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.
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