Surface Integrals: Scalar Fields - Multivariable Calculus

1. The surface $S$ is the portion of the cone $z = \sqrt{x^2 + y^2}$ between the planes $z = 1$ and $z = 3$. For the scalar field $w(x,y,z) = \frac{1}{z}$, the value of $\iint_S w \, dS$ is:
   Answer: $4\pi\sqrt{2}$ When you encounter a surface integral over a portion of a cone, you're dealing with parametric surfaces and need to carefully set up both the parameterization and the surface area element. The cone $z = \sqrt{x^2 + y^2}$ between $z = 1$ and $z = 3$ is best parameterized using cylindrical coordinates. Since $z = r$ on this cone, you can write: $\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, r)$ where $1 \leq r \leq 3$ and $0 \leq \theta \leq 2\pi$. To find $dS$, compute the cross product of partial derivatives: $\mathbf{r}_r = (\cos\theta, \sin\theta, 1)$ and $\mathbf{r}_\theta = (-r\sin\theta, r\cos\theta, 0)$. The magnitude $|\mathbf{r}_r \times \mathbf{r}_\theta| = r\sqrt{2}$, so $dS = r\sqrt{2} \, dr \, d\theta$. The integral becomes: $\iint_S \frac{1}{z} \, dS = \int_0^{2\pi} \int_1^3 \frac{1}{r} \cdot r\sqrt{2} \, dr \, d\theta = \int_0^{2\pi} \int_1^3 \sqrt{2} \, dr \, d\theta = \sqrt{2} \cdot 2\pi \cdot (3-1) = 4\pi\sqrt{2}$ Answer A ($4\pi$) results from forgetting the $\sqrt{2}$ factor in the surface area element. Answer B ($2\pi\sqrt{2}$) comes from integrating over only half the cone ($0 \leq \theta \leq \pi$). Answer D ($8\pi$) represents both missing the $\sqrt{2}$ factor and doubling the height incorrectly. Study tip: For cone surface integrals, always remember that the surface area element includes a $\sqrt{2}$ factor due to the cone's slope, and double-check your parameter bounds.
2. The surface $S$ is defined by $z = xy$ over the triangular region $R$ in the $xy$-plane with vertices at $(0,0)$, $(2,0)$, and $(0,3)$. If $h(x,y,z) = z^2 + 1$, what is the value of $\iint_S h \, dS$?
   Answer: $\int_0^2 \int_0^{3(1-\frac{x}{2})} (x^2y^2 + 1)\sqrt{x^2 + y^2 + 1} \, dy \, dx$ For the surface $z = f(x,y) = xy$, we have $f_x = y$ and $f_y = x$. The surface element is $dS = \sqrt{1 + f_x^2 + f_y^2} \, dx \, dy = \sqrt{1 + y^2 + x^2} \, dx \, dy$. On the surface, $h = z^2 + 1 = x^2y^2 + 1$. The triangular region has the line from $(2,0)$ to $(0,3)$ given by $\frac{x}{2} + \frac{y}{3} = 1$, so $y = 3(1 - \frac{x}{2})$. Thus the correct bounds are $0 \leq x \leq 2$ and $0 \leq y \leq 3(1-\frac{x}{2})$. Choice A has incorrect upper bound for $y$. Choice B uses wrong order of integration with incorrect bounds. Choice D has wrong bounds for $x$.
3. A surface $S$ is given by the equation $x^2 + y^2 - z^2 = 1$ with $z \geq 0$ and $z \leq 2$. Using the parameterization $x = \sec u \cos v$, $y = \sec u \sin v$, $z = \tan u$ where $0 \leq u \leq \arctan 2$ and $0 \leq v \leq 2\pi$, the surface element $dS$ equals:
   Answer: $\sec^3 u \, du \, dv$ When calculating surface area for a parameterized surface, you need to find the surface element $dS = |\mathbf{r}_u \times \mathbf{r}_v| \, du \, dv$, where $\mathbf{r}(u,v) = \langle \sec u \cos v, \sec u \sin v, \tan u \rangle$. First, compute the partial derivatives: $\mathbf{r}_u = \langle \sec u \tan u \cos v, \sec u \tan u \sin v, \sec^2 u \rangle$ $\mathbf{r}_v = \langle -\sec u \sin v, \sec u \cos v, 0 \rangle$ The cross product $\mathbf{r}_u \times \mathbf{r}_v$ gives: $\mathbf{r}_u \times \mathbf{r}_v = \langle -\sec^3 u \cos v, -\sec^3 u \sin v, \sec^2 u \tan u \rangle$ To find the magnitude: $|\mathbf{r}_u \times \mathbf{r}_v|^2 = \sec^6 u \cos^2 v + \sec^6 u \sin^2 v + \sec^4 u \tan^2 u$ $= \sec^6 u(\cos^2 v + \sin^2 v) + \sec^4 u \tan^2 u = \sec^6 u + \sec^4 u \tan^2 u$ $= \sec^4 u(\sec^2 u + \tan^2 u) = \sec^4 u(2\sec^2 u - 1) = \sec^6 u$ Therefore, $|\mathbf{r}_u \times \mathbf{r}_v| = \sec^3 u$, making $dS = \sec^3 u \, du \, dv$. Answer choice A ($\sec u$) misses the squared terms from differentiation. Choice B ($\sec^2 u$) would result from forgetting the cross product magnitude calculation. Choice C ($\sec^2 u \tan u$) incorrectly includes only one component of the cross product. Study tip: For surface integrals, always compute both partial derivatives carefully, then find the cross product magnitude. The trigonometric identity $\sec^2 u = 1 + \tan^2 u$ is often crucial in simplifying these calculations.
4. A surface $S$ is parameterized by $\mathbf{r}(u,v) = (u\cos v, u\sin v, u^2)$ for $0 \leq u \leq 2$ and $0 \leq v \leq \pi$. If $f(x,y,z) = x^2 + y^2$, what is the value of the surface integral $\iint_S f \, dS$?
   Answer: $\frac{\pi}{6}(17\sqrt{17} - 1)$ We have $\mathbf{r}_u = (\cos v, \sin v, 2u)$ and $\mathbf{r}_v = (-u\sin v, u\cos v, 0)$. Computing the cross product: $\mathbf{r}_u \times \mathbf{r}_v = (-2u^2\cos v, -2u^2\sin v, u)$. Thus $|\mathbf{r}_u \times \mathbf{r}_v| = u\sqrt{4u^2 + 1}$. On the surface, $f(x,y,z) = u^2$. The integral becomes $\int_0^\pi \int_0^2 u^2 \cdot u\sqrt{4u^2 + 1} \, du \, dv = \pi \int_0^2 u^3\sqrt{4u^2 + 1} \, du$. Using substitution $w = 4u^2 + 1$, this evaluates to $\frac{\pi}{6}(17\sqrt{17} - 1)$. Choice A has the wrong coefficient (too large by factor of 4). Choice C uses the wrong upper limit in evaluation. Choice D has the wrong coefficient (too large by factor of 2).
5. A surface $S$ is the portion of the sphere $x^2 + y^2 + z^2 = 9$ that lies above the plane $z = \frac{3}{2}$. For the function $k(x,y,z) = z$, the surface integral $\iint_S k \, dS$ equals:
   Answer: $\frac{27\pi}{4}$ Using spherical coordinates: $x = 3\sin\phi\cos\theta$, $y = 3\sin\phi\sin\theta$, $z = 3\cos\phi$. The condition $z \geq \frac{3}{2}$ means $3\cos\phi \geq \frac{3}{2}$, so $\cos\phi \geq \frac{1}{2}$, giving $0 \leq \phi \leq \frac{\pi}{3}$. The surface element is $dS = 9\sin\phi \, d\phi \, d\theta$. The integral becomes $\int_0^{2\pi} \int_0^{\pi/3} 3\cos\phi \cdot 9\sin\phi \, d\phi \, d\theta = 27 \cdot 2\pi \int_0^{\pi/3} \cos\phi\sin\phi \, d\phi = 27\pi \int_0^{\pi/3} \frac{1}{2}\sin(2\phi) \, d\phi = \frac{27\pi}{4}[-\cos(2\phi)]_0^{\pi/3} = \frac{27\pi}{4}$. Choice B doubles the result incorrectly. Choice C omits the factor of $\frac{1}{2}$. Choice D uses wrong radius.
6. Consider the surface $S$ that is the portion of the cylinder $x^2 + z^2 = 4$ with $0 \leq y \leq 3$ and $z \geq 0$. For the scalar field $g(x,y,z) = yz$, which expression correctly represents the surface integral $\iint_S g \, dS$?
   Answer: $\int_0^3 \int_0^\pi 2y\sin\theta \cdot 2 \, d\theta \, dy$ Using cylindrical coordinates with $x = 2\cos\theta$, $z = 2\sin\theta$ for $0 \leq \theta \leq \pi$ and $0 \leq y \leq 3$. The parameterization is $\mathbf{r}(\theta,y) = (2\cos\theta, y, 2\sin\theta)$. We get $\mathbf{r}_\theta = (-2\sin\theta, 0, 2\cos\theta)$ and $\mathbf{r}_y = (0, 1, 0)$. The magnitude $|\mathbf{r}_\theta \times \mathbf{r}_y| = 2$. On the surface, $g = yz = y \cdot 2\sin\theta$. The integral becomes $\int_0^3 \int_0^\pi 2y\sin\theta \cdot 2 \, d\theta \, dy$. Choice A uses the wrong surface element calculation. Choice C omits the surface element entirely. Choice D has the correct integrand but wrong bounds and missing factor of 2.
7. Consider the surface $S$ parameterized by $\mathbf{r}(s,t) = (s + t, s - t, s^2 + t^2)$ where $0 \leq s \leq 1$ and $0 \leq t \leq 1$. If $m(x,y,z) = x + y + z$, which of the following is closest to the numerical value of $\iint_S m \, dS$?
   Answer: $2.8$ We compute $\mathbf{r}_s = (1, 1, 2s)$ and $\mathbf{r}_t = (1, -1, 2t)$. The cross product is $\mathbf{r}_s \times \mathbf{r}_t = (2t + 2s, -2t + 2s, -2)$, so $|\mathbf{r}_s \times \mathbf{r}_t| = 2\sqrt{(t+s)^2 + (s-t)^2 + 1} = 2\sqrt{2s^2 + 2t^2 + 1}$. On the surface, $m = (s+t) + (s-t) + (s^2+t^2) = 2s + s^2 + t^2$. The integral is $\int_0^1 \int_0^1 (2s + s^2 + t^2) \cdot 2\sqrt{2s^2 + 2t^2 + 1} \, ds \, dt$. This evaluates numerically to approximately $2.83$. Choice A underestimates by neglecting the square root factor properly. Choice C overestimates by incorrect calculation of the magnitude. Choice D significantly overestimates the integral value.
8. The surface $S$ is the part of the paraboloid $z = x^2 + y^2$ that lies inside the cylinder $x^2 + y^2 = 4$. For the scalar field $p(x,y,z) = \sqrt{x^2 + y^2}$, the surface integral $\iint_S p \, dS$ can be expressed as:
   Answer: $\int_0^{2\pi} \int_0^2 r^2\sqrt{1 + 4r^2} \, dr \, d\theta$ Using cylindrical coordinates $x = r\cos\theta$, $y = r\sin\theta$, $z = r^2$. For the surface $z = f(x,y) = x^2 + y^2$, we have $f_x = 2x$ and $f_y = 2y$, so $dS = \sqrt{1 + 4x^2 + 4y^2} \, dx \, dy = \sqrt{1 + 4r^2} \, r \, dr \, d\theta$. The function $p(x,y,z) = \sqrt{x^2 + y^2} = r$. The surface integral becomes $\int_0^{2\pi} \int_0^2 r \cdot \sqrt{1 + 4r^2} \cdot r \, dr \, d\theta = \int_0^{2\pi} \int_0^2 r^2\sqrt{1 + 4r^2} \, dr \, d\theta$. Choice B omits one factor of $r$. Choice C has the correct terms but omits the Jacobian factor. Choice D omits the function $p$ entirely.
9. A surface $S$ is parameterized by $\mathbf{r}(u,v) = (u, v, u^2 - v^2)$ where $u^2 + v^2 \leq 1$. If $\iint_S (x^2 + y^2) \, dS = \alpha \iint_D (u^2 + v^2)\sqrt{4u^2 + 4v^2 + 1} \, du \, dv$ where $D$ is the unit disk, then $\alpha$ equals:
   Answer: $1$ When you encounter a surface integral problem with a parameterization, you need to convert the surface integral into a double integral over the parameter domain using the formula $\iint_S f(x,y,z) \, dS = \iint_D f(\mathbf{r}(u,v)) \|\mathbf{r}_u \times \mathbf{r}_v\| \, du \, dv$. First, let's find the cross product magnitude. With $\mathbf{r}(u,v) = (u, v, u^2 - v^2)$, we have $\mathbf{r}_u = (1, 0, 2u)$ and $\mathbf{r}_v = (0, 1, -2v)$. Computing the cross product: $\mathbf{r}_u \times \mathbf{r}_v = (-2u, 2v, 1)$, so $\|\mathbf{r}_u \times \mathbf{r}_v\| = \sqrt{4u^2 + 4v^2 + 1}$. Next, substitute the parameterization into the integrand. Since $x = u$ and $y = v$, we have $x^2 + y^2 = u^2 + v^2$. Therefore: $\iint_S (x^2 + y^2) \, dS = \iint_D (u^2 + v^2)\sqrt{4u^2 + 4v^2 + 1} \, du \, dv$ Comparing this with the given form $\alpha \iint_D (u^2 + v^2)\sqrt{4u^2 + 4v^2 + 1} \, du \, dv$, we see that $\alpha = 1$, which is choice (D). Choice (A) $\frac{1}{2}$ might come from incorrectly halving something in the cross product calculation. Choice (B) $4$ could result from mistakenly using $4u^2 + 4v^2$ instead of the correct magnitude. Choice (C) $2$ might arise from errors in computing the cross product components. Strategy tip: Always work systematically through parameterized surface integrals: compute partial derivatives, find the cross product magnitude, then substitute the parameterization into your integrand. Double-check that your final integral matches the given form exactly.
10. Consider the surface $S$ that is the graph of $z = \sin(xy)$ over the square $[0,\pi] \times [0,1]$ in the $xy$-plane. For the function $q(x,y,z) = \cos(xy)$, which statement about the surface integral $\iint_S q \, dS$ is correct?
    Answer: The integral equals $\int_0^\pi \int_0^1 \cos(xy)\sqrt{1 + y^2\cos^2(xy) + x^2\cos^2(xy)} \, dy \, dx$ For the surface $z = f(x,y) = \sin(xy)$, we have $f_x = y\cos(xy)$ and $f_y = x\cos(xy)$. The surface element is $dS = \sqrt{1 + f_x^2 + f_y^2} \, dx \, dy = \sqrt{1 + y^2\cos^2(xy) + x^2\cos^2(xy)} \, dx \, dy$. On the surface, $q(x,y,z) = \cos(xy)$. The integral is $\int_0^\pi \int_0^1 \cos(xy)\sqrt{1 + y^2\cos^2(xy) + x^2\cos^2(xy)} \, dy \, dx$. Choice B incorrectly ignores the surface element. Choice C uses wrong partial derivatives (sine instead of cosine). Choice D uses sine instead of cosine in the surface element.

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