Cylindrical Coordinates - Calculus 3

1. The region in cylindrical coordinates defined by $1 \leq r \leq 3$, $\frac{\pi}{4} \leq \theta \leq \frac{3\pi}{4}$, and $0 \leq z \leq r$ has what volume? Give your answer in simplest form.
   Answer: $\frac{13\pi}{3}$ The volume integral in cylindrical coordinates is $\int_{\pi/4}^{3\pi/4} \int_1^3 \int_0^r r \, dz \, dr \, d\theta$. First: $\int_0^r r \, dz = r^2$. Second: $\int_1^3 r^2 \, dr = \frac{r^3}{3}\Big|_1^3 = \frac{27-1}{3} = \frac{26}{3}$. Third: $\int_{\pi/4}^{3\pi/4} \frac{26}{3} \, d\theta = \frac{26}{3} \cdot \left(\frac{3\pi}{4} - \frac{\pi}{4}\right) = \frac{26}{3} \cdot \frac{\pi}{2} = \frac{13\pi}{3}$.
2. The equation $r^2 + z^2 = 16$ in cylindrical coordinates represents which surface?
   Answer: A sphere of radius 4 centered at the origin Converting to Cartesian coordinates: $r^2 = x^2 + y^2$, so the equation becomes $x^2 + y^2 + z^2 = 16$, which is a sphere of radius 4 centered at the origin. Choice A would be $r = 4$ (constant $r$). Choice C would have the form $z^2 = k(x^2 + y^2)$ or $z^2 = kr^2$. Choice D would have the form $z = k(x^2 + y^2)$ or $z = kr^2$.
3. Two points have cylindrical coordinates $A(3, \frac{\pi}{3}, 4)$ and $B(5, \frac{5\pi}{6}, 1)$. What is the distance between these points?
   Answer: $\sqrt{43}$ Using the distance formula in cylindrical coordinates: $d^2 = r_1^2 + r_2^2 - 2r_1r_2\cos(\theta_2-\theta_1) + (z_2-z_1)^2$. Here: $d^2 = 3^2 + 5^2 - 2(3)(5)\cos\left(\frac{5\pi}{6}-\frac{\pi}{3}\right) + (1-4)^2 = 9 + 25 - 30\cos\left(\frac{\pi}{2}\right) + 9 = 34 - 30(0) + 9 = 43$. Therefore, $d = \sqrt{43}$.
4. The cylindrical coordinate equation $r = 2\cos(\theta)$ represents which curve in the $xy$-plane?
   Answer: A circle centered at $(1, 0)$ with radius 1 Multiply both sides by $r$: $r^2 = 2r\cos(\theta)$. Converting to Cartesian: $x^2 + y^2 = 2x$. Rearranging: $x^2 - 2x + y^2 = 0$. Completing the square: $(x-1)^2 - 1 + y^2 = 0$, so $(x-1)^2 + y^2 = 1$. This is a circle centered at $(1, 0)$ with radius 1. Choice B would correspond to $r = 2\sin(\theta)$. Choice C would be $r = 4\cos(\theta)$. Choice D would be $\theta = \text{constant}$ or $y = mx$.
5. A point $P$ has cylindrical coordinates $(4, \frac{2\pi}{3}, -3)$. After applying the transformation that rotates the point $\frac{\pi}{6}$ radians counterclockwise about the $z$-axis and then reflects it across the $xy$-plane, what are the new cylindrical coordinates?
   Answer: $(4, \frac{5\pi}{6}, 3)$ Starting with $(r, \theta, z) = (4, \frac{2\pi}{3}, -3)$. First, rotating $\frac{\pi}{6}$ counterclockwise about the $z$-axis adds $\frac{\pi}{6}$ to $\theta$: $\theta_{new} = \frac{2\pi}{3} + \frac{\pi}{6} = \frac{4\pi + \pi}{6} = \frac{5\pi}{6}$. The $r$ coordinate remains unchanged. Then, reflecting across the $xy$-plane changes $z$ to $-z$: $z_{new} = -(-3) = 3$. Choice B uses $\frac{\pi}{2}$ instead of $\frac{5\pi}{6}$. Choice C doesn't reflect across the $xy$-plane. Choice D subtracts $\frac{\pi}{6}$ instead of adding it.
6. A point moves along a curve given by $r(t) = 2 + \sin(t)$, $\theta(t) = t$, $z(t) = t$ in cylindrical coordinates. What is the Cartesian $y$-coordinate of the point when $t = \frac{3\pi}{2}$?
   Answer: $-1$ At $t = \frac{3\pi}{2}$: $r = 2 + \sin(\frac{3\pi}{2}) = 2 + (-1) = 1$, $\theta = \frac{3\pi}{2}$, $z = \frac{3\pi}{2}$. Converting to Cartesian: $y = r\sin(\theta) = 1 \cdot \sin(\frac{3\pi}{2}) = 1 \cdot (-1) = -1$. Choice A gives the $z$-coordinate. Choice B would result from using $\sin(\frac{3\pi}{2}) = 1$ incorrectly. Choice D would result from using $\sin(\frac{3\pi}{2}) = 0$ incorrectly.
7. A solid region is bounded by $r = 2$, $r = 4\cos(\theta)$, $z = 0$, and $z = r$ in cylindrical coordinates. For what values of $\theta$ is this region well-defined?
   Answer: $-\frac{\pi}{3} \leq \theta \leq \frac{\pi}{3}$ When working with regions defined in cylindrical coordinates, you need to ensure that all boundary conditions can be satisfied simultaneously. This means finding where the given curves actually intersect and create a meaningful enclosed region. The key insight is determining where the two radial boundaries $r = 2$ (a circle of radius 2) and $r = 4\cos(\theta)$ (a circle in Cartesian coordinates) intersect. For the region to be well-defined, we need $4\cos(\theta) \geq 2$, which gives us $\cos(\theta) \geq \frac{1}{2}$. Since $\cos(\theta) = \frac{1}{2}$ when $\theta = \pm\frac{\pi}{3}$, and cosine is greater than $\frac{1}{2}$ between these values, the region exists for $-\frac{\pi}{3} \leq \theta \leq \frac{\pi}{3}$. Within this range, $r$ varies from the inner boundary $r = 2$ to the outer boundary $r = 4\cos(\theta)$, and $z$ extends from 0 to $r$. Answer A ($-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$) is too restrictive—while this range works, it doesn't capture the full region where the boundaries intersect meaningfully. Answer B ($-\frac{\pi}{6} \leq \theta \leq \frac{\pi}{6}$) is even more restrictive for the same reason. Answer D ($0 \leq \theta \leq \frac{\pi}{2}$) extends too far—for $\theta > \frac{\pi}{3}$, we'd have $4\cos(\theta) < 2$, making the radial bounds inconsistent. Study tip: Always check where boundary curves intersect by setting up inequalities. For cylindrical coordinate regions, ensure the radial bounds make geometric sense throughout the angular range.
8. A point starts at cylindrical coordinates $(2, 0, 3)$ and moves such that $r$ increases linearly at 1 unit per second, $\theta$ increases at $\frac{\pi}{4}$ radians per second, and $z$ decreases at 0.5 units per second. What are the Cartesian coordinates after 2 seconds?
   Answer: $(0, 4, 2)$ When you encounter a problem involving changing cylindrical coordinates, you need to track how each coordinate evolves over time, then convert to Cartesian coordinates using the standard transformation formulas. Let's trace each coordinate after 2 seconds. Starting from $(r, \theta, z) = (2, 0, 3)$: - $r$ increases at 1 unit/second: $r = 2 + 1(2) = 4$ - $\theta$ increases at $\frac{\pi}{4}$ radians/second: $\theta = 0 + \frac{\pi}{4}(2) = \frac{\pi}{2}$ - $z$ decreases at 0.5 units/second: $z = 3 - 0.5(2) = 2$ So after 2 seconds, the cylindrical coordinates are $(4, \frac{\pi}{2}, 2)$. Now convert to Cartesian using $x = r\cos\theta$, $y = r\sin\theta$, $z = z$: - $x = 4\cos\left(\frac{\pi}{2}\right) = 4(0) = 0$ - $y = 4\sin\left(\frac{\pi}{2}\right) = 4(1) = 4$ - $z = 2$ This gives us $(0, 4, 2)$, which is answer D. Answer A $(4, 0, 2)$ incorrectly assumes $\theta$ remains 0. Answer B $(-2\sqrt{2}, 2\sqrt{2}, 2)$ suggests $r = 4$ and $\theta = \frac{3\pi}{4}$, indicating a sign error or wrong angle calculation. Answer C $(2\sqrt{2}, 2\sqrt{2}, 2)$ implies $r = 4$ and $\theta = \frac{\pi}{4}$, suggesting the student used 1 second instead of 2 seconds for the angular motion. Study tip: Always update all coordinates first using their rates of change, then convert the entire result to Cartesian. Don't try to track Cartesian coordinates directly when motion is described in cylindrical terms.
9. The Jacobian for the transformation from cylindrical coordinates $(r, \theta, z)$ to Cartesian coordinates $(x, y, z)$ is used in triple integrals. If we instead consider the transformation from $(u, v, w)$ where $u = r^2$, $v = \theta$, $w = z$, what is the Jacobian for transforming from $(u, v, w)$ to $(x, y, z)$?
   Answer: $\frac{1}{2}$ We have $x = \sqrt{u}\cos(v)$, $y = \sqrt{u}\sin(v)$, $z = w$. The Jacobian matrix is $J = \begin{pmatrix} \frac{\cos(v)}{2\sqrt{u}} & -\sqrt{u}\sin(v) & 0 \\ \frac{\sin(v)}{2\sqrt{u}} & \sqrt{u}\cos(v) & 0 \\ 0 & 0 & 1 \end{pmatrix}$. The determinant is $\frac{\cos(v)}{2\sqrt{u}} \cdot \sqrt{u}\cos(v) + \sqrt{u}\sin(v) \cdot \frac{\sin(v)}{2\sqrt{u}} = \frac{\cos^2(v)}{2} + \frac{\sin^2(v)}{2} = \frac{1}{2}$.

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