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Math Help - Mass of a solid

  1. #1
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    Mass of a solid

    Right now I'm learning about the Change of Variables formula and applications of integration, and I'm having trouble with this question:


    Find the mass of the solid bounded by the cylinder x^2+y^2=2x and the cone z^2=x^2+y^2 if the density is delta=sqrt(x^2+y^2).

    I have the idea that I should switch to cylindrical coordinates. So I would let x=rcostheta, y=rsintheta, and z=z. I get that delta=r, but I am having trouble figuring out what my limits of integration will be. Any help would be appreciated.
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  2. #2
    MHF Contributor

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    Yes, I agree that, because of the circular symmetry, cylindrical coordinates should simplify the integration. I assume that by "the solid bounded by" the cylinder and cone, they mean the region between the two "nappes" of the cone, inside the cylinder.

    The easy part- \theta will range from 0 to 2\pi. Slightly harder: Since the cone, in cylindrical coordinates, is z^2= r^2 and the cylinder by r= 2cos(\theta) ( x^2+ y^2= r^2 and 2x= 2rcos(\theta): r^2= 2rcos(\theta) and we can cancel an "r"), the cone and cylinder intersect when z^2= 4cos^2(\theta). For each [itex]\theta[/itex], z ranges from -2 cos(\theta) to 2 cos(\theta). Finally, for each z and \theta, r ranges from the cone out to the cylinder. The cone is r= |z| (r is positive, of course) and the cylinder r= 2cos(\theta) as before.

    Your density function is \delta= \sqrt{x^2+ y^2}= r and the differential of volume is r drd\theta dz.

    Putting all of that together, your integral is
    \int_{\theta= 0}^{2\pi}\int_{z= -2cos(\theta)}^{2cos(\theta)}\int_{r= |z|}^{2cos(\theta)} r (r drdzd\theta)
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