10. Multidimensional integrals

Double integrals

Given a function f\colon \mathbb{R}^2\to \mathbb{R} and a set D\subseteq \mathbb{R}^2, the integral

    \[\int\int_{D} f(x,y)\,dx\,dy\]

is simply the volume under the graph of f over the given set.

Assume that we would like to calculate the volume of a pyramid with a base in form of a isosceles right triangle (with boundary of axes and line y=-x+1) and height 1, so in point (x,y) the height is given by the following formula h(x,y)=-x-y+1. So the volume is

    \[\int\int_D h(x,y)\,dx\,dy\]

where D is the triangle mentioned above.

How to calculate this integral? It will be described below.

Fubini Theorem

Fubini theorem states that such an integral is the integral over x of the integral over y, and i the same as integral over y of the integral over x, in other words (for functions nice enough):

    \[\int\int_{D} f(x,y)\,dx\,dy = \int_{y_0}^{y_1}\left(\int_{x_0(y)}^{x_1(y)} f(x,y)\,dx\right)\,dy=\]

    \[=\int_{x_0}^{x_1}\left(\int_{y_0(x)}^{y_1(x)} f(x,y)\,dy\right)\,dx,\]

where y_0,y_1, x_0(y), x_1(y) and x_0,x_1, y_0(y), y_1(y) are such that

    \[D=\{(x,y)\colon y_0\leq y\leq y_1, x_0(y)\leq x\leq  x_1(y)\}=\]

    \[=\{(x,y)\colon x_0\leq x\leq x_1, y_0(x)\leq y\leq  y_1(x)\}.\]

Obviously you have to treat the outer variable as a constant parameter when calculating the inner integral.

Therefore, returning to our example of the pyramid, the section at x has area P_x=\int_0^{1-x} h(x,y)\, dy, so the volume is:

    \[\int_0^1\left(\int_0^{1-x} h(x,y)\,dy\right)\, dx= \int_0^1\left(-xy-y^2/2+y\right)|_0^{1-x}\,dx=\]

    \[=\int_0^1\frac{1}{2}(x-1)^2\,dx=(x^3/6-x^2/2+x/2)|_0^1=\frac{1}{6},\]

which is consistent with our knowledge about geometry.