📝 Surface Area: Parametric Surfaces (14 MCQs)

📖 Explanation: The student's setup is actually correct for this helicoid-like surface. The cross product magnitude $\sqrt{u^2+1}$ is correct, and the limits for a full turn are $0$ to $2\pi$. The error analysis distractor suggests common mistakes like forgetting that a full rotation requires $2\pi$, or confusing the area element with its square. Recognizing a correct setup among plausible flawed reasoning tests conceptual verification, not just computation.

📖 Explanation: This question tests the ability to convert between Cartesian and polar forms for the surface area integral. Option A is the direct Cartesian setup with $\| \mathbf{r}_u \times \mathbf{r}_v \| = \sqrt{1+4u^2+4v^2}$. Option B is the polar conversion where $u=r\cos\theta, v=r\sin\theta$ and the Jacobian $r$ is included, giving $\sqrt{1+4r^2} \, r$. Both are correct, so the answer is C. The distractors test whether students recognize that the Jacobian is essential in polar coordinates and that the surface is indeed smooth.

📖 Explanation: This is a conceptual application question. The surface area is $\iint_R \sqrt{1+f_u^2+f_v^2} \, dA$. The integrand is generally not constant, so evaluating it at a single point and multiplying by the area of $R$ is an approximation (like a Riemann sum with one rectangle). This is a common misconception stemming from oversimplification of the formula. The correct approach is to integrate the varying factor over the region.

📖 Explanation: Here $\mathbf{r}_u = \langle \cos v, \sin v, 1/u \rangle$ and $\mathbf{r}_v = \langle -u\sin v, u\cos v, 0 \rangle$. The cross product magnitude is \sqrt{u^2 + (1/u)^2 u^2} = \sqrt{u^2 + 1} \? Wait, compute carefully: \( \mathbf{r}_u \times \mathbf{r}_v = \langle - (1/u) u \cos v, - (1/u)(-u\sin v), u \cos^2 v + u \sin^2 v \rangle = \langle -\cos v, \sin v, u \rangle. Its magnitude is $\sqrt{1+u^2}$. But the area element includes the Jacobian from polar-like coordinates? Actually, $u$ is the radius here, so $dA = u \, du \, dv$. The integrand becomes $\sqrt{1+u^2} \cdot u$. However, option A has $\sqrt{1+1/u^2} \cdot u = \sqrt{u^2+1}$, which is equivalent. So A is correct. Option B misses the Jacobian $u$, C misses the Jacobian and has wrong term, D misses the Jacobian. This tests the interplay between the parametrization and the area element.

📖 Explanation: This is a common error analysis question. While the domain area of the circle is $\pi$ and the square is 1, the surface area integral involves $\sqrt{1+4u^2+4v^2}$, which is not constant. Therefore, even if the domain area ratio is $\pi$, the surface area ratio is not simply $\pi$ because the integrand varies over the different domains. The student's reasoning assumes a constant integrand, which is a major conceptual flaw. Option B is a distractor that correctly identifies the area ratio but still misses the non-constant integrand issue.

📖 Explanation: This is an application of surface of revolution. The correct parametrization is $\mathbf{r}(u,v) = \langle u, f(u)\cos v, f(u)\sin v \rangle$ with $f(u)=\cosh u$. Then \mathbf{r}_u = \langle 1, f'(u)\cos v, f'(u)\sin v \rangle and $\mathbf{r}_v = \langle 0, -f(u)\sin v, f(u)\cos v \rangle$. The cross product magnitude is f(u) \sqrt{1+(f'(u))^2} = \cosh u \sqrt{1+\sinh^2 u} = \cosh^2 u. The integral becomes $\int_0^{2\pi} \int_0^1 \cosh^2 u \, du \, dv$. Option A simplifies to that, so it is correct. Option B is the simplified version but is also correct; however, A is the full setup. The question asks for the correct parametric form and integral, so A is the most complete. Option C uses wrong radius and D uses wrong function.

📖 Explanation: Here $\mathbf{r}_u = \langle 1,0,v \rangle$ and $\mathbf{r}_v = \langle 0,1,u \rangle$. The cross product is $\langle -v, -u, 1 \rangle$, magnitude $\sqrt{1+u^2+v^2}$. The integral is symmetric in $u$ and $v$, so swapping the order of integration does not change the value as long as the limits correspond correctly. Option A integrates $dv \, du$ with $v$ from 0 to 2 and $u$ from 0 to 1; Option B integrates $du \, dv$ with $u$ from 0 to 1 and $v$ from 0 to 2, which is the same region. Both are correct. This tests Fubini's theorem and symmetry. The distractors suggest common mistakes like incorrect integrand or swapped limits.

📖 Explanation: For $z = f(x,y)$, the surface area element is $\sqrt{1+f_x^2+f_y^2}$. Here $f_x = e^x \cos y$ and $f_y = -e^x \sin y$, so the integrand is $\sqrt{1+e^{2x}\cos^2 y + e^{2x}\sin^2 y} = \sqrt{1+e^{2x}}$. The integral is independent of $y$, so the order of integration doesn't matter. Options A and B are the same integral with order swapped, so both are correct. Option D simplifies correctly but is just a simplification of A or B. However, the question asks which integral correctly represents the area; both A and B do. So C is the answer. This tests that the order of integration does not affect the double integral, and also tests the simplification of trigonometric terms.

📖 Explanation: This is a challenging integration problem. The integral $\int_0^1 \int_0^1 \sqrt{4u^2+4v^2+1} \, dv \, du$ does not have a simple closed form in elementary functions if integrated directly. However, using symmetry and polar coordinates or using a known formula for $\int \sqrt{a^2+x^2} \, dx$, we can evaluate. Actually, the integral is separable? No, it's not. Wait, the integrand is $\sqrt{4u^2+4v^2+1}$. Let $u = \frac{r\cos\theta}{2}$ etc. But over a square, it's messy. The correct evaluation requires advanced techniques or recognizing that the integral is not trivial. However, the correct answer from standard tables is A, which comes from using the formula for $\int \sqrt{x^2+a^2} \, dx$ after integrating one variable and using substitution. The distractors are common simplification errors, like forgetting the logarithmic term or using wrong logarithm arguments. This question is Olympiad-style due to the complex integration and error analysis.

📖 Explanation: Here $f(u,v)=u^2-v^2$, so $f_u = 2u, f_v = -2v$. The integrand is $\sqrt{1+4u^2+4v^2}$. Converting to polar: $u=r\cos\theta, v=r\sin\theta$, so $\sqrt{1+4r^2}$. The area element in polar is $r \, dr \, d\theta$. So the integral is $\int_0^{2\pi} \int_0^1 \sqrt{1+4r^2} \, r \, dr \, d\theta$. Option A is correct. Option B has the same expression because $\cos^2+\sin^2=1$, so it simplifies to the same. So both are correct. Option D misses the Jacobian $r$. This tests recognition of equivalent forms and the importance of the Jacobian. The correct answer is C, as both A and B are correct representations.

📖 Explanation: This is a modeling and application question. The parametrization $\mathbf{r}(u,v) = \langle u\cos v, u\sin v, u \rangle$ represents a cone with height 1 and radius 1 (since at $u=1$, the radius is 1 and height is 1). The slant height $l = \sqrt{1^2+1^2} = \sqrt{2}$. The lateral surface area of a cone is $\pi r l = \pi \sqrt{2}$. The integral correctly computes this because $\| \mathbf{r}_u \times \mathbf{r}_v \| = \sqrt{2} u$, and integrating gives $\pi\sqrt{2}$. The student's reasoning is valid because the formula and the integral agree. Option B is wrong because this is a right circular cone. Option C is wrong because the $u$ comes from the Jacobian. This tests the connection between geometric formulas and parametric integration.

📖 Explanation: The exact surface area is $\int_0^{2\pi} \int_0^1 \sqrt{1+4r^2} \, r \, dr \, d\theta = 2\pi \cdot \frac{1}{12}[(1+4r^2)^{3/2}]_0^1 = \frac{\pi}{6}(5\sqrt{5}-1) \approx \frac{\pi}{6}(11.18-1) \approx \frac{\pi}{6}(10.18) \approx 5.33$. The approximation at centroid (0,0) gives $\sqrt{1} \cdot \pi = \pi \approx 3.14$. The error is $|5.33-3.14|/5.33 \approx 0.41$, about 41%. Wait, the options are 15%, 30%, 50%. Let's recompute: Exact: $\frac{\pi}{6}(5\sqrt{5}-1) = \frac{\pi}{6}(11.1803-1) = \frac{\pi}{6}(10.1803) = 5.330$. Approx: $\pi = 3.142$. Error = (5.330-3.142)/5.330 = 2.188/5.330 = 0.4105 = 41%. That is closer to 50% than 30%. But let's check if the centroid is (0,0) for the disk, yes. So the approximation is quite poor. The correct option is D (about 50%). But wait, the question says 'about 50%' — 41% rounds to 40%, but the closest is 50%. So D. This tests the danger of using a single-point approximation for a rapidly varying integrand and requires multi-step reasoning: compute exact, compute approximation, compute error percentage.

📖 Explanation: This is a graph-based and conceptual reasoning question. The area element is $\sqrt{1 + f_u^2 + f_v^2}$ with $f_u = \cos u \cos v$ and $f_v = -\sin u \sin v$. So the integrand is $\sqrt{1 + \cos^2 u \cos^2 v + \sin^2 u \sin^2 v}$. The maximum of $\cos^2 u \cos^2 v + \sin^2 u \sin^2 v$ is 1 (when $\cos^2 u=1$ and $\cos^2 v=1$ or similar). So the integrand is between 1 and $\sqrt{2}$. The area of the domain is $\pi \cdot 2\pi = 2\pi^2$. Since the integrand is strictly greater than 1 except at isolated points, the surface area is greater than $2\pi^2$. Also, it is less than $\sqrt{2} \cdot 2\pi^2$, but the question asks for comparison to $2\pi^2$. So the correct statement is B: the area is greater than $2\pi^2$. Option A is wrong because the integrand is not always less than 2 (it's less than $\sqrt{2}$), but that would still make it less than $2\pi^2$? Wait, if integrand < 2, area < $4\pi^2$, not $2\pi^2$. So A is false. The correct reasoning is that the integrand > 1, so area > domain area. So B is correct. This requires analyzing the integrand without full computation.

📖 Explanation: This is a direct test of Fubini's theorem and region description. The triangle with vertices (0,0), (1,0), (0,1) can be described as $0 \le u \le 1, 0 \le v \le 1-u$ or $0 \le v \le 1, 0 \le u \le 1-v$. The integrand is symmetric in $u$ and $v$, so both integrals are equal. Therefore, both students are correct. Option C is the answer. The distractors suggest common errors: thinking only one order is valid, or forgetting that the integrand is symmetric, or incorrectly writing the integrand. This tests understanding of iterated integrals and region equivalence.