# 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}?

Volume and Triple Integrals
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 $z=$ $\sqrt{{x}^{2}+{y}^{2}}$?
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falwsay
Explanation:
If you have triple integral of type $\iiint FDV$ 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