Area in Polar Coordinates - Multivariable Calculus

1. Find the area of the region in the first quadrant bounded by the spiral $r=\theta$, the polar axis, and the line $\theta = \pi/2$.
   Answer: $\frac{\pi^3}{48}$ When you encounter area problems in polar coordinates, you need to recognize that the area formula is $A = \frac{1}{2}\int_{\alpha}^{\beta} r^2 \, d\theta$, where the region is bounded by rays at angles $\alpha$ and $\beta$. Here, you're finding the area swept out by the spiral $r = \theta$ from the polar axis ($\theta = 0$) to the line $\theta = \pi/2$. Setting up the integral: $A = \frac{1}{2}\int_{0}^{\pi/2} \theta^2 \, d\theta$ To evaluate this, use the power rule: $\int \theta^2 \, d\theta = \frac{\theta^3}{3}$ So: $A = \frac{1}{2} \left[ \frac{\theta^3}{3} \right]_0^{\pi/2} = \frac{1}{2} \cdot \frac{1}{3} \left[ \theta^3 \right]_0^{\pi/2} = \frac{1}{6} \left( \frac{\pi^3}{8} - 0 \right) = \frac{\pi^3}{48}$ This confirms answer D is correct. Answer A ($\frac{\pi^2}{8}$) results from incorrectly using $r = \theta$ directly instead of $r^2 = \theta^2$ in the area formula. Answer B ($\frac{\pi^3}{24}$) comes from forgetting the $\frac{1}{2}$ factor in the polar area formula. Answer C ($\frac{\pi^2}{16}$) combines both errors: using $r$ instead of $r^2$ and making an integration mistake. Remember: polar area always involves $r^2$, and don't forget the $\frac{1}{2}$ coefficient. These are the most common mistakes in polar area problems.
2. The total area enclosed by the curve $r^2 = 9\cos(2\theta)$ is:
   Answer: $9$ This is a lemniscate (figure-eight curve). The curve exists only when $\cos(2\theta) \geq 0$, which occurs for $-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$ and $\frac{3\pi}{4} \leq \theta \leq \frac{5\pi}{4}$. Since $r = 3\sqrt{\cos(2\theta)}$, the total area is $A = 2 \cdot \frac{1}{2}\int_{-\pi/4}^{\pi/4} 9\cos(2\theta) d\theta = 9\int_{-\pi/4}^{\pi/4} \cos(2\theta) d\theta = 9\left[\frac{\sin(2\theta)}{2}\right]_{-\pi/4}^{\pi/4} = 9 \cdot \frac{1}{2}[\sin(\pi/2) - \sin(-\pi/2)] = 9 \cdot \frac{1}{2}[1 - (-1)] = 9$. Choice B incorrectly uses the standard area formula without accounting for the $r^2$ form. Choice C doubles the answer incorrectly. Choice D uses wrong bounds or formula.
3. Find the area of the region defined by the polar inequalities $1 \le r \le 1+\cos\theta$.
   Answer: $2 + \frac{\pi}{4}$ The region is bounded by the circle $r=1$ and the cardioid $r=1+\cos\theta$. The inequality $1 \le 1+\cos\theta$ implies $\cos\theta \ge 0$. This condition holds for $\theta \in [-\pi/2, \pi/2]$. So, these are the bounds for our integration. The area is the difference between the area of the outer curve and the inner curve over this interval. $A = \frac{1}{2} \int_{-\pi/2}^{\pi/2} ((1+\cos\theta)^2 - 1^2) d\theta$ By symmetry, we can integrate from $0$ to $\pi/2$ and double the result. $A = \int_{0}^{\pi/2} ((1+2\cos\theta+\cos^2\theta) - 1) d\theta = \int_{0}^{\pi/2} (2\cos\theta + \cos^2\theta) d\theta$ Using the identity $\cos^2\theta = \frac{1+\cos(2\theta)}{2}$: $A = \int_{0}^{\pi/2} (2\cos\theta + \frac{1}{2} + \frac{1}{2}\cos(2\theta)) d\theta$ $A = [2\sin\theta + \frac{\theta}{2} + \frac{1}{4}\sin(2\theta)]_{0}^{\pi/2}$ $A = (2\sin(\pi/2) + \frac{\pi/2}{2} + \frac{1}{4}\sin(\pi)) - (0) = 2(1) + \frac{\pi}{4} + 0 = 2 + \frac{\pi}{4}$ Distractor C results from incorrectly integrating over $[0, 2\pi]$ and making an error, which gives the area of the cardioid minus the area of the circle over the whole domain. The integral of $2\cos\theta$ over $[0, 2\pi]$ is 0, leaving $\frac{1}{2} \int_0^{2\pi} \cos^2\theta d\theta = \pi/2$.
4. The region bounded by $r = 1 + \cos\theta$ (cardioid) has area:
   Answer: $\frac{3\pi}{2}$ The area is $A = \frac{1}{2}\int_0^{2\pi} (1 + \cos\theta)^2 d\theta = \frac{1}{2}\int_0^{2\pi} (1 + 2\cos\theta + \cos^2\theta) d\theta$. Using $\cos^2\theta = \frac{1+\cos(2\theta)}{2}$: $A = \frac{1}{2}\int_0^{2\pi} \left(1 + 2\cos\theta + \frac{1+\cos(2\theta)}{2}\right) d\theta = \frac{1}{2}\int_0^{2\pi} \left(\frac{3}{2} + 2\cos\theta + \frac{\cos(2\theta)}{2}\right) d\theta$. Evaluating: $A = \frac{1}{2}\left[\frac{3\theta}{2} + 2\sin\theta + \frac{\sin(2\theta)}{4}\right]_0^{2\pi} = \frac{1}{2} \cdot \frac{3 \cdot 2\pi}{2} = \frac{3\pi}{2}$. Choice B forgets the $\cos^2\theta$ term. Choice C omits the $\frac{1}{2}$ factor. Choice D uses incorrect bounds or makes a computational error.
5. The area of one petal of the rose curve $r = a \cos(3\theta)$ is equal to $\pi$. What is the value of the positive constant $a$?
   Answer: $2\sqrt{3}$ The rose curve $r = a \cos(3\theta)$ has 3 petals. One petal is traced as $3\theta$ goes from $-\pi/2$ to $\pi/2$, which means $\theta$ goes from $-\pi/6$ to $\pi/6$. The area $A$ of one petal is given by: $A = \frac{1}{2} \int_{-\pi/6}^{\pi/6} (a \cos(3\theta))^2 d\theta = \frac{a^2}{2} \int_{-\pi/6}^{\pi/6} \cos^2(3\theta) d\theta$ Using the identity $\cos^2(x) = \frac{1+\cos(2x)}{2}$: $A = \frac{a^2}{2} \int_{-\pi/6}^{\pi/6} \frac{1+\cos(6\theta)}{2} d\theta = \frac{a^2}{4} \left[ \theta + \frac{1}{6}\sin(6\theta) \right]_{-\pi/6}^{\pi/6}$ $A = \frac{a^2}{4} \left[ (\frac{\pi}{6} + \frac{1}{6}\sin(\pi)) - (-\frac{\pi}{6} + \frac{1}{6}\sin(-\pi)) \right] = \frac{a^2}{4} (\frac{\pi}{6} + \frac{\pi}{6}) = \frac{a^2}{4} (\frac{2\pi}{6}) = \frac{\pi a^2}{12}$ We are given that the area is $\pi$. So, $\frac{\pi a^2}{12} = \pi$. This simplifies to $a^2 = 12$, and since $a$ is a positive constant, $a = \sqrt{12} = 2\sqrt{3}$.
6. The area enclosed by one petal of the rose curve $r = 5\sin(2\theta)$ is:
   Answer: $\frac{25\pi}{8}$ One petal occurs when $\sin(2\theta) \geq 0$, which happens for $0 \leq \theta \leq \frac{\pi}{2}$ (first petal). The area is $A = \frac{1}{2}\int_0^{\pi/2} (5\sin(2\theta))^2 d\theta = \frac{25}{2}\int_0^{\pi/2} \sin^2(2\theta) d\theta$. Using $\sin^2(2\theta) = \frac{1-\cos(4\theta)}{2}$: $A = \frac{25}{4}\int_0^{\pi/2} (1-\cos(4\theta)) d\theta = \frac{25}{4}\left[\theta - \frac{\sin(4\theta)}{4}\right]_0^{\pi/2} = \frac{25}{4} \cdot \frac{\pi}{2} = \frac{25\pi}{8}$. Choice B forgets the factor of $\frac{1}{2}$ in the area formula. Choice C uses incorrect bounds $[0, \pi/4]$. Choice D compounds both errors from choices B and has an additional factor error.
7. Find the area of the region enclosed by the inner loop of the limaçon $r = 1 - 2\sin\theta$.
   Answer: $\pi - \frac{3\sqrt{3}}{2}$ The inner loop of a limaçon is traced when $r$ is negative. First, we find the values of $\theta$ for which $r=0$: $1 - 2\sin\theta = 0 \implies \sin\theta = 1/2$. In the interval $[0, 2\pi]$, this occurs at $\theta = \pi/6$ and $\theta = 5\pi/6$. For $\theta$ between $\pi/6$ and $5\pi/6$, $\sin\theta > 1/2$, so $r = 1-2\sin\theta < 0$. This interval traces the inner loop. The area $A$ is: $A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (1 - 2\sin\theta)^2 d\theta$ $A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (1 - 4\sin\theta + 4\sin^2\theta) d\theta$ Using the identity $\sin^2\theta = \frac{1-\cos(2\theta)}{2}$: $A = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (1 - 4\sin\theta + 4\frac{1-\cos(2\theta)}{2}) d\theta = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (3 - 4\sin\theta - 2\cos(2\theta)) d\theta$ $A = \frac{1}{2} [3\theta + 4\cos\theta - \sin(2\theta)]_{\pi/6}^{5\pi/6}$ $A = \frac{1}{2} [(3\frac{5\pi}{6} + 4\cos(\frac{5\pi}{6}) - \sin(\frac{5\pi}{3})) - (3\frac{\pi}{6} + 4\cos(\frac{\pi}{6}) - \sin(\frac{\pi}{3}))]$ $A = \frac{1}{2} [(\frac{5\pi}{2} - 4\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})) - (\frac{\pi}{2} + 4\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2})] = \frac{1}{2} [(\frac{5\pi}{2} - \frac{3\sqrt{3}}{2}) - (\frac{\pi}{2} + \frac{3\sqrt{3}}{2})] = \frac{1}{2} [2\pi - 3\sqrt{3}] = \pi - \frac{3\sqrt{3}}{2}$
8. Find the area enclosed by the curve $r = 2 + \cos(3\theta)$ for $0 \leq \theta \leq 2\pi$.
   Answer: $\frac{9\pi}{2}$ The area is given by $A = \frac{1}{2}\int_0^{2\pi} r^2 \, d\theta = \frac{1}{2}\int_0^{2\pi} (2 + \cos(3\theta))^2 \, d\theta$. Expanding: $(2 + \cos(3\theta))^2 = 4 + 4\cos(3\theta) + \cos^2(3\theta)$. Using $\cos^2(3\theta) = \frac{1 + \cos(6\theta)}{2}$, we get $A = \frac{1}{2}\int_0^{2\pi} \left(4 + 4\cos(3\theta) + \frac{1 + \cos(6\theta)}{2}\right) d\theta = \frac{1}{2}\left[4\theta + \frac{4\sin(3\theta)}{3} + \frac{\theta}{2} + \frac{\sin(6\theta)}{12}\right]_0^{2\pi} = \frac{1}{2}(8\pi + \pi) = \frac{9\pi}{2}$. Choice B results from forgetting the $\cos^2$ term. Choice C comes from using $\cos^2(3\theta) = \frac{1}{2}$ instead of the correct identity. Choice D adds an extra $\pi$ term incorrectly.
9. Find the area of one loop of the lemniscate given by the equation $r^2 = 9\cos(2\theta)$.
   Answer: $\frac{9}{2}$ When you encounter a lemniscate (figure-eight curve) in polar coordinates, you're working with area calculations that require careful attention to the curve's symmetry and domain. For the lemniscate $r^2 = 9\cos(2\theta)$, you first need to determine where the curve exists. Since $r^2 \geq 0$, we need $\cos(2\theta) \geq 0$, which occurs when $-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$ for one complete loop. The area formula in polar coordinates is $A = \frac{1}{2}\int_{\alpha}^{\beta} r^2 \, d\theta$. Since $r^2 = 9\cos(2\theta)$, the area of one loop becomes: $A = \frac{1}{2}\int_{-\pi/4}^{\pi/4} 9\cos(2\theta) \, d\theta = \frac{9}{2}\int_{-\pi/4}^{\pi/4} \cos(2\theta) \, d\theta$ Evaluating: $A = \frac{9}{2} \left[\frac{\sin(2\theta)}{2}\right]_{-\pi/4}^{\pi/4} = \frac{9}{4}[\sin(\pi/2) - \sin(-\pi/2)] = \frac{9}{4}[1 - (-1)] = \frac{9}{2}$ Option A ($\frac{9\pi}{2}$) incorrectly includes $\pi$ in the final answer, likely from confusing this with a circular area formula. Option B ($9$) doubles the correct answer, possibly from calculating both loops. Option C ($\frac{9\pi}{4}$) combines both errors—including $\pi$ and using incorrect integration limits. Remember: lemniscates have restricted domains where $r^2 \geq 0$, and the area formula directly uses $r^2$, making the integration straightforward once you identify the correct limits.
10. What is the area of a single petal of the rose curve given by the polar equation $r = 4\cos(2\theta)$?
    Answer: $2\pi$ The curve $r = 4\cos(2\theta)$ is a rose curve with $2n=4$ petals. To find the area of one petal, we need to determine the bounds of integration that trace it out once. A petal is traced when $r$ goes from $0$ to a maximum and back to $0$. Let's consider the petal symmetric about the polar axis. $r=0$ when $\cos(2\theta)=0$, which occurs when $2\theta = \pm \pi/2, \pm 3\pi/2, \dots$. This implies $\theta = \pm \pi/4, \pm 3\pi/4, \dots$. The petal on the positive polar axis is traced for $\theta$ from $-\pi/4$ to $\pi/4$. The area $A$ is given by: $A = \frac{1}{2} \int_{-\pi/4}^{\pi/4} r^2 d\theta = \frac{1}{2} \int_{-\pi/4}^{\pi/4} (4\cos(2\theta))^2 d\theta$ $A = 8 \int_{-\pi/4}^{\pi/4} \cos^2(2\theta) d\theta$ Using the identity $\cos^2(x) = \frac{1+\cos(2x)}{2}$, we get: $A = 8 \int_{-\pi/4}^{\pi/4} \frac{1+\cos(4\theta)}{2} d\theta = 4 \left[ \theta + \frac{1}{4}\sin(4\theta) \right]_{-\pi/4}^{\pi/4}$ $A = 4 \left[ (\frac{\pi}{4} + \frac{1}{4}\sin(\pi)) - (-\frac{\pi}{4} + \frac{1}{4}\sin(-\pi)) \right] = 4(\frac{\pi}{4} - (-\frac{\pi}{4})) = 4(\frac{\pi}{2}) = 2\pi$

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