The question:

Find the volume of the solid enclosed by the sphere $x^2 + y^2 + z^2 - 6z = 0$ , and the hemisphere $x^2 + y^2 + z^2 = 49 , z ≥ 0$

I set up the triple integral

$\int_0^{2\pi}\int_0^\pi\int_{6cos(\phi)}^7 \rho^2sin(\phi)~d\rho~d\phi~d\theta$

And evaluated it to get $\frac{1372\pi}{3}$, but I am unsure about the integration limits I used. This is one of my first spherical problems and I am a little confused about how to find the $\phi$ and $\rho$ limits.

  • $\begingroup$ The smaller sphere fits entirely under the hemisphere without intersection (radius of 3 with center 3 units above the $ \ xy-$ plane in a hemisphere of radius 7), so integration isn't really called for. Are the coefficients you show for the equations of the surfaces correct? $\endgroup$ – colormegone Apr 3 '14 at 18:42
  • $\begingroup$ @RecklessReckoner The coefficients are correct. So this is looking like a decoy question that is just a simple calculation? $\endgroup$ – user2847441 Apr 3 '14 at 18:44
  • $\begingroup$ @RecklessReckoner I get the same answer using the formula for the volume of a sphere, so I guess it must be right. Thanks $\endgroup$ – user2847441 Apr 3 '14 at 18:46
  • $\begingroup$ I also graphed it to be sure. The sphere is completely contained inside the hemisphere, so the volume enclosed by both surfaces is just the volume of the sphere. If this is a "trick" question, they may be checking to see if you visualized the situation correctly. As for your question, the geometrical situation will influence your choice of integration limits (with no intersection of the surfaces, you wouldn't integrate for $ \ \rho \ $ all the way out to 7 ) . $\endgroup$ – colormegone Apr 3 '14 at 18:46

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