Answer: $\iint_R \sqrt{f_x^2 + f_y^2 + 1} \, dA$

Identify the surface area formula: For a surface $z = f(x, y)$, the surface area element is $dS = \sqrt{f_x^2 + f_y^2 + 1} \, dA$, which accounts for the tilt of the surface relative to the $xy$-plane. Why Option C is correct: Option C correctly includes the $+1$ under the square root, which comes from the cross product of the tangent vectors $\mathbf{r}_x \times \mathbf{r}_y$ when $\mathbf{r}(x,y) = \langle x, y, f(x,y) \rangle$. Why distractors fail: Option A is missing the $+1$ under the radical. Option B has no square root, which would give incorrect units and magnitude. Option D incorrectly sums partial derivatives linearly rather than using the Pythagorean form.

Answer: $\sqrt{4x^2 + 4y^2 + 1}$

Apply the surface area formula: The surface area integrand is $\sqrt{f_x^2 + f_y^2 + 1}$. With $f_x = 2x$ and $f_y = 2y$, this gives $\sqrt{(2x)^2 + (2y)^2 + 1} = \sqrt{4x^2 + 4y^2 + 1}$. Why Option B is correct: Option B correctly squares the partial derivatives and adds 1 under the square root. Why distractors fail: Option A forgets the $+1$. Option C has the correct expression under the radical but omits the square root itself. Option D fails to square the partial derivatives.

Answer: $4$

Recall the relationship between curvature and radius of curvature: The radius of curvature $\rho$ is the reciprocal of the curvature: $\rho = 1/\kappa$. Compute the radius: $\rho = 1/0.25 = 4$. The osculating circle at that point has radius 4. Why distractors fail: Option A ($0.25$) confuses curvature with radius. Option B ($2$) might result from taking $\sqrt{1/\kappa}$ or another incorrect manipulation. Option D ($0.5$) could come from computing $2\kappa$ instead of $1/\kappa$.

Answer: $L = \int_a^b \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} \, dt$

Recall the arc length formula: The arc length of a space curve $\mathbf{r}(t)$ from $t = a$ to $t = b$ is $L = \int_a^b |\mathbf{r}'(t)| \, dt$, where $|\mathbf{r}'(t)|$ is the magnitude of the derivative vector. Why the correct answer works: Since $\mathbf{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle$, its magnitude is $\sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}$, giving the formula in Option B. Why distractors fail: Option A omits the $z'(t)^2$ term, which is the 2D arc length formula. Option C sums the derivatives without squaring and taking a square root, which does not compute a magnitude. Option D uses the position components $x(t), y(t), z(t)$ instead of their derivatives.

Answer: The rate at which the unit tangent vector changes direction per unit arc length

Definition of curvature: Curvature $\kappa$ is defined as $\kappa = \left| \frac{d\mathbf{T}}{ds} \right|$, where $\mathbf{T}$ is the unit tangent vector and $s$ is arc length. It measures how sharply the curve bends. Why Option B is correct: Option B directly states the definition: curvature is the rate of change of the unit tangent vector with respect to arc length. Why distractors fail: Option A describes speed $|\mathbf{r}'(t)|$, not curvature. Option C describes arc length, a different geometric quantity. Option D describes slope, which is a concept from single-variable calculus and does not capture bending in 3D.

Answer: $10\pi$

Compute the derivative: $\mathbf{r}'(t) = \langle -3\sin t, \, 3\cos t, \, 4 \rangle$. Find the speed: $|\mathbf{r}'(t)| = \sqrt{9\sin^2 t + 9\cos^2 t + 16} = \sqrt{9 + 16} = \sqrt{25} = 5$. Integrate to find arc length: $L = \int_0^{2\pi} 5 \, dt = 5 \cdot 2\pi = 10\pi$. Option C is correct. Why distractors fail: Option A ($6\pi$) may result from using only the $xy$-components (radius 3). Option B ($8\pi$) could come from incorrectly computing the speed as $\sqrt{9+16} = 4$ or a similar arithmetic error. Option D ($5\pi$) uses the correct speed but integrates over $[0, \pi]$ instead of $[0, 2\pi]$.

Answer: $\int_0^{2\pi} \int_0^2 \int_0^r r \, dz \, dr \, d\theta$

Convert to cylindrical coordinates: In cylindrical coordinates, $x^2 + y^2 = r^2$, so the cone $z = \sqrt{x^2+y^2}$ becomes $z = r$, and the cylinder $x^2+y^2 = 4$ becomes $r = 2$. Set up limits of integration: The solid has $0 \leq \theta \leq 2\pi$, $0 \leq r \leq 2$, and $0 \leq z \leq r$. The volume element in cylindrical coordinates is $r \, dz \, dr \, d\theta$. Why distractors fail: Option B uses $z = r^2$ as the upper bound, which corresponds to a paraboloid, not a cone. Option C omits the Jacobian factor $r$ from the cylindrical volume element. Option D uses $r = 4$ as the upper limit for $r$, but $x^2 + y^2 = 4$ means $r = 2$, not $r = 4$.

Answer: Compute $\iint_R \sqrt{4x^2 + 4y^2 + 1} \, dA$ over the disk $x^2 + y^2 \leq 4$ for the curved cap, and add $\pi(2)^2 = 4\pi$ for the flat base.

Identify the two surfaces: The solid is bounded by the paraboloid cap $z = 4 - x^2 - y^2$ (for $z \geq 0$, i.e., $x^2+y^2 \leq 4$) and the flat disk at $z = 0$ with radius 2. The total surface area is the sum of both. Compute the cap surface area: For $z = 4 - x^2 - y^2$, the partial derivatives are $f_x = -2x$, $f_y = -2y$. The surface area of the cap is $\iint_R \sqrt{4x^2 + 4y^2 + 1} \, dA$ over the disk $x^2 + y^2 \leq 4$. Add the base area: The base is a flat disk of radius 2 in the $z = 0$ plane, so its area is $\pi(2)^2 = 4\pi$. The total surface area is the sum of the two. Why distractors fail: Option A claims the base is already included in the surface area integral, but that integral only accounts for the paraboloid cap. Option B confuses volume and surface area — differentiation of volume does not yield surface area in general. Option D doubles the cap area, which does not equal the cap plus a flat disk.

Answer: The student is correct; $\frac{2}{3}\pi a^3$ is the volume of a hemisphere, which is half of the full sphere volume $\frac{4}{3}\pi a^3$.

Check the limits of integration: The integral has $z$ ranging from $0$ to $\sqrt{a^2 - r^2}$, which traces the upper hemisphere of the sphere $x^2+y^2+z^2 = a^2$. This is indeed only the top half. Verify the computation: The volume of a full sphere is $\frac{4}{3}\pi a^3$, so a hemisphere has volume $\frac{2}{3}\pi a^3$. The student's integral and result are both correct. Why distractors fail: Option A is wrong because the student's integral is correctly set up for a hemisphere. Option C gives an incorrect volume ($\frac{1}{3}\pi a^3$ is the volume of a cone of height $a$ and radius $a$). Option D is wrong because the integrand $r \, dz \, dr \, d\theta$ is the cylindrical volume element, not a surface area element.

Answer: The integrand equals 1 only when $f_x = f_y = 0$, but otherwise $\sqrt{f_x^2 + f_y^2 + 1} > 1$, so the surface area integral is at least as large as $\iint_R 1 \, dA$.

Analyze the integrand: The integrand is $\sqrt{f_x^2 + f_y^2 + 1}$. Since $f_x^2 \geq 0$ and $f_y^2 \geq 0$, the expression under the radical is always $\geq 1$, so $\sqrt{f_x^2 + f_y^2 + 1} \geq 1$. Compare to the area of R: The area of $R$ is $\iint_R 1 \, dA$. Since the surface area integrand is always $\geq 1$, we have $SA = \iint_R \sqrt{f_x^2 + f_y^2 + 1} \, dA \geq \iint_R 1 \, dA = \text{Area}(R)$. Why distractors fail: Option B is vague and does not provide a valid mathematical justification. Option C is a nonsensical comparison between different types of integrals. Option D is false because partial derivatives can be zero or negative.

Answer: The colleague's suggestion would simplify the integrand to 1, but finding the arc length reparametrization requires evaluating the same integral first, so it does not help.

Verify the engineer's integral: $\mathbf{r}'(t) = \langle 2t, 3t^2, 1 \rangle$, so $|\mathbf{r}'(t)| = \sqrt{4t^2 + 9t^4 + 1}$. The integral $L = \int_0^1 \sqrt{4t^2 + 9t^4 + 1} \, dt$ is correctly set up. Analyze the reparametrization suggestion: Arc length reparametrization uses $s(t) = \int_0^t |\mathbf{r}'(u)| \, du$. After reparametrizing, $|\mathbf{r}'(s)| = 1$, so $L = \int_0^L 1 \, ds$. However, finding the function $s(t)$ requires evaluating the original integral, creating a circular dependency for the purpose of computing $L$. Why distractors fail: Option A incorrectly computes the speed without squaring the derivative components. Option C is technically true as a formal identity but misleading — you cannot use $L = \int_0^L 1 \, ds$ to compute $L$ since $L$ is unknown. Option D is wrong because the integrand is $|\mathbf{r}'(t)|$, which is an arc length integrand.