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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?

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

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  • $\begingroup$ does the inner integral in the second statement have any geometrical meaning or what does it mean for the integral to have an upper bound as a function of x $\endgroup$ – john bishop May 22 '18 at 10:28
  • $\begingroup$ @johnbishop That integral sums infinitely many little square of dimension $dt \times dx$ within the specified bounds for $t$ and $x$. these bounds are set by $f$. $\endgroup$ – Kevin May 22 '18 at 10:35
  • $\begingroup$ @johnbishop More generally however, this type of relation is a direct consequence of Fubini's theorem. This theorem permits the computation of this integral recursively in terms of simple and / or double integrals over certain intervals or, resp. two-dimensional domains. $\endgroup$ – Kevin May 22 '18 at 10:41
  • $\begingroup$ oh ok i think i need to research a bit more, thanks for taking the time! $\endgroup$ – john bishop May 22 '18 at 10:51

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