📝 Convert double integral to polar coordinates (14 MCQs)

📖 Explanation: The equation $x^2 + y^2 = 2y$ converts to $r^2 = 2r\sin\theta$, giving $r = 2\sin\theta$. Since $r \ge 0$, we need $\sin\theta \ge 0$, which is $0 \le \theta \le \pi$. The range $-\pi/2$ to $\pi/2$ includes negative $\sin\theta$ values, which would make $r$ negative, an invalid representation. This is a common error where students treat the circle as if centered at the origin and fail to enforce $r \ge 0$ when determining angular bounds.

📖 Explanation: The region is a sector of a circle of radius 2 bounded by the x-axis ($y=0$) and the line $y=x$ (angle $\pi/4$). The first integral is a triangle under $y=x$ up to $x=\sqrt{2}$, and the second is the circular segment from $x=\sqrt{2}$ to 2. Together, they perfectly fill the sector with $0 \le \theta \le \pi/4$ and $0 \le r \le 2$. Option B is the area under $y=x$ but outside the circle, which is incorrect. This tests the student's ability to geometrically interpret piecewise rectangular limits as a single polar region.

📖 Explanation: The region is the first-quadrant quarter of the unit circle: $0 \le \theta \le \pi/2$ and $0 \le r \le 1$. The integral becomes $\int_{0}^{\pi/2} \int_{0}^{1} e^{r^2} r \, dr \, d\theta$. Let $u = r^2$, $du = 2r dr$. Inner integral is $\frac{1}{2}(e - 1)$. Multiply by $\theta$-length $\pi/2$ gives $\frac{\pi}{4}(e - 1)$. Option B is a common mistake where the inner integral is not halved, while C and D confuse the exponent or fail to integrate correctly.

📖 Explanation: The original integral has $x$ from -1 to 1 and $y$ from 0 to $\sqrt{1-x^2}$. For each $x$, $y$ is positive, so the region is the upper half of the unit circle. However, the angular range for the upper half circle (centered at origin) is indeed $0$ to $\pi$ in standard polar coordinates. Wait, let's re-evaluate: In polar, upper half means $y \ge 0$ so $\sin\theta \ge 0$, hence $0 \le \theta \le \pi$. The student's conversion is actually correct if they include $r$. The flaw in the question is that the student's claim is correct; the question tests if you can spot that there is no flaw. However, the correct answer is D because many students confuse the upper and right halves. The right half is $-\pi/2$ to $\pi/2$, and the upper half is $0$ to $\pi$. The given limits are for the upper half, so the conversion is correct. Since the question asks for the flaw, and there is none, the correct distractor is that the student is correct (option B). But the instruction says 'identifies the flaw', so I must provide an option that points out a misconception. Let's correct: The region is the upper semicircle, so $0$ to $\pi$ is correct. The only possible flaw is if they forget the Jacobian $r$. But the option B says 'the student is correct'. Since the question asks for a flaw, and the conversion is correct, the student has no flaw. To satisfy the question, I'll make option B the correct answer and explain that the student is indeed correct. However, the question says 'which correctly identifies the flaw' – so I'll rephrase. Actually, I'll make the correct answer 'No flaw; the conversion is correct' as option B.

📖 Explanation: The original integral has $x$ from 0 to $\sqrt{4-y^2}$ and $y$ from 0 to 2. This is the first-quadrant quarter of a circle of radius 2 centered at the origin. In polar, $0 \le \theta \le \pi/2$ and $0 \le r \le 2$. The integrand $\sqrt{x^2+y^2} = r$, and the Jacobian is $r$, so the integrand becomes $r^2$. The integral is $\int_{0}^{\pi/2} \int_{0}^{2} r^2 \, dr \, d\theta = \int_{0}^{\pi/2} [r^3/3]_0^2 \, d\theta = \int_{0}^{\pi/2} (8/3) \, d\theta = 4\pi/3$. Wait, $8/3 * \pi/2 = 4\pi/3$. But option C says $8\pi/3$ – that's a common error of forgetting the $\theta$ integration gives $\pi/2$ not 1. So the correct value is $4\pi/3$. I need to adjust options. Let's redo: The correct integral is $\int_{0}^{\pi/2} \int_{0}^{2} r^2 \, dr \, d\theta = (8/3)(\pi/2) = 4\pi/3$. So option C with correct value $4\pi/3$ should be correct. I'll set option C as correct with value $4\pi/3$.

📖 Explanation: The region is bounded by $y=0$, $x=0$ to 1, and $y = \sqrt{2x - x^2}$. Squaring gives $x^2 + y^2 = 2x$, or $r^2 = 2r\cos\theta$, so $r = 2\cos\theta$. Since $y \ge 0$, $\sin\theta \ge 0$, and $x \ge 0$ gives $\cos\theta \ge 0$. This restricts $\theta$ to the first quadrant, i.e., $0 \le \theta \le \pi/2$. This is a circle centered at (1,0) with radius 1, but only the upper half. Option B has $\pi/4$ which would be a sector, not the full half-circle. Option C is a circle centered at (0,1). Option D has wrong $r$ and $\theta$ range.

📖 Explanation: The region is the triangle under the line $y=x$ in the first quadrant, from $x=0$ to 1, and $y=0$ to $x$. In polar, $y=x$ gives $\theta = \pi/4$. The vertical line $x=1$ gives $r\cos\theta = 1$, so $r = \sec\theta$. The lower bound $y=0$ is $\theta=0$. Thus, $0 \le \theta \le \pi/4$ and $0 \le r \le \sec\theta$. The student's conversion is exactly correct. This tests the ability to recognize a triangular region and correctly map the line $x=1$ to $r = \sec\theta$, a common source of error where students might use $\csc\theta$ for horizontal lines.

📖 Explanation: The circle is centered at the origin with radius 2. The inequality $x \ge y$ means the region is the half-plane below the line $y = x$, i.e., angles where $\cos\theta \ge \sin\theta$, which is $-\pi/4 \le \theta \le 3\pi/4$. This is a semicircle. Option B is the opposite half-plane ($x \le y$). Option C is the right half-plane. Option D incorrectly tries to shift the circle. This requires combining polar angle inequalities with a circular radial bound, a classic multi-step reasoning problem.

📖 Explanation: The circle $r = 2\sin\theta$ is centered at (0,1) with radius 1. The circle $r=1$ is centered at the origin. To find the intersection, set $2\sin\theta = 1$ => $\sin\theta = 1/2$ => $\theta = \pi/6$ and $5\pi/6$. The region outside $r=1$ and inside the larger circle requires $r$ from 1 to $2\sin\theta$, and $\theta$ between these intersection angles. Option B misses the intersection bounds and would include areas where $2\sin\theta < 1$. Option C subtracts the entire unit circle, which is incorrect because the unit circle is not fully inside the region for all $\theta$. Option D compounds the error. This tests the ability to find intersection points and set up the radial limits correctly.

📖 Explanation: In polar coordinates, $dA = r \, dr \, d\theta$ and $x^2 + y^2 = r^2$. Therefore, the integrand becomes $r^2 \cdot r = r^3$. The student only used $r^2$, forgetting the Jacobian. The angular range for a standard cardioid $r = 1 + \cos\theta$ is indeed $0$ to $2\pi$ to trace the full curve; there is no inner loop. So the only error is the missing $r$. This is a very common misconception where students forget to multiply by the Jacobian when converting the area element.

📖 Explanation: The region is in the first quadrant, so $0 \le \theta \le \pi/2$. The radial bounds are $1 \le r \le 2$. The integrand becomes $\frac{1}{1+r^2}$, with Jacobian $r$. So the integral is $\int_{0}^{\pi/2} \int_{1}^{2} \frac{r}{1+r^2} \, dr \, d\theta$. Inner integral: let $u=1+r^2$, $du = 2r dr$ -> $\frac{1}{2} [\ln(1+r^2)]_1^2 = \frac{1}{2} (\ln 5 - \ln 2) = \frac{1}{2} \ln(5/2)$. Multiply by $\pi/2$ gives $\frac{\pi}{4} \ln(5/2)$, not $\frac{\pi}{2}\ln 2$. Wait, I need to correct. Actually, $\int \frac{r}{1+r^2} dr = \frac{1}{2}\ln(1+r^2)$. From 1 to 2: $\frac{1}{2}(\ln 5 - \ln 2) = \frac{1}{2}\ln(5/2)$. Times $\pi/2$ = $\frac{\pi}{4}\ln(5/2)$. But option A says $\frac{\pi}{2}\ln 2$. That is incorrect. Let me recalc: If the region were between 0 and 1, it would be $\frac{1}{2}\ln 2$. But here it's 1 to 2. So the correct is $\frac{\pi}{4}\ln(5/2)$. I'll set that as option A and adjust others. So correct answer is A as $\frac{\pi}{4}\ln(5/2)$.

📖 Explanation: For the lemniscate $r^2 = 4\cos 2\theta$, the curve exists only where $\cos 2\theta \ge 0$, i.e., $-\pi/4 \le \theta \le \pi/4$ and $3\pi/4 \le \theta \le 5\pi/4$. In the first quadrant, $\theta$ ranges from 0 to $\pi/4$. At $\theta = \pi/4$, $\cos(\pi/2) = 0$, so $r=0$. The student's limits are actually correct: $0 \le \theta \le \pi/4$ and $0 \le r \le \sqrt{4\cos 2\theta}$. But the area would be positive. The question states the area is negative, which might happen if the student integrated from $\pi/4$ to 0 (reversed limits) or if they used $\theta$ from 0 to $\pi/2$ where $\cos 2\theta$ becomes negative, leading to imaginary $r$. The most plausible reason is that they extended $\theta$ to $\pi/2$ and got a negative radicand. But option B says the correct upper limit is $\pi/4$, which is true, but that doesn't cause negativity. I'll rephrase: The student might have used $0$ to $\pi/2$, so the inner integral has $\sqrt{\cos 2\theta}$ which is imaginary for $\theta > \pi/4$, leading to a non-real or negative area if they force it. So the correct error is using an angular range where $\cos 2\theta$ is negative. I'll set option B as 'The student used the range $0$ to $\pi/2$, which includes angles where $\cos 2\theta < 0$, making the integrand imaginary, and the correct range is $0$ to $\pi/4$.'

📖 Explanation: The region is the first-quadrant quarter circle of radius 2. The integrand $(x^2+y^2)^{3/2} = (r^2)^{3/2} = r^3$. The Jacobian is $r$, so the integrand becomes $r^4$. The integral is $\int_{0}^{\pi/2} \int_{0}^{2} r^4 \, dr \, d\theta = \int_{0}^{\pi/2} [r^5/5]_0^2 \, d\theta = \int_{0}^{\pi/2} (32/5) \, d\theta = 16\pi/5$. Option A uses $r^3$ forgetting the Jacobian. Option C also uses $r^3$ but with a different constant. Option D uses the wrong angular range (upper half instead of first quadrant) and also has an incorrect constant. This tests the correct application of the Jacobian and the exponent simplification.

📖 Explanation: The circle $x^2+y^2=2x$ is centered at (1,0) with radius 1. It lies entirely in the right half-plane ($x \ge 0$). The line $y=x$ has angle $\pi/4$. The region above the line $y=x$ means $\theta \ge \pi/4$. Since the circle only exists for $-\pi/2 \le \theta \le \pi/2$ (as $2\cos\theta \ge 0$), and we are above the line, the valid $\theta$ range is $\pi/4 \le \theta \le \pi/2$. So the student's claim is correct. This requires visualizing the circle, understanding its angular extent, and correctly intersecting it with the half-plane condition. Option B misunderstands the center's effect on angular limits. Option C incorrectly extends into the second quadrant where the circle doesn't exist. Option D is a distractor about symmetry.