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Surface Integrals of Scalar Fields - Calculus 3

  1. Lateral surface of cylinder x2x^2+y2y^2=R2R^2, 0≤z≤h; find ∬_S x2x^2 dS.
    Answer: πR^3h On the cylinder, use x = R cosθ and surface element dS = R dθ dz. Then x2x^2 dS = R3R^3 cos⁡2\cos^2θ dθ dz. Integrate θ from 0 to 2π and z from 0 to h. The θ integral of cos⁡2\cos^2θ is π, so the result is R3R^3 h π = πR^3h. The tempting wrong value 2πR^3h comes from treating x2x^2 as R2R^2 everywhere, but cos⁡2\cos^2θ averages to 1/2, not 1.
  2. Sphere x2x^2+y2y^2+z2z^2=a2a^2; compute ∬_S z2z^2 dS.
    Answer: 4πa4a^4/3 On the sphere, x2x^2 + y2y^2 + z2z^2 = a2a^2, so the integral of this sum over the surface is a2a^2 times the surface area 4πa2a^2, giving 4πa4a^4. By symmetry, the integrals of x2x^2, y2y^2, and z2z^2 are equal, so each is one third of that total: 4πa4a^4/3. The tempting wrong value 4πa4a^4 treats z2z^2 as constant a2a^2 everywhere, which ignores the symmetry division.
  3. Consider the surface SS parameterized by r⃗(u,v)=⟨ucos⁡v,usin⁡v,v⟩\vec{r}(u,v) = \langle u\cos v, u\sin v, v \rangle where 1≤u≤21 \leq u \leq 2 and 0≤v≤π0 \leq v \leq \pi. The value of ∬S(x2+y2) dS\iint_S (x^2 + y^2) \, dS is:
    Answer: π3(152+7)\frac{\pi}{3}(15\sqrt{2} + 7) When you encounter a surface integral like this, you're computing the integral of a function over a curved surface. The key is recognizing that you need to find the surface area element dSdS and express everything in terms of the parameters uu and vv. First, find the partial derivatives: r⃗u=⟨cos⁡v,sin⁡v,0⟩\vec{r}_u = \langle \cos v, \sin v, 0 \rangle and r⃗v=⟨−usin⁡v,ucos⁡v,1⟩\vec{r}_v = \langle -u\sin v, u\cos v, 1 \rangle. The cross product gives r⃗u×r⃗v=⟨sin⁡v,−cos⁡v,u⟩\vec{r}_u \times \vec{r}_v = \langle \sin v, -\cos v, u \rangle, so ∣r⃗u×r⃗v∣=sin⁡2v+cos⁡2v+u2=1+u2|\vec{r}_u \times \vec{r}_v| = \sqrt{\sin^2 v + \cos^2 v + u^2} = \sqrt{1 + u^2}. Therefore, dS=1+u2 du dvdS = \sqrt{1 + u^2} \, du \, dv. Next, express x2+y2x^2 + y^2 in terms of parameters. Since x=ucos⁡vx = u\cos v and y=usin⁡vy = u\sin v, we have x2+y2=u2x^2 + y^2 = u^2. The integral becomes: ∬Su2 dS=∫0π∫12u21+u2 du dv\iint_S u^2 \, dS = \int_0^\pi \int_1^2 u^2 \sqrt{1 + u^2} \, du \, dv The vv-integral gives π\pi. For the uu-integral, use substitution u=tan⁡θu = \tan \theta or integration by parts to get ∫12u21+u2 du=13(152+7)\int_1^2 u^2\sqrt{1 + u^2} \, du = \frac{1}{3}(15\sqrt{2} + 7). The final answer is π⋅13(152+7)=π3(152+7)\pi \cdot \frac{1}{3}(15\sqrt{2} + 7) = \frac{\pi}{3}(15\sqrt{2} + 7), which is choice B. Choice A has the wrong coefficient π6\frac{\pi}{6}, likely from incorrect integration. Choice C reverses the terms inside the parentheses. Choice D doubles the coefficient to 2π3\frac{2\pi}{3}, possibly from integrating over 2π2\pi instead of π\pi. Remember: always verify your surface area element calculation and double-check parameter substitutions—these are the most common error sources in surface integrals.
  4. The surface SS is the part of the cylinder x2+y2=4x^2 + y^2 = 4 between the planes z=0z = 0 and z=3z = 3. If f(x,y,z)=zx2+y2f(x,y,z) = z\sqrt{x^2 + y^2}, then ∬Sf dS\iint_S f \, dS equals:
    Answer: 36π36\pi When evaluating surface integrals over cylindrical surfaces, you'll typically use cylindrical coordinates and parametrize the surface to find the surface area element dSdS. For the cylinder x2+y2=4x^2 + y^2 = 4, we can parametrize using r(θ,z)=(2cos⁡θ,2sin⁡θ,z)\mathbf{r}(\theta, z) = (2\cos\theta, 2\sin\theta, z) where 0≤θ≤2π0 \leq \theta \leq 2\pi and 0≤z≤30 \leq z \leq 3. The partial derivatives are rθ=(−2sin⁡θ,2cos⁡θ,0)\mathbf{r}_\theta = (-2\sin\theta, 2\cos\theta, 0) and rz=(0,0,1)\mathbf{r}_z = (0, 0, 1). Their cross product gives rθ×rz=(2cos⁡θ,2sin⁡θ,0)\mathbf{r}_\theta \times \mathbf{r}_z = (2\cos\theta, 2\sin\theta, 0), so ∣rθ×rz∣=2|\mathbf{r}_\theta \times \mathbf{r}_z| = 2, meaning dS=2 dθ dzdS = 2\,d\theta\,dz. On the cylinder surface, x2+y2=2\sqrt{x^2 + y^2} = 2, so f(x,y,z)=zx2+y2=2zf(x,y,z) = z\sqrt{x^2 + y^2} = 2z. The integral becomes: ∬Sf dS=∫02π∫032z⋅2 dz dθ=4∫02π∫03z dz dθ\iint_S f \, dS = \int_0^{2\pi} \int_0^3 2z \cdot 2 \, dz \, d\theta = 4 \int_0^{2\pi} \int_0^3 z \, dz \, d\theta Evaluating: 4∫02π[z22]03dθ=4∫02π92dθ=18∫02πdθ=36π4 \int_0^{2\pi} \left[\frac{z^2}{2}\right]_0^3 d\theta = 4 \int_0^{2\pi} \frac{9}{2} d\theta = 18 \int_0^{2\pi} d\theta = 36\pi Choice A (72π72\pi) likely comes from forgetting to take the square root when computing ∣rθ×rz∣|\mathbf{r}_\theta \times \mathbf{r}_z|. Choice C (18π18\pi) results from using dS=dθ dzdS = d\theta\,dz instead of dS=2 dθ dzdS = 2\,d\theta\,dz. Choice D (54π54\pi) might arise from incorrectly computing the zz-integral as z3/3z^3/3 instead of z2/2z^2/2. Remember: always carefully compute the surface area element dSdS by finding the magnitude of the cross product of parametric partial derivatives—this is where most errors occur in surface integrals.
  5. The surface SS consists of the part of the paraboloid z=x2+y2z = x^2 + y^2 that lies below the plane z=4z = 4. If we evaluate ∬Sx2+y2 dS\iint_S \sqrt{x^2 + y^2} \, dS using cylindrical coordinates, the integral becomes:
    Answer: ∫02π∫02r21+4r2 dr dθ\int_0^{2\pi} \int_0^2 r^2\sqrt{1 + 4r^2} \, dr \, d\theta When evaluating surface integrals over parametric surfaces, you need to master three key components: setting up the correct coordinate system, finding the surface area element dSdS, and properly transforming the integrand. For the paraboloid z=x2+y2z = x^2 + y^2 below z=4z = 4, cylindrical coordinates are ideal. Converting to cylindrical: x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta, so z=r2z = r^2. The constraint z≤4z \leq 4 becomes r2≤4r^2 \leq 4, giving us 0≤r≤20 \leq r \leq 2 and 0≤θ≤2π0 \leq \theta \leq 2\pi. To find dSdS, parametrize the surface as r⃗(r,θ)=(rcos⁡θ,rsin⁡θ,r2)\vec{r}(r,\theta) = (r\cos\theta, r\sin\theta, r^2). Computing the partial derivatives and their cross product gives ∣r⃗r×r⃗θ∣=r1+4r2|\vec{r}_r \times \vec{r}_\theta| = r\sqrt{1 + 4r^2}, so dS=r1+4r2 dr dθdS = r\sqrt{1 + 4r^2} \, dr \, d\theta. The integrand x2+y2\sqrt{x^2 + y^2} becomes r2cos⁡2θ+r2sin⁡2θ=r\sqrt{r^2\cos^2\theta + r^2\sin^2\theta} = r. Therefore: ∬Sx2+y2 dS=∫02π∫02r⋅r1+4r2 dr dθ=∫02π∫02r21+4r2 dr dθ\iint_S \sqrt{x^2 + y^2} \, dS = \int_0^{2\pi} \int_0^2 r \cdot r\sqrt{1 + 4r^2} \, dr \, d\theta = \int_0^{2\pi} \int_0^2 r^2\sqrt{1 + 4r^2} \, dr \, d\theta This confirms answer D. A and B both miss the factor of rr from the integrand transformation. B also uses the wrong upper limit r=4r = 4 instead of r=2r = 2. C incorrectly writes 4r2+1\sqrt{4r^2 + 1} instead of 1+4r2\sqrt{1 + 4r^2} in the surface element. Strategy tip: Always transform both the integrand AND the surface element when changing coordinates. Double-check your limits by substituting back into the original constraint.
  6. For S: z=x+y, x,y∈[0,1]; find ∬_S x2x^2 dS.
    Answer: √3/3 Since z=x+y, the partial derivatives are 1 and 1, so dS = sqrt(1+1+1) dx dy = sqrt(3) dx dy. Thus the integral is sqrt(3) times the integral of x2x^2 over the unit square: sqrt(3) (1/3) 1 = sqrt(3)/3. The tempting 1/3 comes from integrating x2x^2 but forgetting the sqrt(3) surface-area factor.
  7. S is plane x+y+z=1 in first octant; find ∬_S xy dS.
    Answer: √3/24 Parameterize z=1-x-y over the triangle x≥0, y≥0, x+y≤1. The surface element is √(1+1+1)=√3 times dx dy. Integrate xy: ∫010^1 ∫01−x0^{1-x} xy dy dx = 1/24. Multiply by √3 to get √3/24. The tempting error is omitting the √3 from dS, which leaves 1/24.
  8. Consider the hemisphere S:x2+y2+z2=9S: x^2 + y^2 + z^2 = 9 with z≥0z \geq 0. The surface integral ∬S1x2+y2+z2 dS\iint_S \frac{1}{x^2 + y^2 + z^2} \, dS is equal to:
    Answer: 2π2\pi On the hemisphere, x2+y2+z2=9x^2 + y^2 + z^2 = 9, so f(x,y,z)=1x2+y2+z2=19f(x,y,z) = \frac{1}{x^2 + y^2 + z^2} = \frac{1}{9}. Using spherical coordinates with ρ=3\rho = 3, 0≤ϕ≤π20 \leq \phi \leq \frac{\pi}{2}, 0≤θ≤2π0 \leq \theta \leq 2\pi, the surface element is dS=ρ2sin⁡ϕ dϕ dθ=9sin⁡ϕ dϕ dθdS = \rho^2 \sin\phi \, d\phi \, d\theta = 9\sin\phi \, d\phi \, d\theta. The integral becomes ∫02π∫0π/219⋅9sin⁡ϕ dϕ dθ=∫02π∫0π/2sin⁡ϕ dϕ dθ=2π⋅1=2π\int_0^{2\pi} \int_0^{\pi/2} \frac{1}{9} \cdot 9\sin\phi \, d\phi \, d\theta = \int_0^{2\pi} \int_0^{\pi/2} \sin\phi \, d\phi \, d\theta = 2\pi \cdot 1 = 2\pi. Choice A incorrectly uses the radius 3 as a factor. Choice B multiplies by both 9 and 2. Choice D uses only the radius without proper integration.
  9. The surface SS is given by z=x2+y2z = \sqrt{x^2 + y^2} for 1≤x2+y2≤41 \leq x^2 + y^2 \leq 4. What is ∬Sz dS\iint_S z \, dS?
    Answer: 14π23\frac{14\pi\sqrt{2}}{3} Using cylindrical coordinates, x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta, z=rz = r where 1≤r≤21 \leq r \leq 2 and 0≤θ≤2π0 \leq \theta \leq 2\pi. We have ∂z∂x=xx2+y2=xr\frac{\partial z}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} = \frac{x}{r} and ∂z∂y=yx2+y2=yr\frac{\partial z}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} = \frac{y}{r}. Thus dS=1+x2r2+y2r2 dx dy=2 dx dy=2r dr dθdS = \sqrt{1 + \frac{x^2}{r^2} + \frac{y^2}{r^2}} \, dx \, dy = \sqrt{2} \, dx \, dy = \sqrt{2} r \, dr \, d\theta. The integral becomes ∫02π∫12r⋅2r dr dθ=2π2∫12r2 dr=2π2⋅73=14π23\int_0^{2\pi} \int_1^2 r \cdot \sqrt{2} r \, dr \, d\theta = 2\pi\sqrt{2} \int_1^2 r^2 \, dr = 2\pi\sqrt{2} \cdot \frac{7}{3} = \frac{14\pi\sqrt{2}}{3}. Choice A omits the 13\frac{1}{3} factor. Choice B is missing 2\sqrt{2}. Choice D has an error in evaluating ∫12r2 dr\int_1^2 r^2 \, dr.
  10. Cone z=√(x2+y2x^2+y^2), 0≤z≤1; find ∬_S z dS.
    Answer: 2π√2/3 Parametrize the cone with radius r and angle θ. The surface element is r√2 dr dθ, and z = r. Integrate r2r^2√2 from r=0 to1 and θ=0 to2π, giving 2π√2/3. The tempting error is 2π/3, which comes from using the projected area element without the slant factor √2.
  11. Consider the surface SS that is the part of the plane 2x+3y+z=62x + 3y + z = 6 that lies in the first octant. If we want to evaluate ∬S(x+y+z) dS\iint_S (x + y + z) \, dS, which of the following represents the correct setup?
    Answer: ∫03∫02−2x3(x+y+6−2x−3y)14 dy dx\int_0^3 \int_0^{2-\frac{2x}{3}} (x + y + 6 - 2x - 3y)\sqrt{14} \, dy \, dx The surface is z=6−2x−3yz = 6 - 2x - 3y. In the first octant, we need x≥0x \geq 0, y≥0y \geq 0, and z≥0z \geq 0. From z≥0z \geq 0: 6−2x−3y≥06 - 2x - 3y \geq 0, so y≤2−2x3y \leq 2 - \frac{2x}{3}. The x-intercept is 3 (when y=z=0y = z = 0), so 0≤x≤30 \leq x \leq 3. For each xx, 0≤y≤2−2x30 \leq y \leq 2 - \frac{2x}{3}. The surface element is dS=1+(∂z∂x)2+(∂z∂y)2 dx dy=1+4+9 dx dy=14 dx dydS = \sqrt{1 + (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2} \, dx \, dy = \sqrt{1 + 4 + 9} \, dx \, dy = \sqrt{14} \, dx \, dy. Choice B has incorrect integration bounds. Choice C uses 13\sqrt{13} instead of 14\sqrt{14}. Choice D extends x beyond the first octant boundary.
  12. A surface SS is parameterized by r⃗(u,v)=⟨ucos⁡v,usin⁡v,u2⟩\vec{r}(u,v) = \langle u\cos v, u\sin v, u^2 \rangle where 0≤u≤20 \leq u \leq 2 and 0≤v≤π0 \leq v \leq \pi. If f(x,y,z)=x2+y2f(x,y,z) = x^2 + y^2, what is the value of ∬Sf dS\iint_S f \, dS?
    Answer: 8π3(1+417)\frac{8\pi}{3}(1 + 4\sqrt{17}) First, we compute r⃗u=⟨cos⁡v,sin⁡v,2u⟩\vec{r}_u = \langle \cos v, \sin v, 2u \rangle and r⃗v=⟨−usin⁡v,ucos⁡v,0⟩\vec{r}_v = \langle -u\sin v, u\cos v, 0 \rangle. The cross product is r⃗u×r⃗v=⟨−2u2cos⁡v,−2u2sin⁡v,u⟩\vec{r}_u \times \vec{r}_v = \langle -2u^2\cos v, -2u^2\sin v, u \rangle, so ∣∣r⃗u×r⃗v∣∣=u4u2+1||\vec{r}_u \times \vec{r}_v|| = u\sqrt{4u^2 + 1}. Since f(x,y,z)=x2+y2=u2f(x,y,z) = x^2 + y^2 = u^2, we have ∬Sf dS=∫0π∫02u2⋅u4u2+1 du dv=π∫02u34u2+1 du\iint_S f \, dS = \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=4u2+1w = 4u^2 + 1, this evaluates to 8π3(1+417)\frac{8\pi}{3}(1 + 4\sqrt{17}). Choice B has incorrect coefficients from algebraic errors. Choice C is missing a term from incomplete integration. Choice D has an error in the cross product calculation.

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