Relationship between volume under the graph and the double integral of the function

Question

How does the volume under the graph of f (x,y) over some rectangular region, x from a to b, and y from c to d relate to the double integral of the function. Is there some fundamental theorem kind of proof like in the single variable case where if the surface is continuous, the derivative of the volume function at some y = k is the cross sectional area ( i.e the area under the line for x between a and b) and so the anti derivative would give me the volume under the graph?

Answer 1

Double and triple integration allows for volume calculations. For example, consider the area under a curve $f(x)>0$ over some domain. The the area $A$ is obviously given by

$$A = \int^a_bf(x)dx$$

Now the following conclusions are equivalent $$\int^a_b\int_0^{f(x)} dtdx = \int^a_bf(x)dx = A$$

hence, the computation of area can be achieved byu both single and double integration.

To compute volume : $$\int\int\int^{f(x,y)}_0dzdxdy = \int\int f(x,y)dxdy$$