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When finding the volume of a solid using triple integrals, is the "integral" always $\int \int \int $ $dV$, regardless of what the "z function" is?

For instance, for the problem: Find the volume of the solid that lies within the sphere $x^2 + y^2 + z^2 = 4$, above the xy-plane, and below the cone $z$ = $\sqrt{x^2 + y^2}$.

Is the integral just $\int \int \int $ $dV$ with the correct bounds of integration, regardless of $z$ = $\sqrt{x^2 + y^2}$? Thanks.

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  • $\begingroup$ $\iiint \; \mathrm{d}V$ just symbolically means 3-dimensional integral over some volume-element $\text{d}V$. You must first determine over what volume-element you're integrating. (your case: some sphere with the lower half cut off and a cone cut out, i.stack.imgur.com/Q5zZr.png) $\endgroup$ – Maximilian Gerhardt May 9 '16 at 10:36
  • $\begingroup$ You can use \iiint to get a proper triple integral (as in the above comment). $\endgroup$ – joriki May 9 '16 at 10:51
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If you have triple integral of type $\iiint F DV$ and region in some volume V. This means you are adding value of F over all points over V. But if F is 1, this means you're finding volume of V

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