📝 Surface area formula double integral (14 MCQs)

📖 Explanation: The expression is the norm of the cross product of the two tangent vectors $\mathbf{r}_x = (1,0,f_x)$ and $\mathbf{r}_y = (0,1,f_y)$. This norm $||\mathbf{r}_x \times \mathbf{r}_y|| = \sqrt{1+f_x^2+f_y^2}$ is the area of the infinitesimal parallelogram on the tangent plane that corresponds to a unit area in the $xy$-plane. Option A is incorrect as the normal vector is not fully specified; the gradient lacks the vertical component; and D misinterprets the geometric meaning.

📖 Explanation: Here $f_x = 2$ and $f_y = 3$. The integrand is $\sqrt{1+4+9} = \sqrt{14}$. The area of the rectangular region $R$ is $1 \times 2 = 2$. Therefore, the surface area is $\sqrt{14} \times 2 = 2\sqrt{14}$. Option B forgets to multiply by the region's area, a common mistake. Options C and D come from incorrectly computing $1 + f_x^2 + f_y^2$.

📖 Explanation: The student's error is in the algebraic simplification: $\sqrt{1+4r^2} \cdot r$ is not equal to $(1+4r^2)$. The correct integral is $2\pi \int_0^1 r\sqrt{1+4r^2} \, dr$, which requires a substitution like $u=1+4r^2$. The Jacobian $r$ is correctly included, and the expression $1 + f_x^2 + f_y^2 = 1 + 4r^2$ is correct. The simplification error is a common but critical mistake that dramatically alters the result.

📖 Explanation: The surface is $z = x^2$, so $f_x = 2x$, $f_y = 0$. The surface area is $\iint_R \sqrt{1 + 4x^2} \, dA$. The projection onto the $xz$-plane is the region where $y$ is constant, and its area is the area of the rectangle $0 \le x \le 1, 0 \le y \le 1$, which is 1. Option A correctly identifies the difference as the surface area integral minus the area of the projection (which is 1). Option B incorrectly subtracts the integral of $f_x$. Option C loses the $y$-dimension. Option D misrepresents the projection area.

📖 Explanation: This requires comparing integrals without direct evaluation. The integrands are $\sqrt{1+f_x^2+f_y^2}$. For A, $f_x=f_y=1$, integrand $\sqrt{3} \approx 1.732$. For B, $f_x=2x, f_y=2y$, max integrand $\sqrt{1+8} = 3$ at (1,1), but average is lower. For C, derivatives vary from 0 to $\pi$, integrand up to $\sqrt{1+2\pi^2} \approx 4.58$. For D, $f_x=f_y=e^{x+y}$, integrand $\sqrt{1+2e^{2(x+y)}}$, which is at least $\sqrt{1+2} \approx 1.732$ and grows exponentially, reaching $\sqrt{1+2e^4} \approx 10.9$ at (1,1). Over the entire square, D's integrand dominates, so its surface area is largest.

📖 Explanation: If $\nabla f = (f_x, f_y) = (0,0)$ everywhere on $R$, then $f$ is constant on $R$. The surface is a horizontal plane. For a horizontal plane $z = c$, the surface area element is $\sqrt{1+0+0} \, dA = dA$. Therefore, the surface area over $R$ is exactly the area of $R$. Option B is incorrect as a non-horizontal plane has greater area. Option C is impossible for a differentiable surface over a region. Option D is false because the condition fully determines the geometry.

📖 Explanation: Since $R_2$ contains $R_1$ (the unit disk is inside the square), the surface area over $R_2$ is the integral over a larger region. The integrand $\sqrt{1+4x^2+4y^2}$ is always positive. Therefore, the integral over the larger region $R_2$ must be strictly greater than the integral over the smaller region $R_1$. So $S_1 < S_2$. Option A is impossible. Option C would only be true if the integrand were zero outside the disk. Option D is incorrect because the containment is independent of the function, as long as the integrand is positive.

📖 Explanation: For $z = \sqrt{x^2+y^2}$, we have $f_x = \frac{x}{\sqrt{x^2+y^2}}$ and $f_y = \frac{y}{\sqrt{x^2+y^2}}$. Then $1 + f_x^2 + f_y^2 = 1 + \frac{x^2}{x^2+y^2} + \frac{y^2}{x^2+y^2} = 2$. In polar coordinates, $dA = r \, dr \, d\theta$. So the integral is $\int_0^{2\pi} \int_0^a \sqrt{2} \, r \, dr \, d\theta$. Option A is the correct setup but is repeated as C. Option B incorrectly uses $\sqrt{1+r^2}$. Option D misses the Jacobian $r$. This tests the common misconception that $f_r$ is used instead of the correct partial derivatives.

📖 Explanation: At $P$, $f_x = f_y = 0$, so the tangent plane is horizontal. The local surface area element $dS = \sqrt{1+0+0} \, dA = dA$. This means locally, the surface is 'flat' and the area scaling factor is 1. While the integrand has a local minimum (since it's 1 at P and ≥1 elsewhere), it doesn't globally minimize the integral over all surfaces. Option B is false because the integral depends on the entire region. Option D is a dangerous misconception; a single point has measure zero and doesn't affect the integral value, but its geometric interpretation is important.

📖 Explanation: The function $1+f_x^2+f_y^2 = 1+4x^2+4y^2 = 1+4r^2$ in polar coordinates. The region is a disk, so its bounds are constants in polar coordinates ($0 \le r \le a, 0 \le \theta \le 2\pi$). This allows the integral to be separated and evaluated using a simple substitution. Option A is less efficient because the region bounds would involve square roots. Option C is incorrect; the integral $\int r\sqrt{1+4r^2} dr$ has a closed form. Option D is inappropriate as spherical coordinates are for 3D volumes, not surface areas of graphs.

📖 Explanation: The new integrand is $\sqrt{1 + k^2 f_x^2 + k^2 f_y^2} = \sqrt{1 + k^2 (f_x^2+f_y^2)}$. This is not a simple multiple of the original integrand $\sqrt{1+f_x^2+f_y^2}$. The scaling factor depends on the magnitude of the gradient. For example, if $f_x=f_y=0$, the area stays the same. If the gradient is large, the area increases roughly by factor $k$. Therefore, the scaling is non-linear and depends on the function's derivatives. Options A and B are false because they assume linearity. Option D is incorrect because changing $f$ changes the surface.

📖 Explanation: A vertical translation of a surface does not change its partial derivatives, i.e., $f_x$ and $f_y$ remain the same. The surface area integrand $\sqrt{1+f_x^2+f_y^2}$ is unchanged. Therefore, the surface area over the same region $R$ is identical. Options A and B mistakenly treat area as a linear function of height. Option C is a common physical intuition that is mathematically incorrect; shifting a plane vertically does not change its slope or area.

📖 Explanation: The surface area integral computes the area of the curved surface (the fabric). The tent's fabric is the graph of $z=f(x,y)$. Option A is incorrect; volume is $\iint f \, dA$. Option C is incorrect; the shadow is the area of $R$. Option D is incorrect; average height is $\frac{1}{\text{Area}(R)} \iint f \, dA$. This question connects the mathematical formula to a real-world context, reinforcing the interpretation of the surface area as the 'material' needed to cover the 3D shape.

📖 Explanation: For $f(x,y) = x^2 y$, we have $f_x = 2xy$ and $f_y = x^2$. Then $1+f_x^2+f_y^2 = 1 + (2xy)^2 + (x^2)^2 = 1 + 4x^2 y^2 + x^4$. The student's integral has the correct integrand and the correct limits for a triangular region $0 \le x \le 1, 0 \le y \le 1-x$. The setup is completely correct. Option A is false because the limits are correct. Option B is false because no factor of 2 is missing. Option C is false because $f_y = x^2$ is correct. This question tests the ability to identify a correct setup amid plausible errors.