📝 Area in polar coordinates double integral (13 MCQs)

📖 Explanation: This is a classic error in polar area. The equation $r=4\cos(2\theta)$ is a rose curve. One loop is traced as $\theta$ goes from $-\pi/4$ to $\pi/4$ because $r \ge 0$ in that interval. The student's limits from 0 to $\pi/4$ only capture half of that loop, missing the symmetrical lower half. The integrand $r$ is correct for area in polar coordinates.

📖 Explanation: This problem requires converting the line $x=1$ into polar form, which is $r\cos\theta = 1$ or $r = \sec\theta$. The circle is $r=2$. The region to the right of the line means $r$ must be greater than $\sec\theta$, and inside the circle means $r$ is less than 2. The intersection of the line and circle occurs at $\cos\theta = 1/2$, so $\theta = \pm \pi/3$. Option A correctly sets up the integral. Option B is the left region, Option C misses the Jacobian $r$, and Option D uses incorrect limits.

📖 Explanation: The student's formula is correct only if $r_{out} \ge r_{in}$ for all $\theta$ in the integration interval. Here, $r=2+\sin\theta$ ranges from 1 to 3. The inner curve is $r=1$. They are equal when $\sin\theta = -1$, i.e., at $\theta = 3\pi/2$. For $\theta$ near $3\pi/2$, $r_{out} = 1$, so the region vanishes. However, for $\theta$ from $0$ to $2\pi$, $r_{out} \ge 1$ always, so the setup is actually correct. The question tests if the student understands the condition for this formula and doesn't falsely reject it. The correct answer is A, but the analysis requires checking inequality over the entire range.

📖 Explanation: This question requires analyzing the graph (provided in a test setting) to determine which curve is farther from the origin on which interval. Intersection occurs when $3\cos\theta = 1+\cos\theta \Rightarrow 2\cos\theta = 1 \Rightarrow \cos\theta = 1/2$, so $\theta = \pm \pi/3$. For $\theta \in [-\pi/3, \pi/3]$, $3\cos\theta \ge 1+\cos\theta$. So the circle is outside the cardioid, and the area inside the circle but outside the cardioid is the integral of the difference of squares over that interval. Option A is correct. Option B would give a negative area. Options C and D are incorrect approaches because simple subtraction of total areas doesn't account for the overlapping region correctly.

📖 Explanation: The equation $r^2 = 4\cos(2\theta)$ is a lemniscate. It has two loops. The area of one loop is found by integrating $\frac{1}{2} r^2$ from $\theta = -\pi/4$ to $\pi/4$ (where $\cos(2\theta) \ge 0$). By symmetry, this is $2 \times \frac{1}{2} \int_{0}^{\pi/4} 4\cos(2\theta) \, d\theta$, which simplifies to $4 \int_{0}^{\pi/4} \cos(2\theta) \, d\theta$. Option A is correct. Option B's upper limit $\pi/2$ is wrong because $\cos(2\theta)$ is negative there. Option C integrates over an interval where $r^2$ is negative. Option D is a double integral but its limits only cover one quadrant, not a full loop.

📖 Explanation: The student's setup is perfectly correct. The circle $r=2\sin\theta$ has a diameter from 0 to 2 on the y-axis and is traced for $\theta \in [0,\pi]$. The circle $r=1$ is centered at the origin. Their intersection is found by $2\sin\theta = 1 \Rightarrow \sin\theta = 1/2$, which gives $\theta = \pi/6$ and $5\pi/6$. For $\theta$ between these values, $2\sin\theta \ge 1$. Therefore, the area outside $r=1$ and inside $r=2\sin\theta$ is exactly given by that double integral. The question tests the student's ability to verify the radial inequality over the entire angular interval and to recognize a correctly set up problem.

📖 Explanation: This is a classic problem requiring identification of the correct curve as the inner boundary. The intersection is found by $\sqrt{3}\cos\theta = \sin\theta \Rightarrow \tan\theta = \sqrt{3} \Rightarrow \theta = \pi/3$. For $\theta \in [0, \pi/3]$, $\sqrt{3}\cos\theta \ge \sin\theta$, so the boundary is the first curve. For $\theta \in [\pi/3, \pi/2]$, $\sin\theta \ge \sqrt{3}\cos\theta$, so the boundary is the second curve. Option A correctly breaks the integral at the intersection and sums the areas under the outer curve. Option B uses wrong intersection and wrong intervals. Option C subtracts incorrectly (would give area between them, not inside both). Option D is conceptually incorrect because the min function would give the inner boundary, not the outer.

📖 Explanation: This problem tests the student's ability to analyze the convergence of a polar integral. The region is defined with $\theta$ going to $\pi/2$, where $\sec\theta \to \infty$ and $\tan\theta \to \infty$. The radial limit $r = \sec\theta \tan\theta$ grows without bound as $\theta \to \pi/2^-$. The area integral is $\frac{1}{2}\int_{0}^{\pi/2} \sec^2\theta \tan^2\theta \, d\theta = \frac{1}{2}[\frac{1}{3}\tan^3\theta]_{0}^{\pi/2}$, which diverges to infinity. Therefore, the area is infinite. This is a higher-order thinking question because it requires recognizing that not all polar regions have finite area, and students must analyze the behavior of the integrand at the limit of integration rather than just applying the formula mechanically.

📖 Explanation: First, find intersections: $4\sin\theta = 2 \Rightarrow \sin\theta = 1/2 \Rightarrow \theta = \pi/6, 5\pi/6$. The circle $r=2$ is centered at origin, radius 2. The curve $r=4\sin\theta$ is a circle centered at (0,2) with radius 2. The region inside both is the overlap of two circles of radius 2 with distance 2 between centers. The area of overlap of two circles with radius R=2 and distance d=2 is $2R^2 \cos^{-1}(d/2R) - (d/2)\sqrt{4R^2 - d^2} = 8\cos^{-1}(1/2) - \sqrt{16-4} = 8(\pi/3) - \sqrt{12} = \frac{8\pi}{3} - 2\sqrt{3}$. In polar, the area is $\int_{\pi/6}^{5\pi/6} \int_{2}^{4\sin\theta} r \, dr \, d\theta$ but careful: that's the area inside the circle $r=4\sin\theta$ and outside $r=2$. For inside both, we need the area inside the smaller radial value. For $\theta \in [\pi/6, 5\pi/6]$, $4\sin\theta \ge 2$, so the inner curve is $r=2$ and outer is $r=4\sin\theta$. The area inside both is the area inside $r=2$ plus the area between them? Actually, the region inside both is the intersection of the two disks. In polar, from $\theta=0$ to $\pi/6$, only the circle $r=2$ exists (since $4\sin\theta < 2$), and from $\pi/6$ to $5\pi/6$, the boundary is $r=2$ for the inner part? Let's think: The region inside both means $r \le 2$ AND $r \le 4\sin\theta$. So for each $\theta$, $r$ goes from 0 to $\min(2, 4\sin\theta)$. For $\theta \in [0, \pi/6] \cup [5\pi/6, \pi]$, $4\sin\theta \le 2$, so $r$ from 0 to $4\sin\theta$. For $\theta \in [\pi/6, 5\pi/6]$, $r$ from 0 to 2. So area = $\frac{1}{2}[\int_{0}^{\pi/6} (4\sin\theta)^2 d\theta + \int_{\pi/6}^{5\pi/6} 4 d\theta + \int_{5\pi/6}^{\pi} (4\sin\theta)^2 d\theta ]$. Compute: $\int (4\sin\theta)^2 d\theta = 16 \int \sin^2\theta d\theta = 8\theta - 4\sin(2\theta)$. Evaluate from 0 to $\pi/6$: $8(\pi/6) - 4\sin(\pi/3) = 4\pi/3 - 2\sqrt{3}$. The middle integral: $4 \times (5\pi/6 - \pi/6) = 4 \times (2\pi/3) = 8\pi/3$. The last integral from $5\pi/6$ to $\pi$: by symmetry, same as first: $4\pi/3 - 2\sqrt{3}$. Sum = $8\pi/3 + 8\pi/3 - 4\sqrt{3} = 16\pi/3 - 4\sqrt{3}$. Then half of that is $8\pi/3 - 2\sqrt{3}$. So option B is correct. This problem requires multi-step reasoning and careful consideration of the min function.

📖 Explanation: The student's integral $\int_{0}^{2\pi} \int_{0}^{1+\cos\theta} \, dr \, d\theta$ integrates the function 1 over the region, which would give the area in Cartesian-like coordinates but without the Jacobian $r$ that is required for polar coordinates. The correct integral is $\int_{0}^{2\pi} \int_{0}^{1+\cos\theta} r \, dr \, d\theta = \frac{1}{2}\int_{0}^{2\pi} (1+\cos\theta)^2 d\theta = \frac{1}{2}\int_{0}^{2\pi} (1 + 2\cos\theta + \cos^2\theta) d\theta = \frac{1}{2}(2\pi + 0 + \pi) = \frac{3\pi}{2}$. Option A correctly identifies the missing Jacobian and gives the correct area. Option B gives half the correct area, which would be the result of an additional factor error. Option C has wrong limits (the cardioid is traced for $0$ to $2\pi$). Option D has the wrong integrand power.

📖 Explanation: The area is $\int_{\pi/6}^{5\pi/6} \int_{a}^{2a\sin\theta} r \, dr \, d\theta = \frac{1}{2}\int_{\pi/6}^{5\pi/6} (4a^2\sin^2\theta - a^2) d\theta = \frac{a^2}{2} \int_{\pi/6}^{5\pi/6} (4\sin^2\theta - 1) d\theta$. Compute $\int_{\pi/6}^{5\pi/6} 4\sin^2\theta \, d\theta = 4 \int_{\pi/6}^{5\pi/6} \frac{1-\cos(2\theta)}{2} d\theta = 2[\theta - \frac{\sin(2\theta)}{2}]_{\pi/6}^{5\pi/6} = 2[ (5\pi/6 - \pi/6) - \frac{1}{2}(\sin(5\pi/3) - \sin(\pi/3)) ] = 2[ (2\pi/3) - \frac{1}{2}(-\sqrt{3}/2 - \sqrt{3}/2) ] = 2[ 2\pi/3 + \sqrt{3}/2 ] = 4\pi/3 + \sqrt{3}$. Then subtract $\int_{\pi/6}^{5\pi/6} 1 \, d\theta = 2\pi/3$. So total integral = $4\pi/3 + \sqrt{3} - 2\pi/3 = 2\pi/3 + \sqrt{3}$. Then area = $\frac{a^2}{2} (2\pi/3 + \sqrt{3}) = a^2(\pi/3 + \sqrt{3}/2)$. Set equal to $\pi/3 + \sqrt{3}/2$, we get $a^2 = 1$, so $a=1$ (since a>0). This is a multi-step problem requiring setting up the integral, evaluating it symbolically, and solving for the parameter. Option A is correct.

📖 Explanation: The curve $r = 2\cos(3\theta)$ is a rose with 3 petals. Each petal is traced when $\cos(3\theta) \ge 0$, which occurs for $\theta \in [-\pi/6, \pi/6]$, $[\pi/2, 5\pi/6]$, and $[7\pi/6, 3\pi/2]$. Method 1 integrates from 0 to $2\pi$, but the integrand $\cos^2(3\theta)$ is non-negative, so it correctly sums the area of all petals. However, it includes the full square, which over the intervals where $\cos(3\theta)$ is negative, $r$ is negative, and the curve is traced in the opposite direction. In polar coordinates, the area enclosed by the curve is given by $\frac{1}{2}\int r^2 d\theta$ over one full tracing, which is from 0 to $2\pi$ for a complete curve. But careful: the curve $r = 2\cos(3\theta)$ is traced exactly once as $\theta$ goes from 0 to $2\pi$. The area enclosed is indeed $\frac{1}{2}\int_{0}^{2\pi} [2\cos(3\theta)]^2 d\theta$ because the formula $\frac{1}{2}\int r^2 d\theta$ gives the area swept by the radius vector, and even if $r$ is negative, $r^2$ is positive, and the curve is traced. Actually, both methods give the same numerical value because $\int_{0}^{2\pi} \cos^2(3\theta) d\theta = \pi$, and $3 \times \int_{-\pi/6}^{\pi/6} \cos^2(3\theta) d\theta = 3 \times (\pi/6) = \pi/2$? Wait, compute: $\int_{-\pi/6}^{\pi/6} \cos^2(3\theta) d\theta = \frac{1}{2}\int_{-\pi/6}^{\pi/6} (1+\cos(6\theta)) d\theta = \frac{1}{2}[\theta + \frac{\sin(6\theta)}{6}]_{-\pi/6}^{\pi/6} = \frac{1}{2}[(\pi/3) + 0] = \pi/6$. Then Method 2: $3 \times \frac{1}{2} \times 4 \times (\pi/6) = 3 \times 2 \times (\pi/6) = \pi$. Method 1: $\frac{1}{2} \times 4 \times \int_{0}^{2\pi} \cos^2(3\theta) d\theta = 2 \times \pi = 2\pi$. Wait, they are not equal. So Method 2 is correct because it sums the area of each petal without including the areas traced when $r$ is negative? Actually, the area of a polar curve is given by $\frac{1}{2}\int r^2 d\theta$ over one complete traversal. For $r = 2\cos(3\theta)$, as $\theta$ goes from 0 to $2\pi$, the curve is traced twice? No, the rose with 3 petals is traced once as $\theta$ goes from 0 to $\pi$? Let's check: $\cos(3\theta)$ has period $2\pi/3$. So $r$ repeats every $2\pi/3$. The curve is traced completely when $\theta$ goes from 0 to $\pi$ because $r$ changes sign, but the full set of petals is traced for $\theta$ from 0 to $2\pi$. Actually, for $r = \cos(n\theta)$, if n is odd, the curve has n petals and is traced once as $\theta$ goes from 0 to $\pi$. If n is even, it has 2n petals and is traced once as $\theta$ goes from 0 to $2\pi$. Here n=3 (odd), so the curve has 3 petals and is traced once for $\theta \in [0, \pi]$. So Method 1 from 0 to $2\pi$ would double count the area. Method 2 correctly sums the three petals, each traced from $-\pi/6$ to $\pi/6$ (or any interval where $\cos(3\theta) \ge 0$). So Method 2 is correct. Therefore, the correct answer is B. This is a conceptual question about the tracing of polar curves and the correct interval for area integration.

📖 Explanation: The limacon $r = 2 + \sin\theta$ has a minimum value of 1 (when $\sin\theta = -1$) and a maximum of 3. The inner circle is $r=1$. Since the minimum of the limacon is exactly 1, the condition $2+\sin\theta \ge 1$ holds for all $\theta \in [0, 2\pi]$. Therefore, for every angle, the radial line from the origin enters the region at $r=1$ and exits at $r=2+\sin\theta$. So the double integral setup is perfectly correct. The student's reasoning is sound. This question tests the student's ability to verify the inequality over the entire domain and not falsely assume an error. The correct answer is A.