Mass of a solid

**Shapeshift** · January 17th 2011, 03:38 PM

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.

**HallsofIvy** · January 17th 2011, 04:55 PM

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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