Is it possible to find the volume of an object bounded by two surfaces in both of these two ways?:
-a triple integral of 1 dV (I know this works)
-a double integral of the top surface - bottom surface dA (over domain of the intersection)
How do you tell if a double integral or triple integral is best for a certain situation?
At least in the case of integral over a cylindrical shape in $\mathbb{R}^3$, it wouldn't matter. I.e., let $B$ the shape of interest, hence, you should be able to describe $B$ in the following way $B=\{(x,y,z)|(x,y)\in C, z_1(x,y) \le z \le z_2(x,y) \}$, where $C$ is the projection of $B$ on the $xy$ plane. Hence, the integral can be expressed as $$ V(B) = \int \int_Ddxdy \int_{z_1(x,y)}^{z_1(x,y)} dz = \int \int_D (z_2(x,y) - z_1(x,y)) dxdy, $$ where the last equality is exactly the second method of double integration.