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Find the volume and centroid of the solid that lies in the intersection of the infinite cone $z=\sqrt{x^2 + y^2}$ and the sphere $x^2 + y^2 + z^2 = 1$. Choose an appropriate coordinate system.

So the second eqn is a perfect sphere with radius one. And since its volume I will probably need a triple integral... but im having a hard time find out what to integrate since I have 2 eqns :s

Also for the coordinate system, im thinking something like spherical?

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You should use spherical coordinates, the cone $z = \sqrt{x^2 + y^2}$ looks like this

http://www.wolframalpha.com/input/?i=z+%3D+sqrt%28xx+%2Byy%29

therefore the volume is only bounded by the top half of the sphere. From an orthographic view you can imagine the volume to look like an ice cream cone with the bottom of the cone at the origin. The cone will intersect the sphere and cast a circle as a projection in the $xy$ plane. To find the $z$ plane of intersection equate the two equations. Your row will be bounded between $0$ and the radius of sphere. Phi will be bounded between $0$ and $\pi/6$ I believe ( you can find phi yourself once you find the $z$ plane through which the cone intersects the sphere). You theta will have bounds $0 ... 2\pi$.

EDIT: Made to value of ROW.

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$\rho$ = rho, not row. We don't really need the z-plane of intersection if the spherical coordinates are to be used. The bound on $\phi$ is best found by equating $\rho\cos\phi$ (which is $z$) to $\rho\sin\phi$ (which is $\sqrt{x^2+y^2}$). –  user31373 Aug 17 '12 at 1:48
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