The Unapologetic Mathematician

Mathematics for the interested outsider

An Example of an Iterated Integral

My description of how to evaluate a multiple integral over some region other than an n-dimensional interval by using iterated integrals might not have been the clearest, so I’m hoping an example will help illustrate what I mean. Let’s calculate the Jordan content of a three-dimensional sphere of radius a, centered at the origin. This will also help cement the fact that Jordan content is closely related to what we mean by “volume”.

So, we have our sphere

\displaystyle S=\left\{(x,y,z)\in\mathbb{R}^3\big\vert x^2+y^2+z^2\leq a^2\right\}

We can put it inside the interval R=[-a,a]\times[-a,a]\times[-a,a], so we know that the Jordan content is

\displaystyle c(S)=\int\limits_S\,d(x,y,z)=\int\limits_R\chi_S(x,y,z)\,d(x,y,z)

Now we want to peel off an integral from the inside. Let’s first integrate over the variable z. Projecting the sphere onto the plane z=0 we’re left with the circle

\displaystyle S_z=\left\{(x,y)\in\mathbb{R}^2\big\vert x^2+y^2\leq a^2\right\}

and we need to write S as the region between the graphs of two functions on this projection. And indeed, we can write

\displaystyle S=\left\{(x,y,z)\in\mathbb{R}^3\big\vert(x,y)\in S_z,-\sqrt{a^2-x^2-y^2}\leq z\leq\sqrt{a^2-x^2-y^2}\right\}

Thus we can write

\displaystyle\int\limits_S\,d(x,y,z)=\int\limits_{S_z}\int\limits_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}}\,dz\,d(x,y)

Next we’ll integrate with respect to y. Projecting S_z onto the line y=0 we find S_{zy}=[-a,a]. And then we find

\displaystyle S_z=\left\{(x,y)\in\mathbb{R}^2\big\vert-a\leq x\leq a,-\sqrt{a^2-x^2}\leq y\leq\sqrt{a^2-x^2}\right\}

which lets us peel off another integral from the inside

\displaystyle\int\limits_S\,d(x,y,z)=\int\limits_{[-a,a]}\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\int\limits_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}}\,dz\,dy\,dx=\int\limits_{-a}^a\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\int\limits_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}}\,dz\,dy\,dx

Finally, we can start evaluating these integrals. The innermost integral goes immediately, using the fundamental theorem of calculus. An antiderivative of 1 with respect to z is z, and so we find

\displaystyle\begin{aligned}\int\limits_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}}\,dz&=z\bigg\vert_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}}\\&=\sqrt{a^2-x^2-y^2}-(-\sqrt{a^2-x^2-y^2})\\&=2\sqrt{a^2-x^2-y^2}\end{aligned}

So we put this into our integral, already in progress:

\displaystyle\int\limits_S\,d(x,y,z)=\int\limits_{-a}^a\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\int\limits_{-\sqrt{a^2-x^2-y^2}}^{\sqrt{a^2-x^2-y^2}}\,dz\,dy\,dx=\int\limits_{-a}^a\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}2\sqrt{a^2-x^2-y^2}\,dy\,dx

Here’s where things start to get difficult. I’ll just tell you (and you can verify) that an antiderivative of 2\sqrt{a^2-x^2-y^2} with respect to y is

\displaystyle y\sqrt{a^2-x^2-y^2}+(a^2-x^2)\arctan\left(\frac{y}{\sqrt{a^2-x^2-y^2}}\right)

So the fundamental theorem of calculus tells us

\displaystyle\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}2\sqrt{a^2-x^2-y^2}\,dy=(a^2-x^2)\left(\arctan\left(\frac{\sqrt{a^2-x^2}}{0^+}\right)-\arctan\left(\frac{-\sqrt{a^2-x^2}}{0^+}\right)\right)

Technically, this is an improper integral (which we haven’t really discussed), so we need to take some limits. The denominators 0^+ are limits as some dummy variable approaches zero from above. Thus we continue evaluating our integral

\displaystyle\int\limits_S\,d(x,y,z)=\int\limits_{-a}^a\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}2\sqrt{a^2-x^2-y^2}\,dy\,dx=\int\limits_{-a}^a\pi(a^2-x^2)\,dx

This one is easy: an antiderivative of a^2-x^2 is a^2x-\frac{x^3}{3}, and so

\displaystyle\begin{aligned}\int\limits_S\,d(x,y,z)&=\int\limits_{-a}^a\pi(a^2-x^2)\,dx\\&=\pi\left(a^3-\frac{a^3}{3}\right)-\pi\left(-a^3-\frac{(-a)^3}{3}\right)\\&=\frac{4}{3}\pi a^3\end{aligned}

And this is exactly the well-known formula for the volume of a sphere of radius a. There are easier ways to get at this formula, of course, but this one manages to illustrate the technique of iterated integrals with variable limits of integration on the inner integrals.

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January 4, 2010 - Posted by | Analysis, Calculus

3 Comments »

  1. It looks like you skipped a step, between “The denominators 0+ are limits as some dummy variable approaches zero from above.” and “Thus we continue evaluating our integral”.

    It is likely that we are supposed to know that arctan(∞) is π/2, but since arctan was introduced as a “I’ll just tell you..”, a statement to that effect would have been nice.

    Comment by Blaise Pascal | January 4, 2010 | Reply

  2. I didn’t mean “I’ll just tell you” to introduce the notion of arctan at all. It was meant to skip the derivation of the antiderivative rather than go through all the mess of trigonometric substitutions. This is perfectly valid, since all we need for the FToC is an antiderivative, wherever it comes from.

    Comment by John Armstrong | January 4, 2010 | Reply

  3. [...] there are other places that so-called “improper integrals” come up, like the example I worked through the other day. In this case, the fundamental theorem of calculus runs into trouble at the endpoints [...]

    Pingback by Improper Integrals II « The Unapologetic Mathematician | January 15, 2010 | Reply


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