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Thread: Find the mass of the cylinder

  1. #1
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    Find the mass of the cylinder

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

    I think we should use the cylincal polar method to do this, however i have problem converting this into the polar form, which is $\displaystyle x=rcos\theta$, $\displaystyle y =rsin\theta$,$\displaystyle z=z$
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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:

    $\displaystyle cos^2(x)- sin^2(x)= sin(2x)$.

    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:

    $\displaystyle cos^2(x)- sin^2(x)= sin(2x)$.

    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 $\displaystyle \int _\,^\!$ $\displaystyle \int _0\,^h\! $ $\displaystyle \int _0\,^{a^2}\! $ $\displaystyle 10z^4+6rsin^2$(2$\displaystyle \theta$ ) $\displaystyle dr dz d\theta$

    is this correct? I'm still not sure with the integrand of $\displaystyle \theta$ 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:

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

    $\displaystyle =10z^{4}+6(r^{2}(\cos^{2}(\theta)-\sin^{2}(\theta)))^{2}=
    10z^{4}+6(r^{2}\sin(2\theta))^{2}=10z^{4}+6r^{4}\s in^{2}(2\theta).$

    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 $\displaystyle r\,dr\,d\theta\,dz,$ not just $\displaystyle dr\,d\theta\,dz.$

    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:

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

    $\displaystyle =10z^{4}+6(r^{2}(\cos^{2}(\theta)-\sin^{2}(\theta)))^{2}=
    10z^{4}+6(r^{2}\sin(2\theta))^{2}=10z^{4}+6r^{4}\s in^{2}(2\theta).$

    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 $\displaystyle r\,dr\,d\theta\,dz,$ not just $\displaystyle dr\,d\theta\,dz.$

    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 $\displaystyle a^2$, z from 0 to h, how about $\displaystyle \theta$. is it 0 to 2pi?
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  7. #7
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    Yes. The condition $\displaystyle x^{2}+y^{2}\le a^{2},$ with no explicit limits on the angle, implies that $\displaystyle 0\le\theta\le 2\pi.$
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  8. #8
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    Quote Originally Posted by Ackbeet View Post
    Yes. The condition $\displaystyle x^{2}+y^{2}\le a^{2},$ with no explicit limits on the angle, implies that $\displaystyle 0\le\theta\le 2\pi.$
    i think i made a mistake finding the integrand of r, it should be in terms of $\displaystyle \theta$, not $\displaystyle 0 to a^2$
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  9. #9
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    Show your work, and I'll take a look.
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