{4}+6(r^{2}(\cos^{2}(\theta)-\sin^{2}(\theta)))^{2}=<br />
10z^{4}+6(r^{2}\sin(2\theta))^{2}=10z^{4}+6r^{4}\s  in^{2}(2\theta).

    You see how I work here: step-by-step. I don't often skip steps.

    The other thing you have to worry about is that the volume differential in cylindrical components is r\,dr\,d\theta\,dz, not just dr\,d\theta\,dz.

    So, putting this all together, what does your integral look like now?
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  6. #6
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    Quote Originally Posted by Ackbeet View Post
    No, I don't think you've quite got it yet. Just plugging in gives:

    \delta=10z^4+6(x^2-y^2)^2=10z^4+6((r\cos(\theta))^2-(r\sin(\theta))^2)^2=10z^{4}+6(r^{2}\cos^{2}(\thet  a)-r^{2}\sin^{2}(\theta))^{2}

    =10z^{4}+6(r^{2}(\cos^{2}(\theta)-\sin^{2}(\theta)))^{2}=<br />
10z^{4}+6(r^{2}\sin(2\theta))^{2}=10z^{4}+6r^{4}\s  in^{2}(2\theta).

    You see how I work here: step-by-step. I don't often skip steps.

    The other thing you have to worry about is that the volume differential in cylindrical components is r\,dr\,d\theta\,dz, not just dr\,d\theta\,dz.

    So, putting this all together, what does your integral look like now?
    yes, r is the jocobian, so what are the integrands here, i got r from 0 to a^2, z from 0 to h, how about \theta. is it 0 to 2pi?
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  7. #7
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    Yes. The condition x^{2}+y^{2}\le a^{2}, with no explicit limits on the angle, implies that 0\le\theta\le 2\pi.
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  8. #8
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    Quote Originally Posted by Ackbeet View Post
    Yes. The condition x^{2}+y^{2}\le a^{2}, with no explicit limits on the angle, implies that 0\le\theta\le 2\pi.
    i think i made a mistake finding the integrand of r, it should be in terms of \theta, not 0 to a^2
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  9. #9
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    Show your work, and I'll take a look.
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