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 $z=$ $\sqrt{{x}^{2}+{y}^{2}}$?

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 $z=$ $\sqrt{{x}^{2}+{y}^{2}}$?