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# Lecture Notes

Below are the lecture notes from my course on multivariable calculus, as well as some handouts I give to students for some of the experiments in the introductory physics lab courses. Note that I update my notes each term, but this website may not link to the latest editions.

### Notes on MTH3015: Calculus III


Week 10: Triple Integrals; Cylindrical and Spherical Co-ordinates

Sujeet Akula

Simple Domains in 3D

To generalize our 2D formalism to 3D, we begin with the simplest case. Recall that we began our discussion in 2D by considering the simple domains in the shape of rectangles. The obvious generalization to 3D is then a rectangular box. Extending the notation that we established last week, we can denote the set of points in a rectangular box as $Q=[a,b]\times[c,d]\times[r,s]$, with corners $(a,c,r)$, $(a,c,s)$, $(a,d,r)$, $(a,d,s)$, $(b,c,r)$, $(b,c,s)$, $(b,d,r)$, and $(b,d,s)$. But, we can do better. Following the same reasoning as in last week, we can construct simple domains in 3D, using simple domains in 2D.

Definition Let $\D$ be a simple domain in 2D of the first or second kind. Let $\E$ be a bounded subset of $\Rthree$. Let $G,H:\D\to\R$ be be continuous functions defined on $\D$, with $G\le H$ for every point in $\D$. $\E$ is a simple domain in 3D if it has the form $$\E = \left\{(x,y,z) : (x,y)\in\D, G(x,y) \le z \le H(x,y) \right\} .$$ Of course, we can construct simple domains in 3D of different kinds by switching the $x$, $y$, and $z$ variables. It's important to keep in mind that we have developed this notion of simple domains so that we are able to evaluate triple integrals as iterated integrals.

3D Jordan Measure

For sets in 3D, it is natural to consider the volume of the set, as an analogy to the area of 2D sets.

Theorem Let $\E$ be a bounded subset of $\Rthree$. Then, $\E$ is Jordan Measurable if and only if the constant function $f(x,y,z)=1$ is integrable over $\E$. In this case, the volume of $\E$ is given by $$\vol(\E)=\iiint_\E 1 .$$ Keep in mind that we still have not developed the means to evaluate this integral. This is to follow.

Triple Integrals

We now present the final piece of the puzzle--the theorem that allows us to evaluate triple integrals, though given our study last week, it is fairly expected.

Theorem Let $f$ be a continuous function defined on a simple domain $\E$ in 3D. Then $f$ is integrable, and its integral is equal to an iterated integral: $$\iiint_\E f = \int_a^b\left(\int_{g(x)}^{h(x)}\left(\int_{G(x,y)}^{H(x,y)} f(x,y,z) dz\right) dy\right) dx .$$ Depending on what kind of simple domain $\E$ is, the iterated integral may take slightly different forms, with different limits, and order of evaluation. The next challenge is to be able to write the domain of integration as a simple domain. This is not a trivial task.

Example Let $\E$ be the solid bounded by the planes $z=1$, $y=1$, $z=y$ and the parabolic cylinder $y=x^2$. Evaluate the triple integral $\iiint_\E z .$ Our first task is to write $\E$ as a simple domain. Let's begin with the 2D projection: $\D = \left\{(x,y) : -1 \le x \le 1, x^2 \le y \le 1\right\} .$ Then $\E$ is given by $\E = \left\{(x,y,z) : -1 \le x \le 1, x^2 \le y \le 1, y \le z \le 1\right\} .$ Now we can use the theorem to write an iterated integral: $\iiint_\E z = \int_{-1}^1\left(\int_{x^2}^1\left(\int_y^1 z dz\right) dy\right) dx$ We have three integrals to evaluate, and we must evaluate them inside-out. \begin{gather*} \int_y^1 z dz = \frac{1}{2}(1-y^2) \\ \int_{x^2}^1\frac{1}{2}(1-y^2) dy = \left[\frac{y}{2}-\frac{y^3}{6}\right]_{y=x^2}^{y=1} = \frac{1}{3} - \frac{1}{2}x^2+\frac{1}{6}x^6 \\ \Rightarrow \iiint_\E z = \int_{-1}^1\left(\frac{1}{3} - \frac{1}{2}x^2+\frac{1}{6}x^6\right) dx = \frac{8}{21} \end{gather*}

Cylindrical and Spherical Integrals

We now turn to evaluating triple integrals as iterated integrals in cylindrical and spherical co-ordinates.

Simple Domain in Cylindrical Co-ordinates

Let's define the form of a simple domain in cylindrical co-ordinates.

Definition Let $\alpha, \beta \in \R$. Let $g,h:\R\to\R$ be continuously defined on the interval $[\alpha,\beta]$, with $g(\theta) \le h(\theta)$ for every $\theta\in[\alpha,\beta]$. (This gives the simple domain in polar co-ordinates from last week.) Let $G,H:\Rtwo\to\R$ be continuously defined by the span of $g$ and $h$ over the interval in $\theta$, with $G(r,\theta) \le z \le H(r,\theta)$. $\E$ is a simple domain in cylindrical co-ordinates if it has the form $$\E = \left\{(r\cos\theta,r\sin\theta,z) : \alpha \le \theta \le \beta, g(\theta) \le r \le h(\theta), G(r,\theta) \le z \le H(r,\theta)\right\} .$$

Iterated Integral in Cylindrical Co-ordinates

Now that we have the form of a simple domain in cylindrical co-ordinates, we can evaluate a triple integral of a function over such a simple domain, using the theorem that follows.

Theorem Let $f:\E\to\R$ be a continuous function defined on a simple domain $\E$ in cylindrical co-ordinates. Then, $$\iiint_\E f = \int_\alpha^\beta\left(\int_{g(\theta)}^{h(\theta)}\left(\int_{G(r,\theta)}^{H(r,\theta)}f(r\cos\theta,r\sin\theta,z)r dz\right) dr\right) d\theta$$ As in the case of the polar integral, we have an extra factor of $r$. So, for cylindrical co-ordinates, we have $dx dy dz \to r dr d\theta dz$.

Example Let $\E$ be the solid that lies above the upper half of the unit disk centered at the origin in the $x$--$y$ plane which is bounded below and above by the planes $z=0$ and $z=1$, respectively. Evaluate the triple integral $\iiint_\E f \text{ , where } f(x,y,z) = z\sqrt{x^2+y^2} .$ Our first challenge is to write $\E$ as a simple domain. The geometry lends itself to cylindrical co-ordinates, since that would allow us to simplify the radical. We can write $\E = \left\{(r\cos\theta,r\sin\theta,z) : 0 \le \theta \le \pi, 0 \le r \le 1, 0 \le z \le 1\right\} .$ Now we can write the integral as an iterated integral and easily evaluate it. \begin{align} \iiint_\E f &= \int_0^\pi\int_0^1\int_0^1 f(r\cos\theta,r\sin\theta,z)r dz dr d\theta \\ &= \int_0^\pi\int_0^1\int_0^1 zr\sqrt{(r\cos\theta)^2 + (r\sin\theta)^2} dz dr d\theta \\ &= \int_0^\pi\int_0^1\int_0^1 zr^2\sqrt{\cos^2\theta + \sin^2\theta} dz dr d\theta \\ &= \int_0^\pi\int_0^1\int_0^1 zr^2 dz dr d\theta \\ &= \int_0^\pi\int_0^1\frac{1}{2}\left[z^2r^2\right]_{z=0}^{z=1} dr d\theta \\ &= \frac{1}{2}\int_0^\pi\int_0^1r^2 dr d\theta \\ &= \frac{1}{6}\int_0^\pi d\theta \\ \therefore \iiint_\E f &= \frac{\pi}{6} \end{align}

Simple Domain in Spherical Co-ordinates

We define a simple domain in spherical co-ordinates.

Definition Let $\alpha, \beta \in \R$. Let $g,h:\R\to\R$ be continuously defined on the interval $[\alpha,\beta]$, with $g(\theta) \le h(\theta)$ for every $\theta\in[\alpha,\beta]$. Let $G,H:\Rtwo\to\R$ be continuously defined by the span of $g$ and $h$ over the interval in $\theta$, with $G(\theta,\phi) \le \rho \le H(\theta,\phi)$. $\E$ is a simple domain in cylindrical co-ordinates if it has the form $$\E = \left\{(\rho\cos\theta\sin\phi,\rho\sin\theta\cos\phi,\rho\cos\phi) : \alpha \le \theta \le \beta, g(\theta) \le \phi \le h(\theta), G(\theta,\phi) \le \rho \le H(\theta,\phi)\right\} .$$

Iterated Integral in Spherical Co-ordinates

Finally, we present the theorem to evaluate triple integrals using spherical co-ordinates.

Theorem Let $f:\E\to\R$ be a continuous function defined on a simple domain $\E$ in spherical co-ordinates. Then, $$\iiint_\E f = \int_\alpha^\beta\left(\int_{g(\theta)}^{h(\theta)}\left(\int_{G(\theta,\phi)}^{H(\theta,\phi)}F(\rho,\phi,\theta)\rho^2\sin\phi d\rho\right) d\phi\right) d\theta ,$$ where, $$F(\rho,\phi,\theta) = f(\rho\cos\theta\sin\phi,\rho\sin\theta\cos\phi,\rho\cos\phi) .$$ Note that this time we get an extra factor of $\rho^2\sin\phi$. So, when converting to spherical co-ordinates we observe that $dx dy dz\to\rho^2\sin\phi d\rho d\phi d\theta$. Sometimes the purely angular part of the differential is called the solid angle, or more precisely, the differential solid angle, and is sometimes written as $d\Omega = \sin\phi d\phi d\theta$.

Example Let $\E$ be the domain between two spheres centered at the origin with radius 1 and 3. Evaluate $\iiint_\E f \text{ , where } f(x,y,z) = z\sqrt{x^2+y^2} .$ We start as usual by writing $\E$ as a simple domain. \begin{gather*} \E = \left\{(\rho\cos\theta\sin\phi,\rho\sin\theta\cos\phi,\rho\cos\phi) : 0 \le \theta \le 2\pi, 0 \le \phi \le \pi, 1 \le \rho \le 3\right\} \\ \iiint_\E f = \int_0^{2\pi}\int_0^\pi\int_1^3 (\rho\cos\phi)^2(\rho^2\sin\phi) d\rho d\phi d\theta \\ \Leftrightarrow \iiint_\E f = \int_0^{2\pi}\int_0^\pi\int_1^3 \rho^4\cos^2\phi\sin\phi d\rho d\phi d\theta \end{gather*} We can now evaluate the iterated integrals from inside-out. \begin{gather*} \int_1^3 \rho^4\cos^2\phi\sin\phi d\rho = \cos^2\phi\sin\phi\left[\frac{\rho^5}{5}\right]_{\rho=1}^{\rho=3} = \frac{242}{5}\cos^2\phi\sin\phi \\ \int_0^\pi\frac{242}{5}\cos^2\phi\sin\phi d\phi = \frac{484}{15} \\ \therefore \iiint_\E f = \int_0^{2\pi}\frac{484}{15} d\theta = \frac{968}{15}\pi \end{gather*}