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Section 1.7 Substitution: polar and cylindrical coordinates

Theorems 1.7.3–1.7.9 result from applying Theorem 1.6.6 to particular transformations, called the polar and cylindrical transformations:

\begin{align*} G\colon \R^2 \amp\rightarrow \R^2 \amp \text{(polar transformation)}\\ (r,\theta)\amp\mapsto (r\cos\theta, r\sin\theta) \\ G\colon \R^3 \amp \rightarrow \R^3 \amp \text{(cylindrical transformation)}\\ (r,\theta,z)\amp\mapsto (r\cos\theta, r\sin\theta,z) \text{.} \end{align*}

Subsection 1.7.1 Polar transformation

We will think of the polar coordinate equations

\begin{align*} x \amp=r\cos\theta \\ y \amp=r\sin\theta \end{align*}

as defining a function

\begin{align*} G\colon \R^2 \amp \rightarrow \R^2\\ (r,\theta)\amp \mapsto (\underset{x}{\underbrace{r\cos\theta}}, \underset{y}{\underbrace{r\sin\theta}}) \end{align*}

from one copy of \(\R^2\text{,}\) called the \(r\theta\)-plane, to another, called the \(xy\)-plane. Earlier in your calculus career you learned to translate back and forth between polar equations and Cartesian (or rectangular) ones. We will now think of these translations as evaluations of the function \(G\) or its inverse to regions of the \(r\theta\)- or \(xy\)-plane. For example, \(G\) takes the rectangle

\begin{equation*} R=[r_1,r_2]\times [\theta_1, \theta_2]=\{(r,\theta)\colon r_1\leq r\leq r_2, \theta_1\leq \theta\leq\theta_2\} \end{equation*}

lying in the \(r\theta\)-plane, and maps it to the set of all points in the \(xy\)-plane lying between the circles of radius \(r=r_1\) and \(r=r_2\text{,}\) centered at the origin, and between the two rays obtained by rotating the positive \(x\)-axis by an angle of \(\theta=\theta_1\) and \(\theta=\theta_2\text{,}\) respectively.

Figure 1.7.1. Action of polar transformation on a rectangle in \(r\theta\)-plane.
From this observation we can then understand how \(G\) transforms an entire rectangular grid in the \(r\theta\)-plane to a polar grid in the \(xy\)-plane.
Figure 1.7.2. Action of polar transformation on a grid in \(r\theta\)-plane.

It is easy to see that \(G\) is continuously differentiable, and hence is a candidate for an application of Theorem 1.6.6. The only thing we need to be careful about is restricting ourselves to regions \(\mathcal{R}\) where \(G\) is one-to-one, and hence defines a transformation.

Remark 1.7.4. Polar transformation.

It is not difficult to show that

\begin{equation} (r\cos\theta_1, r\sin\theta_1)=(s\cos\theta_2,s\sin\theta_2)\iff \begin{cases} r=s \text{ and } \theta_2=\theta_1+2\pi n, n\in \Z\\ \text{ or } \\ r=-s \text{ and } \theta_2=-\theta_1+2\pi n, n\in \Z \end{cases}\text{.}\tag{1.7.4} \end{equation}

It follows that the polar function \(G(r,\theta)=(r\cos\theta, r\sin\theta)\) is one-to-one on a given set as long as it doesn't contain pairs of points satisfying one of the conditions in (1.7.4). Below you find two common types of sets where \(G\) is one-to-one:

\begin{align*} X \amp = \{(r,\theta)\colon r > 0, \alpha \leq \theta < \alpha+2\pi\}\\ Y \amp = \{(r,\theta)\colon r\ne 0, \alpha\leq \theta < \alpha+\pi\}\\ \amp \text{.} \end{align*}

Observe that \(r\) values are allowed to be positive or negative in the set \(Y\text{,}\) and so the range of \(\theta\) needs to be restricted accordingly.

Example 1.7.5. Area between limaçon and circle.

Let \(\mathcal{S}\subseteq \R^2\) be the region outside the circle of radius 2, centered at the origin, and inside the limaçon with polar equation

\begin{equation*} r=1+2\sin\theta\text{.} \end{equation*}

Compute the area of \(\mathcal{S}\text{.}\)

Solution.

The \(r\theta\)-coordinates of points in \(\mathcal{S}\) satisfy \(r=1+2\sin\theta\geq 2\text{.}\) A little algebra shows this implies \(\pi/6\leq \theta\leq 5\pi/6\text{.}\) Thus the polar transformation \(G(r,\theta)=(r\cos\theta, r\sin\theta)\) maps the \(r\theta\)-region

\begin{equation*} \mathcal{R}=\{(r,\theta)\colon \pi/6\leq \theta\leq 5\pi/6, 2\leq r\leq 1+2\sin\theta\} \end{equation*}

onto \(\mathcal{S}\text{.}\) Using Theorem 1.7.3, we have

\begin{align*} \operatorname{area} \mathcal{S} \amp = \iint\limits_{\mathcal{S}}1\, dA \\ \amp =\iint_{\mathcal{R}}r\, dA \amp \text{(polar subst.)}\\ \amp =\int_{\pi/6}^{5\pi/6}\int_2^{1+2\sin\theta}r\, dr\, d\theta\\ \amp =\frac{1}{2}\int_{\pi/6}^{5\pi/6}4\sin\theta+4\sin^2\theta-3\, d\theta\\ \amp =\frac{1}{2}\int_{\pi/6}^{5\pi/6}4\sin\theta-2\cos 2\theta-1\, d\theta \amp (\sin^2\theta=\frac{1}{2}(1-\cos 2\theta))\\ \amp =\frac{1}{2}(-4\cos\theta-\theta-\sin 2\theta)\Bigr\vert_{\pi/6}^{5\pi/6}\\ \amp =\frac{5\sqrt{3}}{2}-\frac{\pi}{3} \text{.} \end{align*}

Example 1.7.6. Average value over semicircle.

Compute the average value of \(f(x,y)=x\) over the semicircular region

\begin{equation*} \mathcal{R}=\{(x,y)\in \R^2\colon x^2+y^2\leq R^2, x\geq 0\}\text{,} \end{equation*}

where \(R\in \R\) is a fixed constant.

Solution.

The half-disk \(\mathcal{R}\) corresponds to the the region

\begin{equation*} \mathcal{S}=\{(r,\theta)\colon -\pi/2\leq \theta\leq \pi/2, 0\leq r\leq R\} \end{equation*}

in the \(r\theta\)-plane via the polar transformation \(G(r,\theta)=(r\cos\theta,r\sin\theta)\text{.}\) We have

\begin{align*} \operatorname{avg}_\mathcal{R}f \amp =\frac{1}{\operatorname{area}\mathcal{R}}\iint\limits_\mathcal{R}x\, dA\\ \amp=\frac{2}{\pi R^2}\iint\limits_{\mathcal{S}}r\cos\theta\cdot r\, dA \amp \text{(polar subst.)} \\ \amp=\frac{2}{\pi R^2}\int_{-\pi/2}^{\pi/2}\cos\theta\, d\theta\int_0^Rr^2\, dr \\ \amp =\frac{2}{\pi R^2}(\sin\theta)\Bigr\vert_{-\pi/2}^{\pi/2}(r^3/3)\Bigr\vert_0^R\\ \amp =\frac{4R}{3\pi}\text{.} \end{align*}

Subsection 1.7.2 Cylindrical transformation

Our discussion in Subsection 1.7.1 generalizes easily to cylindrical coordinates:

\begin{align*} x \amp=r\cos\theta \\ y \amp =r\sin\theta\\ z \amp = z \text{.} \end{align*}

Figure 1.7.7. Cylindrical coordinates
As before, we now think of these equations as defining a function

\begin{align*} G\colon \R^3 \amp \rightarrow \R^3\\ (r,\theta,z) \amp \mapsto (\underset{x}{\underbrace{r\cos\theta}}, \underset{y}{\underbrace{r\sin\theta}}, z) \end{align*}

from \(r\theta z\)-space to \(xyz\)-space. Recalling some of our translation exercises from cylindrical to and from Cartesian coordinates we see that \(G\) maps the box

\begin{equation*} B=[r_1,r_2]\times[\theta_1,\theta_2]\times[z_1,z_2] \end{equation*}

lying in \(r\theta z\)-space, to a certain region of \(xyz\)-space trapped between two vertical circular cylinders of radius \(r_1\) and \(r_2\text{.}\)

Figure 1.7.8. Action of cylindrical transformation on box in \(r\theta z\)-space.

Example 1.7.11. Volume of sphere: cylindrical coordinates.

Consider the solid spherical ball

\begin{equation*} \mathcal{S}=\{(x,y,z)\in\R^3\colon 0\leq x^2+y^2+z^2\leq R^2\}\text{,} \end{equation*}

where \(R\in \R\) is a fixed constant. Compute the volume of \(\mathcal{S}\) using cylindrical coordinates.

Solution.

The projection of \(\mathcal{S}\) onto the \(xy\)-plane is the disk \(x^2+y^2\leq R^2\text{,,}\) which is described in cylindrical coordinates as \(0\leq \theta\leq 2\pi, 0\leq r\leq R\text{.}\) Next, looking at the vertical line passing through a point \((r,\theta,0)\text{,}\) we see we enter and leave \(\mathcal{S}\) at the lower and upper surfaces of the sphere. In cylindrical coordinates the defining equation \(x^2+y^2+z^2=R^2\) becomes \(r^2+z^2=R^2\text{,}\) and thus the lower and upper surfaces of the sphere have cylindrical equations \(z=-\sqrt{R^2-r^2}\) and \(z=\sqrt{R^2-r^2}\text{,}\) repsectively. We conclude that the cylindrical description of \(\mathcal{S}\) is

\begin{equation*} \{(r,\theta, z)\colon 0\leq \theta\leq 2\pi, 0\leq r\leq R, -\sqrt{R^2-r^2}\leq z\leq \sqrt{R^2-r^2}\text{.} \end{equation*}

Finally, we compute

\begin{align*} \operatorname{vol}\mathcal{S} \amp = \iiint\limits_\mathcal{S}1\, dV \\ \amp =\int_0^{2\pi}\int_0^R, \int_{-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}}r \, dz\, dr\, d\theta \amp \text{(cylind. subst.)}\\ \amp = 2\pi\int_0^R2r\sqrt{R^2-r^2}\, dr\\ \amp =-\frac{4\pi}{3}(R^2-r^2)^{3/2}\Bigr\vert_0^R\\ \amp =\frac{4}{3}\pi R^3\text{.} \end{align*}

We've just derived the volume formula for the sphere of radius \(R\text{!}\)

Example 1.7.12. Average value over hemisphere: cylindrical coordinates.

Consider the solid hemisphere

\begin{equation*} \mathcal{S}=\{(x,y,z)\in\R^3\colon 0\leq x^2+y^2+z^2\leq R^2, z\geq 0\}\text{,} \end{equation*}

where \(R\in \R\) is a fixed constant. Find the average height (distance to \(xy\)-plane) of points in \(\mathcal{S}\text{.}\)

Solution.

First observe that the function in question is just \(f(x,y,z)=z\text{.}\) The cylindrical description of \(\mathcal{S}\) is similar to the previous example, except now \(z\)-values are bounded below by zero:

\begin{equation*} \{(r,\theta,z)\colon 0\leq \theta\leq 2\pi, 0\leq r\leq R, 0\leq z\leq \sqrt{R^2-r^2}\text{.} \end{equation*}

Now compute

\begin{align*} \operatorname{avg}_{\mathcal{S}}f \amp =\frac{1}{\operatorname{vol} \mathcal{S}}\iiint\limits_\mathcal{S}z\, dA \\ \amp =\frac{3}{2\pi R^3}\int_0^{2\pi}\int_0^R\int_0^{\sqrt{R^2-r^2}}zr\, dz\, dr\, d\theta \amp \text{(cylind. subst.)}\\ \amp = \frac{3}{R^3}\int_0^R\frac{1}{2}r(R^2-r^2)\, dr \\ \amp =\frac{3}{2R^3}\left(\frac{1}{2}R^2r^2-\frac{1}{4}r^4\right)\Bigr\vert_0^R\\ \amp = \frac{3}{2R^3}\left(\frac{1}{2}R^4-\frac{1}{4}R^4\right)\\ \amp = \frac{3}{8}R\text{.} \end{align*}