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Thread: Mass of a solid

  1. #1
    Junior Member
    Nov 2008

    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

    Apr 2005
    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- $\displaystyle \theta$ will range from $\displaystyle 0$ to $\displaystyle 2\pi$. Slightly harder: Since the cone, in cylindrical coordinates, is $\displaystyle z^2= r^2$ and the cylinder by $\displaystyle r= 2cos(\theta)$ ($\displaystyle x^2+ y^2= r^2$ and $\displaystyle 2x= 2rcos(\theta)$: $\displaystyle r^2= 2rcos(\theta)$ and we can cancel an "r"), the cone and cylinder intersect when $\displaystyle z^2= 4cos^2(\theta)$. For each [itex]\theta[/itex], z ranges from $\displaystyle -2 cos(\theta)$ to $\displaystyle 2 cos(\theta)$. Finally, for each z and $\displaystyle \theta$, r ranges from the cone out to the cylinder. The cone is r= |z| (r is positive, of course) and the cylinder $\displaystyle r= 2cos(\theta)$ as before.

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

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