12. Applications of multidimensional integrals

Integration with change of variables

If D\colon A\to B is a diffeomorphism, the following theorem holds

    \[\int\int_{B}F(x,y)\,dx\,dy=\int\int_{A}F(D(t,u))\cdot |\det D'|\, dt\,du.\]

Polar coordinates

The most common use of the above method are polar coordinates which are defined by the diffeomorphism \Phi(r,\theta)=(r\cos \theta,r\sin\theta)=(x,y). Thus, x^2+y^2=r^2(\cos^2\theta+\sin^2\theta)=r^2.
Moreover,

    \[\Phi'=\left[\begin{array}{cc}\cos \theta& -r\sin\theta \\ \sin \theta & r\cos\theta\end{array}\right],\]

so

    \[\det \Phi'= r(\cos^2\theta+\sin^2\theta)=r.\]

E.g. let us calculate

    \[\int\int_{\R^2}e^{-x^2-y^2}\,dx\,dy.\]

We get:

    \[\int\int_{\R^2}e^{-x^2-y^2}\,dx\,dy=\int_{(0,\infty)\times (0,2\pi)}|r|e^{r^2}\,dr\,d\theta=\]

    \[=\int_{0}^{\infty}\int_0^{2\pi} re^{r^2}\,d\theta\,dr=\int_{0}^{\infty} \theta\cdot re^{r^2}|_{2}^{2\pi}\,dr=\]

    \[=\pi\int_{0}^{\infty} 2re^{r^2}\,dr=\pi \lim_{a\to \infty}\int_{0}^{a} 2re^{r^2}\,dr=\pi \lim_{a\to \infty} e^{r^2}|_0^2=\pi(1-0)=\pi.\]

Volume

If we want to calculate the volume of a set V we can do it using one of the following three methods:

    \[V=\int\int\int_{V} 1\,dx\,dy\,dz=\int\int_{S} H(x,y)\,dx\,dy=\int_a^b S(x)\,dx,\]

where S is the basis H(x,y) is the height at x,y and S(x) is the area of the section for given x.

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. Therefore 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.

Length of a curve

We can also calculate a length of a curve using integrals. Given a parametrization of a curve (x(t),y(t)), its length between t=a and t=b, is given by the following formula:

    \[\int_a^b\sqrt{(x'(t))^2+(y'(t)})^2\,dt\]

Let us check this formula for the line x=y between x=0 and x=1. Such a line has parametrization x(t)=t, y(t)=t^2. Therefore its length is:

    \[L=\int_0^1\sqrt{1^2+1^2}\,dt=\sqrt{2}t|_0^1=\sqrt{2}.\]