Multiple Integrals

If you can do single integrals, you can do multiple integrals: just do more than one integral, holding variables other than the current variable of integration constant. For example,

$$\begin{array}{rl}{\int }_0^1{\int }_0^y(x-y{)}^{2}dxdy& ={\int }_0^1{\left[\frac{(x-y{)}^3}{3}\right]}_0^{y}dy\\ & ={\int }_0^1\left(\frac{(y-y{)}^3}{3}-\frac{(0-y{)}^3}{3}\right)dy\\ & ={\int }_0^1\frac{y^3}{3}dy\\ & ={\left[\frac{y^4}{12}\right]}_0^1\\ & =\frac{1}{12}\end{array}$$

Change of Order of Integration

We can also integrate in the other order, $dydx$ rather than $dxdy$, as long as we are careful about the limits of integration. Since we’re integrating over all $(x,y)$ with $x$ and $y$ between $0$ and $1$ such that $x\le y$, to integrate the other way we write

$$\begin{array}{rl}{\int }_0^1{\int }_x^1(x-y{)}^{2}dydx& ={\int }_0^1{\left[-\frac{(x-y{)}^3}{3}\right]}_{y=x}^{y=1}dx\\ & ={\int }_0^1\left(-\frac{(x-1{)}^3}{3}+\frac{(x-x{)}^3}{3}\right)dx\\ & ={\int }_0^1\frac{(1-x{)}^3}{3}dx\\ & ={\left[-\frac{(1-x{)}^4}{12}\right]}_0^1\\ & =\frac{1}{12}\end{array}$$

Change of Variables

In making a change of variables with multiple integrals, a Jacobian is needed. Let’s state a two-dimensional version, for concreteness. Suppose we make a change of variables (transformation) from $(x,y)$ to $(u,v)$, say with $x=g(u,v),y=h(u,v)$. Then

$$\iint f(x,y)dxdy=\iint f(g(u,v),h(u,v))\left|\frac{\partial (x,y)}{\partial (u,v)}\right|dudv$$

over the appropriate limits of integration, where $\left|\frac{d(x,y)}{d(u,v)}\right|$ is the absolute value of the determinant of the Jacobian (We assume that the partial derivatives exists and are continuous, and that the determinant is nonzero).

Example

For example, let’s find the are of a circle of radius $1$. To find the are of a region, we just need to integrate $1$ over that region (So any difficulty comes from the limits of integration; the function we’re integration is just the constant $1$). So the area is

$${\iint }_{x^2+y^2\le 1}1dxdy={\int }_{-1}^1{\int }_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}1dxdy=2{\int }_{-1}^1\sqrt{1-y^2}dy$$

Note that the limits of the inner variable $(x)$ of the double integral can depend on the outer variable $(y)$, while the outer limits are constants. The last integral can be done with a trig substitution, but instead let’s simplify the problem by transforming to polar coordinates:

$$x=r\cos \theta ,y=r\sin \theta$$

where $r$ is the distance from $(x,y)$ to the origin and $\theta \in [0,2\pi )$ is the angle. The Jacobian of the transformation is

$$\frac{\partial (x,y)}{\partial (r,\theta )}=\left|\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right|$$

so the absolute value of the determinant is $$\begin{gathered} r( \\ co{s}^2\theta +{\sin }^2\theta )=r \end{gathered}$$. That is, $dxdy$ becomes $rdrd\theta$. So the area of the circle is

$${\int }_0^{2\pi }{\int }_0^{1}rdrd\theta ={\int }_0^{2\pi }\frac{1}{2}d\theta =\pi$$

For a circle of radius $r$, it follows immediately that the area of $\pi r^2$ since we can imagine converting our units of measurement to the unit for which the radius is $1$.

This may seem like a lot of work just to get such a familiar result, but it served a illustration and with similar methods, we can get the volume of a ball in any number of dimensions! It turns out that the volume of a ball of radius $1$ in $n$ dimensions is $\frac{{\pi }^{n/2}}{\Gamma (\frac{n}{2}+1)}$ where $\Gamma$ is the Gamma Function.