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MTH 254 Calculus IV Winter 2026

Richard Zhao

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Worksheet Assigned Problems for Section 12.7

The problems listed below are assigned to be included in your problem set portfolio. Note that a specific selection of these problems will also form the written homework assignments. I recommend working through all of them to ensure a solid grasp of the material. Reach out to Richard for help if you get stuck or have any questions.
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The solutions will be posted after the written homework due dates. If you have any questions about your work, talk to Richard and he is happy to discuss the process with you.
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12.7.3.

Convert the cylindrical coordinates \(\lp 0,\dfrac{\pi}{5},\dfrac{1}{2} \rp\) to rectangular coordinates.
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Solution.
We have \(r = 0\text{,}\) \(\theta = \dfrac{\pi}{5}\text{,}\) and \(z = \dfrac{1}{2}\text{.}\) Thus,
\begin{align*} x \amp= r\cos(\theta) = 0 \cdot \cos\left(\dfrac{\pi}{5}\right) = 0\\ y \amp= r\sin(\theta) = 0 \cdot \sin\left(\dfrac{\pi}{5}\right) = 0\\ z \amp= z = \dfrac{1}{2} \end{align*}
Hence, the rectangular coordinates are \(\lp 0,0,\dfrac{1}{2} \rp\text{.}\)
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12.7.9.

Convert the rectangular coordinates \(\lp \dfrac{5}{\sqrt{2}},\dfrac{5}{\sqrt{2}},2 \rp\) to cylindrical coordinates.
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Solution.
We have \(x = \dfrac{5}{\sqrt{2}}\text{,}\) \(y = \dfrac{5}{\sqrt{2}}\text{,}\) and \(z = 2\text{.}\) Thus,
\begin{align*} r \amp= \sqrt{x^2 + y^2} = \sqrt{\left(\dfrac{5}{\sqrt{2}}\right)^2 + \left(\dfrac{5}{\sqrt{2}}\right)^2} = 5 \end{align*}
Since the point \((x,y) = \lp \frac{5}{\sqrt{2}}, \frac{5}{\sqrt{2}} \rp\) is in the first quadrant, we know that \(0 \leq \theta \leq \frac{\pi}{2}\text{.}\) Therefore,
\begin{equation*} \tan(\theta) = \dfrac{y}{x} = \dfrac{\frac{5}{\sqrt{2}}}{\frac{5}{\sqrt{2}}} = 1 \qquad \implies \qquad \theta = \dfrac{\pi}{4} \end{equation*}
The cooresponding cylindrical coordinates are
\begin{equation*} (r,\theta,z) = \lp 5,\frac{\pi}{4}, 2 \rp \end{equation*}
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12.7.13.

Describe the set \(\left\{(x,y,z) \mid y^2 + z^2 \leq 4, \quad x = 0 \right\}\) in cylindrical coordinates.
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Solution.
The projection of the points in this set onto the \(xy\)-plane are points on the \(y\)-axis, thus \(\theta = \frac{\pi}{2}\) or \(\theta = \frac{3\pi}{2}\text{.}\) Therefore,
\begin{equation*} y = r\sin\lp \frac{\pi}{2} \rp = r\cdot 1 = r \qquad \text{or} \qquad y = r\sin\lp \frac{3\pi}{2} \rp = -r \end{equation*}
In both cases, \(y^2 = r^2\text{,}\) thus the inequality \(y^2 + z^2 \leq 4\) becomes \(r^2 + z^2 \leq 4\text{.}\) In cylindrical coordinates, we obtain the following inequality
\begin{equation*} r^2 + z^2 \leq 4\, , \qquad \theta = \frac{\pi}{2} \quad \text{or} \quad \theta = \dfrac{3\pi}{2} \end{equation*}
Hence, the set in cylindrical coordinates is
\begin{equation*} \left\{(r,\theta,z) \mid r^2 + z^2 \leq 4\, , \quad \theta = \frac{\pi}{2} \quad \text{or} \quad \theta = \dfrac{3\pi}{2} \right\} \end{equation*}
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12.7.21.

Sketch the set \(\left\{(r,\theta,z) \mid 1 \leq r \leq 3 , \quad 0 \leq z \leq 4 \right\}\) described in cylindrical coordinates.
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Solution.
The region \(1 \leq r \leq 3\text{,}\) \(0 \leq z \leq 4\) is a hollow cylinder with inner radius 1, outer radius 3, and height 4. The sketch is shown below.
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Figure 12.7.25.
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12.7.29.

Find an equation of the form \(r = f(\theta,z)\) in cylindrical coordinates for the surface \(\dfrac{x^2}{yz} = 1\text{.}\)
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Solution.
We rewrite the equation in the form
\begin{equation*} \frac{x}{\frac{y}{x}\, z} = 1 \end{equation*}
Substituting \(x = r\cos(\theta)\) and \(\frac{y}{x} = \tan(\theta)\text{,}\) we get
\begin{align*} \frac{r\cos(\theta)}{\lp \tan(\theta) \rp z} \amp= 1 \\ r \amp= \frac{z\tan(\theta)}{\cos(\theta)} \end{align*}
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12.7.37.

Convert the spherical coordinates \(\lp 6,\dfrac{\pi}{6},\dfrac{5\pi}{6} \rp\) to rectangular coordinates.
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Solution.
Since \(\rho = 6\text{,}\) \(\theta = \frac{\pi}{6}\text{,}\) and \(\phi = \frac{5\pi}{6}\text{,}\) we get
\begin{align*} x \amp = \rho \sin(\phi) \cos(\theta) = 6 \sin\left(\frac{5\pi}{6}\right) \cos\left(\frac{\pi}{6}\right) = 6 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \\ y \amp = \rho \sin(\phi) \sin(\theta) = 6 \sin\left(\frac{5\pi}{6}\right) \sin\left(\frac{\pi}{6}\right) = 6 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{2} \\ z \amp = \rho \cos(\phi) = 6 \cos\left(\frac{5\pi}{6}\right) = 6 \cdot -\frac{\sqrt{3}}{2} = -3\sqrt{3} \end{align*}
Hence, the rectangular coordinates are
\begin{equation*} (x,y,z) = \lp \frac{3\sqrt{3}}{2}, \frac{3}{2}, -3\sqrt{3} \rp \end{equation*}
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12.7.43.

Convert the rectangular coordinates \(\lp \dfrac{1}{2},\dfrac{\sqrt{3}}{2},\sqrt{3} \rp\) to spherical coordinates.
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Solution.
We have \(x = \frac{1}{2}\text{,}\) \(y = \frac{\sqrt{3}}{2}\text{,}\) and \(z = \sqrt{3}\text{.}\) Thus,
\begin{equation*} \rho = \sqrt{x^2 + y^2 + z^2} = \sqrt{\lp \frac{1}{2} \rp^2 + \lp \frac{\sqrt{3}}{2} \rp^2 + \lp \sqrt{3} \rp^2} = 2 \end{equation*}
The angular coordinate \(\theta\) satisfies \(0 \leq \theta \leq \frac{\pi}{2}\text{,}\) since the point \((x,y) = \lp \frac{1}{2}, \frac{\sqrt{3}}{2} \rp\) is in the first quadrant. Then we have
\begin{equation*} \tan(\theta) = \frac{y}{x} = \sqrt{3} \qquad \implies \qquad \theta = \frac{\pi}{3} \end{equation*}
The angle of declination \(\phi\) satisfies
\begin{equation*} \cos(\phi) = \frac{z}{\rho} = \frac{\sqrt{3}}{2} \qquad \implies \qquad \phi = \frac{\pi}{6} \end{equation*}
Hence, the spherical coordinates are
\begin{equation*} (\rho, \theta, \phi) = \lp 2, \frac{\pi}{3}, \frac{\pi}{6} \rp \end{equation*}
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12.7.45.

Convert the cylindrical coordinates \(\lp 2,0,2 \rp\) to spherical coordinates.
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Solution.
We are given that \(r = 2\text{,}\) \(\theta = 0\text{,}\) and \(z = 2\text{.}\) Using the conversion formulas, we have
\begin{align*} \rho \amp= \sqrt{x^2 + y^2 + z^2} = \sqrt{r^2 + z^2} = \sqrt{4 + 4} = 2\sqrt{2} \\ \theta \amp= \theta = 0 \\ \phi \amp= \cos^{-1}\left(\frac{z}{\rho}\right) = \cos^{-1}\left(\frac{2}{2\sqrt{2}}\right) = \frac{\pi}{4} \end{align*}
Hence, the spherical coordinates are
\begin{equation*} (\rho, \theta, \phi) = \lp 2\sqrt{2}, 0, \frac{\pi}{4} \rp \end{equation*}
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12.7.47.

Convert the spherical coordinates \(\lp 4,0,\dfrac{\pi}{4} \rp\) to cylindrical coordinates.
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Solution.
We are given that \(\rho = 4\text{,}\) \(\theta = 0\text{,}\) and \(\phi = \frac{\pi}{4}\text{.}\) To find \(r\text{,}\) we use the formulas \(x = r\cos(\theta)\) and \(x = \rho\cos(\theta)\sin(\phi)\) to get
\begin{equation*} r\cos(\theta) = \rho\cos(\theta)\sin(\phi) \end{equation*}
and so
\begin{equation*} r = \rho\sin(\phi) = 4\sin\lp \frac{\pi}{4} \rp = 2\sqrt{2} \end{equation*}
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Clearly, \(\theta = 0\text{,}\) and as for \(z\text{,}\)
\begin{equation*} z = \rho\cos(\phi) = 4 \cos\lp \frac{\pi}{4} \rp = 2\sqrt{2} \end{equation*}
So, in cylindrical coordinates, our point is \(\lp 2\sqrt{2}, 0, 2\sqrt{2} \rp\text{.}\)
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12.7.53.

Describe the set \(\left\{(x,y,z) \mid y^2 + z^2 \leq 4, \quad x = 0 \right\}\) in spherical coordinates.
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Solution.
We substitute \(y = \rho\sin(\theta)\sin(\phi)\) and \(z = \rho\cos(\phi)\) in the given inequality. This gives
\begin{equation*} \rho^2 \sin^2(\theta) \sin^2(\phi) + \rho^2 cos^2(\phi) \leq 4 \end{equation*}
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The equality \(x = 0\) determines that \(\theta = \frac{\pi}{2}\) or \(\theta = \frac{3\pi}{2}\) (and the origin). In both cases, \(\sin^2(\theta) = 1\text{.}\) Hence, we get
\begin{align*} \rho^2 \sin^2(\phi) + \rho^2 \cos^2(\phi) \amp\leq 4 \\ \rho^2 \left( \sin^2(\phi) + \cos^2(\phi) \right) \amp\leq 4 \\ \rho^2 \amp\leq 4 \\ \rho \amp\leq 2 \end{align*}
We obtain the following description
\begin{equation*} \left\{\lp \rho, \theta, \phi \rp \mid 0 \leq \rho \leq 2, \, \theta = \frac{\pi}{2} \text{ or } \theta = \frac{3\pi}{2} \right\} \end{equation*}
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12.7.61.

Sketch the set of points \(\left\{(\rho,\theta,\phi) \mid \rho \leq 2, \quad 0 \leq \theta \leq \dfrac{\pi}{2}, \quad \dfrac{\pi}{2} \leq \phi \leq \pi \right\}\) described in spherical coordinates.
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Solution.
This set is the part of the ball of radius \(2\) which is below the first quadrant of the \(xy\)-plane, as shown in the following figure:
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Figure 12.7.26.
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12.7.71.

Find an equation of the form \(\rho = f(\theta, \phi)\) in spherical coordinates for the surface \(x^2 - y^2 = 4\text{.}\)
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Solution.
We substitute \(x = \rho\cos(\theta)\sin(\phi)\) and \(y = \rho\sin(\theta)\sin(\phi)\) to obtain
\begin{equation*} 4 = \rho^2\cos^2(\theta)\sin^2(\phi) - \rho^2\sin^2(\theta)\sin^2(\phi) = \rho^2\sin^2(\phi)\lp \cos^2(\theta) - \sin^2(\theta) \rp \end{equation*}
Using the identify \(\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)\text{,}\) we get
\begin{align*} 4 \amp = \rho^2\sin^2(\phi)\cos(2\theta) \\ \implies \qquad \rho^2 \amp = \frac{4}{\sin^2(\phi)\cos(2\theta)} \end{align*}
We take the square root of both sides. Siince \(0 \lt \phi \lt \pi\text{,}\) we have \(\sin(\phi) = 0\text{.}\) Hence,
\begin{equation*} \rho = \frac{2}{\sin(\phi)\sqrt{\cos(2\theta)}} \end{equation*}
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12.7.81.

An apple modeled by taking all the points in and on a sphere of radius \(2\) inches is cored with a vertical cylinder of radius \(1\) in. Use inequalities in cylindrical coordinates to describe the set of all points that remain in the apple once the core is removed.
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Solution.
The sphere together with its interior is, in rectangular coordinates, the set of points with \(x^2 + y^2 + z^2 \leq 4\text{.}\) In cylindrical coordinates, this is the set of points with \(r^2 + z^2 \leq 4\text{.}\) So we can parametrize the sphere and its interior as
\begin{equation*} -2 \leq z \leq 2 \qquad \text{and} \qquad 0 \leq r \leq \sqrt{4 - z^2} \end{equation*}
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The vertical cylinder together with its interior is parametrized by \(r \leq 1\text{.}\) The cylinder intersects the sphere when \(1^2 + z^2 = 4\text{,}\) or \(z = \pm \sqrt{3}\text{.}\) When \(|z| \gt \sqrt{3}\text{,}\) all points in the sphere lie inside the cylinder, so are gone when the cylinder is removed. So removing the cylinder from the sphere gives a set of points parametrized as
\begin{equation*} -\sqrt{3} \lt z \lt \sqrt{3} \qquad \text{and} \qquad 1 \leq r \leq \sqrt{4 - z^2} \end{equation*}
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