Multivariable Calculus Quiz: Surface Parameterization And Area
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Surface Parameterization And AreaQuestion 1 of 17

A cone is parametrized by r(u,v)=(ucos⁡v,usin⁡v,2u)\mathbf{r}(u,v) = (u\cos v, u\sin v, 2u) where 1≤u≤31 \leq u \leq 3 and 0≤v≤2π0 \leq v \leq 2\pi. What is the total surface area of this portion of the cone?

16π516\pi\sqrt{5}
4π54\pi\sqrt{5}
12π512\pi\sqrt{5}
8π58\pi\sqrt{5}
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Surface Parameterization And Area

Practice Surface Parameterization And Area in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Surface Parameterization And Area, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A cone is parametrized by r(u,v)=(ucos⁡v,usin⁡v,2u)\mathbf{r}(u,v) = (u\cos v, u\sin v, 2u) where 1≤u≤31 \leq u \leq 3 and 0≤v≤2π0 \leq v \leq 2\pi. What is the total surface area of this portion of the cone?

  1. 16π516\pi\sqrt{5}
  2. 4π54\pi\sqrt{5}
  3. 12π512\pi\sqrt{5}
  4. 8π58\pi\sqrt{5} (correct answer)
Explanation: When you encounter a parametric surface and need to find its area, you're working with the fundamental formula: surface area equals the double integral of the magnitude of the cross product of the partial derivatives of the position vector. Start by finding the partial derivatives of r(u,v)=(ucos⁡v,usin⁡v,2u)\mathbf{r}(u,v) = (u\cos v, u\sin v, 2u). You get ru=(cos⁡v,sin⁡v,2)\mathbf{r}_u = (\cos v, \sin v, 2) and rv=(−usin⁡v,ucos⁡v,0)\mathbf{r}_v = (-u\sin v, u\cos v, 0). Next, compute their cross product: ru×rv=(−2ucos⁡v,−2usin⁡v,u)\mathbf{r}_u \times \mathbf{r}_v = (-2u\cos v, -2u\sin v, u). The magnitude is ∣ru×rv∣=4u2cos⁡2v+4u2sin⁡2v+u2=4u2+u2=u5|\mathbf{r}_u \times \mathbf{r}_v| = \sqrt{4u^2\cos^2 v + 4u^2\sin^2 v + u^2} = \sqrt{4u^2 + u^2} = u\sqrt{5}. The surface area integral becomes ∫02π∫13u5 du dv=5∫02π∫13u du dv=5⋅2π⋅[u22]13=5⋅2π⋅82=8π5\int_0^{2\pi} \int_1^3 u\sqrt{5} \, du \, dv = \sqrt{5} \int_0^{2\pi} \int_1^3 u \, du \, dv = \sqrt{5} \cdot 2\pi \cdot \left[\frac{u^2}{2}\right]_1^3 = \sqrt{5} \cdot 2\pi \cdot \frac{8}{2} = 8\pi\sqrt{5}. Choice A (16π516\pi\sqrt{5}) likely comes from incorrectly calculating ∫13u du\int_1^3 u \, du as 8 instead of 4. Choice B (4π54\pi\sqrt{5}) probably results from using 2u2u instead of u5u\sqrt{5} as the integrand. Choice C (12π512\pi\sqrt{5}) might stem from computational errors in the cross product or integral evaluation. Remember: for parametric surfaces, always compute ∣ru×rv∣|\mathbf{r}_u \times \mathbf{r}_v| carefully—the cross product magnitude often involves recognizing perfect squares under the radical.

Question 2

Let SS be the portion of the cylinder x2+z2=9x^2+z^2=9 bounded by the planes y=0y=0 and y=4−xy=4-x. The area of SS can be found by evaluating ∬Dg(θ,y) dy dθ\iint_D g(\theta,y) \,dy \,d\theta for some function gg and domain DD. Using the parameterization x=3cos⁡θ,z=3sin⁡θx=3\cos\theta, z=3\sin\theta, which integral represents the area of SS?

  1. ∫02π∫04−3cos⁡θ3 dy dθ\int_0^{2\pi} \int_0^{4-3\cos\theta} 3 \,dy \,d\theta (correct answer)
  2. ∫−π/2π/2∫04−3cos⁡θ3 dy dθ\int_{-\pi/2}^{\pi/2} \int_0^{4-3\cos\theta} 3 \,dy \,d\theta
  3. ∫02π∫04−3cos⁡θ9 dy dθ\int_0^{2\pi} \int_0^{4-3\cos\theta} 9 \,dy \,d\theta
  4. ∫−π/2π/2∫043 dy dθ\int_{-\pi/2}^{\pi/2} \int_0^{4} 3 \,dy \,d\theta
Explanation: First, we find the surface area element dSdS. The parameterization is r⃗(θ,y)=⟨3cos⁡θ,y,3sin⁡θ⟩\vec{r}(\theta, y) = \langle 3\cos\theta, y, 3\sin\theta \rangle. The partial derivatives are r⃗θ=⟨−3sin⁡θ,0,3cos⁡θ⟩\vec{r}_\theta = \langle -3\sin\theta, 0, 3\cos\theta \rangle and r⃗y=⟨0,1,0⟩\vec{r}_y = \langle 0, 1, 0 \rangle. Their cross product is r⃗θ×r⃗y=⟨−3cos⁡θ,0,−3sin⁡θ⟩\vec{r}_\theta \times \vec{r}_y = \langle -3\cos\theta, 0, -3\sin\theta \rangle. The magnitude is ∣∣r⃗θ×r⃗y∣∣=9cos⁡2θ+9sin⁡2θ=9=3||\vec{r}_\theta \times \vec{r}_y|| = \sqrt{9\cos^2\theta + 9\sin^2\theta} = \sqrt{9} = 3. So, dS=3 dy dθdS = 3 \,dy \,d\theta. The bounds on yy are 0≤y≤4−x0 \le y \le 4-x. Substituting x=3cos⁡θx=3\cos\theta, we get 0≤y≤4−3cos⁡θ0 \le y \le 4-3\cos\theta. Since the surface is not restricted to positive or negative xx values, the cylinder makes a full revolution, so the bounds for θ\theta are 0≤θ≤2π0 \le \theta \le 2\pi. The condition y≥0y \ge 0 means 4−3cos⁡θ≥04-3\cos\theta \ge 0, which is always true since the maximum of 3cos⁡θ3\cos\theta is 3. Thus, the integral is ∫02π∫04−3cos⁡θ3 dy dθ\int_0^{2\pi} \int_0^{4-3\cos\theta} 3 \,dy \,d\theta.

Question 3

A surface of revolution is formed by rotating the curve y=ex/2y=e^{x/2} for 0≤x≤20 \le x \le 2 about the x-axis. Which of the following expressions represents the surface area element dSdS for this surface, using parameters xx and θ\theta (the angle of revolution)?

  1. dS=1+14ex dx dθdS = \sqrt{1+\frac{1}{4}e^x} \,dx \,d\theta
  2. dS=ex/21+14ex dx dθdS = e^{x/2}\sqrt{1+\frac{1}{4}e^x} \,dx \,d\theta (correct answer)
  3. dS=ex/21+12ex dx dθdS = e^{x/2}\sqrt{1+\frac{1}{2}e^x} \,dx \,d\theta
  4. dS=12ex/21+ex dx dθdS = \frac{1}{2}e^{x/2}\sqrt{1+e^x} \,dx \,d\theta
Explanation: A parameterization for a surface of revolution of y=f(x)y=f(x) about the x-axis is r⃗(x,θ)=⟨x,f(x)cos⁡θ,f(x)sin⁡θ⟩\vec{r}(x,\theta) = \langle x, f(x)\cos\theta, f(x)\sin\theta \rangle. Here, f(x)=ex/2f(x)=e^{x/2}. So, r⃗(x,θ)=⟨x,ex/2cos⁡θ,ex/2sin⁡θ⟩\vec{r}(x,\theta) = \langle x, e^{x/2}\cos\theta, e^{x/2}\sin\theta \rangle. The partial derivatives are r⃗x=⟨1,12ex/2cos⁡θ,12ex/2sin⁡θ⟩\vec{r}_x = \langle 1, \frac{1}{2}e^{x/2}\cos\theta, \frac{1}{2}e^{x/2}\sin\theta \rangle and r⃗θ=⟨0,−ex/2sin⁡θ,ex/2cos⁡θ⟩\vec{r}_\theta = \langle 0, -e^{x/2}\sin\theta, e^{x/2}\cos\theta \rangle. The cross product r⃗x×r⃗θ\vec{r}_x \times \vec{r}_\theta is ⟨12ex,−ex/2cos⁡θ,−ex/2sin⁡θ⟩\langle \frac{1}{2}e^x, -e^{x/2}\cos\theta, -e^{x/2}\sin\theta \rangle. The magnitude is ∣∣r⃗x×r⃗θ∣∣=(12ex)2+(−ex/2cos⁡θ)2+(−ex/2sin⁡θ)2=14e2x+ex(cos⁡2θ+sin⁡2θ)=14e2x+ex=ex(14ex+1)=ex/21+14ex||\vec{r}_x \times \vec{r}_\theta|| = \sqrt{(\frac{1}{2}e^x)^2 + (-e^{x/2}\cos\theta)^2 + (-e^{x/2}\sin\theta)^2} = \sqrt{\frac{1}{4}e^{2x} + e^x(\cos^2\theta+\sin^2\theta)} = \sqrt{\frac{1}{4}e^{2x} + e^x} = \sqrt{e^x(\frac{1}{4}e^x+1)} = e^{x/2}\sqrt{1+\frac{1}{4}e^x}. Thus, dS=ex/21+14ex dx dθdS = e^{x/2}\sqrt{1+\frac{1}{4}e^x} \,dx \,d\theta.

Question 4

A surface SS is parameterized by r⃗(u,v)=⟨u,v,2uv⟩\vec{r}(u,v) = \langle u, v, 2uv \rangle. A new parameterization of SS is introduced with u=s+tu = s+t and v=s−tv = s-t, giving R⃗(s,t)\vec{R}(s,t). What is the magnitude of the normal vector for the new parameterization, ∣∣R⃗s×R⃗t∣∣||\vec{R}_s \times \vec{R}_t||?

  1. 21+4s2+4t22\sqrt{1+4s^2+4t^2}
  2. 1+4(s+t)2+4(s−t)2\sqrt{1+4(s+t)^2+4(s-t)^2}
  3. 21+8s2+8t22\sqrt{1+8s^2+8t^2} (correct answer)
  4. 21+4s2−4t22\sqrt{1+4s^2-4t^2}
Explanation: First, substitute the new parameters into the original parameterization: R⃗(s,t)=⟨s+t,s−t,2(s+t)(s−t)⟩=⟨s+t,s−t,2(s2−t2)⟩\vec{R}(s,t) = \langle s+t, s-t, 2(s+t)(s-t) \rangle = \langle s+t, s-t, 2(s^2-t^2) \rangle. Now, compute the partial derivatives with respect to ss and tt: R⃗s=⟨1,1,4s⟩\vec{R}_s = \langle 1, 1, 4s \rangle and R⃗t=⟨1,−1,−4t⟩\vec{R}_t = \langle 1, -1, -4t \rangle. Next, compute the cross product: R⃗s×R⃗t=∣i^j^k^114s1−1−4t∣=i^(−4t−(−4s))−j^(−4t−4s)+k^(−1−1)=⟨4s−4t,4s+4t,−2⟩\vec{R}_s \times \vec{R}_t = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 4s \\ 1 & -1 & -4t \end{vmatrix} = \hat{i}(-4t - (-4s)) - \hat{j}(-4t - 4s) + \hat{k}(-1 - 1) = \langle 4s-4t, 4s+4t, -2 \rangle. Finally, find the magnitude of this vector: ∣∣R⃗s×R⃗t∣∣=(4s−4t)2+(4s+4t)2+(−2)2=16(s2−2st+t2)+16(s2+2st+t2)+4=16s2−32st+16t2+16s2+32st+16t2+4=32s2+32t2+4=4(1+8s2+8t2)=21+8s2+8t2||\vec{R}_s \times \vec{R}_t|| = \sqrt{(4s-4t)^2 + (4s+4t)^2 + (-2)^2} = \sqrt{16(s^2-2st+t^2) + 16(s^2+2st+t^2) + 4} = \sqrt{16s^2-32st+16t^2 + 16s^2+32st+16t^2 + 4} = \sqrt{32s^2+32t^2+4} = \sqrt{4(1+8s^2+8t^2)} = 2\sqrt{1+8s^2+8t^2}.

Question 5

A surface SS is parameterized by r⃗(u,v)\vec{r}(u,v) over a domain DD. The surface area is given by A(S)=∬D∣∣r⃗u×r⃗v∣∣ du dvA(S) = \iint_D ||\vec{r}_u \times \vec{r}_v|| \,du \,dv. Consider a re-parameterization given by R⃗(s,t)=r⃗(s2,t2)\vec{R}(s,t) = \vec{r}(s^2, t^2). Using the chain rule, how does the new surface area scaling factor, ∣∣R⃗s×R⃗t∣∣||\vec{R}_s \times \vec{R}_t||, relate to the original scaling factor, ∣∣r⃗u×r⃗v∣∣||\vec{r}_u \times \vec{r}_v||?

  1. ∣∣R⃗s×R⃗t∣∣=4st ∣∣r⃗u×r⃗v∣∣||\vec{R}_s \times \vec{R}_t|| = 4st \, ||\vec{r}_u \times \vec{r}_v|| (correct answer)
  2. ∣∣R⃗s×R⃗t∣∣=2(s+t) ∣∣r⃗u×r⃗v∣∣||\vec{R}_s \times \vec{R}_t|| = 2(s+t) \, ||\vec{r}_u \times \vec{r}_v||
  3. ∣∣R⃗s×R⃗t∣∣=(s2+t2) ∣∣r⃗u×r⃗v∣∣||\vec{R}_s \times \vec{R}_t|| = (s^2+t^2) \, ||\vec{r}_u \times \vec{r}_v||
  4. ∣∣R⃗s×R⃗t∣∣=4s2t2 ∣∣r⃗u×r⃗v∣∣||\vec{R}_s \times \vec{R}_t|| = 4s^2t^2 \, ||\vec{r}_u \times \vec{r}_v||
Explanation: We apply the chain rule for vector functions. Let u(s,t)=s2u(s,t) = s^2 and v(s,t)=t2v(s,t) = t^2. Then R⃗s=∂r⃗∂u∂u∂s+∂r⃗∂v∂v∂s=r⃗u(2s)+r⃗v(0)=2sr⃗u\vec{R}_s = \frac{\partial \vec{r}}{\partial u}\frac{\partial u}{\partial s} + \frac{\partial \vec{r}}{\partial v}\frac{\partial v}{\partial s} = \vec{r}_u (2s) + \vec{r}_v (0) = 2s \vec{r}_u. Similarly, R⃗t=∂r⃗∂u∂u∂t+∂r⃗∂v∂v∂t=r⃗u(0)+r⃗v(2t)=2tr⃗v\vec{R}_t = \frac{\partial \vec{r}}{\partial u}\frac{\partial u}{\partial t} + \frac{\partial \vec{r}}{\partial v}\frac{\partial v}{\partial t} = \vec{r}_u (0) + \vec{r}_v (2t) = 2t \vec{r}_v. In both cases, the partials r⃗u\vec{r}_u and r⃗v\vec{r}_v are evaluated at (u,v)=(s2,t2)(u,v)=(s^2, t^2). Now we compute the cross product: R⃗s×R⃗t=(2sr⃗u)×(2tr⃗v)=(2s)(2t)(r⃗u×r⃗v)=4st(r⃗u×r⃗v)\vec{R}_s \times \vec{R}_t = (2s \vec{r}_u) \times (2t \vec{r}_v) = (2s)(2t) (\vec{r}_u \times \vec{r}_v) = 4st (\vec{r}_u \times \vec{r}_v). Taking the magnitude of both sides gives ∣∣R⃗s×R⃗t∣∣=∣4st∣ ∣∣r⃗u×r⃗v∣∣||\vec{R}_s \times \vec{R}_t|| = |4st| \, ||\vec{r}_u \times \vec{r}_v||. Assuming s,ts, t are positive, this simplifies to 4st ∣∣r⃗u×r⃗v∣∣4st \, ||\vec{r}_u \times \vec{r}_v||.

Question 6

A torus (donut shape) is generated by rotating a circle of radius rr centered at (R,0,0)(R,0,0) in the xzxz-plane about the zz-axis, where 0<r<R0 < r < R. A standard parameterization is r⃗(θ,ϕ)=⟨(R+rcos⁡ϕ)cos⁡θ,(R+rcos⁡ϕ)sin⁡θ,rsin⁡ϕ⟩\vec{r}(\theta, \phi) = \langle (R+r\cos\phi)\cos\theta, (R+r\cos\phi)\sin\theta, r\sin\phi \rangle. What is the surface area element dSdS for this torus?

  1. dS=(R+rcos⁡ϕ) dθ dϕdS = (R+r\cos\phi) \,d\theta \,d\phi
  2. dS=R(R+rcos⁡ϕ) dθ dϕdS = R(R+r\cos\phi) \,d\theta \,d\phi
  3. dS=rR dθ dϕdS = rR \,d\theta \,d\phi
  4. dS=r(R+rcos⁡ϕ) dθ dϕdS = r(R+r\cos\phi) \,d\theta \,d\phi (correct answer)
Explanation: When you encounter a parameterized surface, finding the surface area element requires computing the magnitude of the cross product of the partial derivatives. This fundamental technique appears frequently in multivariable calculus. For the torus parameterization r⃗(θ,ϕ)=⟨(R+rcos⁡ϕ)cos⁡θ,(R+rcos⁡ϕ)sin⁡θ,rsin⁡ϕ⟩\vec{r}(\theta, \phi) = \langle (R+r\cos\phi)\cos\theta, (R+r\cos\phi)\sin\theta, r\sin\phi \rangle, you need to find r⃗θ\vec{r}_\theta and r⃗ϕ\vec{r}_\phi: r⃗θ=⟨−(R+rcos⁡ϕ)sin⁡θ,(R+rcos⁡ϕ)cos⁡θ,0⟩\vec{r}_\theta = \langle -(R+r\cos\phi)\sin\theta, (R+r\cos\phi)\cos\theta, 0 \rangle r⃗ϕ=⟨−rsin⁡ϕcos⁡θ,−rsin⁡ϕsin⁡θ,rcos⁡ϕ⟩\vec{r}_\phi = \langle -r\sin\phi\cos\theta, -r\sin\phi\sin\theta, r\cos\phi \rangle Computing their cross product r⃗θ×r⃗ϕ\vec{r}_\theta \times \vec{r}_\phi and finding its magnitude gives ∣r⃗θ×r⃗ϕ∣=r(R+rcos⁡ϕ)|\vec{r}_\theta \times \vec{r}_\phi| = r(R+r\cos\phi). Therefore, dS=r(R+rcos⁡ϕ) dθ dϕdS = r(R+r\cos\phi) \,d\theta \,d\phi. Answer A omits the factor of rr, which would arise if you mistakenly used just the distance from the z-axis without accounting for the circle's radius contribution. Answer B incorrectly includes an extra factor of RR, perhaps from confusing this with a different geometric setup. Answer C gives a constant rRrR, which ignores the varying distance (R+rcos⁡ϕ)(R+r\cos\phi) from the z-axis as the generating circle moves around. Remember: for any parameterized surface, the surface area element always equals the magnitude of the cross product of partial derivatives. Don't try to shortcut this calculation—work through the cross product systematically to avoid missing important geometric factors.

Question 7

The area of a small patch of a parameterized surface r⃗(u,v)\vec{r}(u,v) is approximated by the area of a parallelogram on its tangent plane, given by dAsurf≈∣∣r⃗u×r⃗v∣∣ du dvdA_{surf} \approx ||\vec{r}_u \times \vec{r}_v|| \,du \,dv. What does the quantity ∣∣r⃗u×r⃗v∣∣||\vec{r}_u \times \vec{r}_v|| represent geometrically?

  1. The ratio of the area of a small surface patch to the area of its projection onto the xyxy-plane.
  2. The local rate at which the parameterization stretches or shrinks area from the uvuv-plane to the surface in R3\mathbb{R}^3. (correct answer)
  3. The magnitude of the curvature of the surface at the point corresponding to (u,v)(u,v).
  4. The length of the normal vector to the surface, which is constant for all parameterizations.
Explanation: The expression ∣∣r⃗u×r⃗v∣∣||\vec{r}_u \times \vec{r}_v|| is the surface area scaling factor, also known as the Jacobian for the surface parameterization. It describes how area is transformed from the parameter domain (the uvuv-plane) to the surface in 3D space. A small rectangle in the uvuv-plane with area du dvdu \,dv is mapped to a small patch of the surface with area dS=∣∣r⃗u×r⃗v∣∣ du dvdS = ||\vec{r}_u \times \vec{r}_v|| \,du \,dv. Therefore, ∣∣r⃗u×r⃗v∣∣||\vec{r}_u \times \vec{r}_v|| is the local ratio of the area on the surface to the corresponding area in the parameter domain, representing the local stretching or shrinking of area.

Question 8

The surface of a cone with height HH and radius RR is given by z=H(1−x2+y2R)z = H(1 - \frac{\sqrt{x^2+y^2}}{R}). A student parameterizes the cone using r⃗(r,θ)=⟨rcos⁡θ,rsin⁡θ,H(1−r/R)⟩\vec{r}(r,\theta) = \langle r\cos\theta, r\sin\theta, H(1-r/R) \rangle for 0≤r≤R0 \le r \le R and 0≤θ≤2π0 \le \theta \le 2\pi. Which of the following is the correct surface area element dSdS for this parameterization?

  1. dS=r2+H2/R2 dr dθdS = \sqrt{r^2 + H^2/R^2} \,dr \,d\theta
  2. dS=(1+H/R)r dr dθdS = (1 + H/R) r \,dr \,d\theta
  3. dS=1+H2/R2 dr dθdS = \sqrt{1 + H^2/R^2} \,dr \,d\theta
  4. dS=R2+H2Rr dr dθdS = \frac{\sqrt{R^2+H^2}}{R} r \,dr \,d\theta (correct answer)
Explanation: When calculating surface area using parametric equations, you need to find the magnitude of the cross product of the partial derivative vectors. This measures how the surface "stretches" compared to the parameter space. Given r⃗(r,θ)=⟨rcos⁡θ,rsin⁡θ,H(1−r/R)⟩\vec{r}(r,\theta) = \langle r\cos\theta, r\sin\theta, H(1-r/R) \rangle, you first compute the partial derivatives: r⃗r=⟨cos⁡θ,sin⁡θ,−H/R⟩\vec{r}_r = \langle \cos\theta, \sin\theta, -H/R \rangle r⃗θ=⟨−rsin⁡θ,rcos⁡θ,0⟩\vec{r}_\theta = \langle -r\sin\theta, r\cos\theta, 0 \rangle Next, find their cross product: r⃗r×r⃗θ=⟨Hrcos⁡θR,Hrsin⁡θR,r⟩\vec{r}_r \times \vec{r}_\theta = \langle \frac{Hr\cos\theta}{R}, \frac{Hr\sin\theta}{R}, r \rangle The magnitude is: ∣r⃗r×r⃗θ∣=H2r2cos⁡2θR2+H2r2sin⁡2θR2+r2=H2r2R2+r2=rH2+R2R2=rR2+H2R|\vec{r}_r \times \vec{r}_\theta| = \sqrt{\frac{H^2r^2\cos^2\theta}{R^2} + \frac{H^2r^2\sin^2\theta}{R^2} + r^2} = \sqrt{\frac{H^2r^2}{R^2} + r^2} = r\sqrt{\frac{H^2 + R^2}{R^2}} = \frac{r\sqrt{R^2+H^2}}{R} Therefore, dS=R2+H2Rr dr dθdS = \frac{\sqrt{R^2+H^2}}{R} r \,dr \,d\theta, which is answer D. Answer A incorrectly has r2r^2 under the square root instead of factoring out rr. Answer B uses (1+H/R)(1 + H/R) instead of the proper Pythagorean relationship and includes an extra rr factor. Answer C completely omits the rr term that comes from the Jacobian of the cylindrical coordinate transformation. Study tip: Remember that surface area elements always involve the magnitude of r⃗r×r⃗θ\vec{r}_r \times \vec{r}_\theta. For surfaces of revolution or cones, expect to see 1+(slope)2\sqrt{1 + (\text{slope})^2} patterns, and don't forget the rr factor when using cylindrical-like coordinates.

Question 9

A surface is described by the vector parameterization r⃗(u,v)=⟨ucos⁡v,usin⁡v,v⟩\vec{r}(u,v) = \langle u \cos v, u \sin v, v \rangle for 0≤u≤20 \le u \le 2 and 0≤v≤π0 \le v \le \pi. This surface is a portion of a helicoid. What is its surface area element, dSdS?

  1. dS=u du dvdS = u \,du \,dv
  2. dS=(1+u2) du dvdS = (1+u^2) \,du \,dv
  3. dS=1+u2 du dvdS = \sqrt{1+u^2} \,du \,dv (correct answer)
  4. dS=u1+u2 du dvdS = u\sqrt{1+u^2} \,du \,dv
Explanation: To find the surface area element dS=∣∣r⃗u×r⃗v∣∣ du dvdS = ||\vec{r}_u \times \vec{r}_v|| \,du \,dv, we first compute the partial derivatives of r⃗(u,v)\vec{r}(u,v). We have 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. Next, we compute their cross product: r⃗u×r⃗v=∣i^j^k^cos⁡vsin⁡v0−usin⁡vucos⁡v1∣=i^(sin⁡v−0)−j^(cos⁡v−0)+k^(ucos⁡2v+usin⁡2v)=⟨sin⁡v,−cos⁡v,u⟩\vec{r}_u \times \vec{r}_v = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \cos v & \sin v & 0 \\ -u \sin v & u \cos v & 1 \end{vmatrix} = \hat{i}(\sin v - 0) - \hat{j}(\cos v - 0) + \hat{k}(u \cos^2 v + u \sin^2 v) = \langle \sin v, -\cos v, u \rangle. Finally, we find the magnitude of this vector: ∣∣r⃗u×r⃗v∣∣=(sin⁡v)2+(−cos⁡v)2+u2=sin⁡2v+cos⁡2v+u2=1+u2||\vec{r}_u \times \vec{r}_v|| = \sqrt{(\sin v)^2 + (-\cos v)^2 + u^2} = \sqrt{\sin^2 v + \cos^2 v + u^2} = \sqrt{1+u^2}. Therefore, the surface area element is dS=1+u2 du dvdS = \sqrt{1+u^2} \,du \,dv.

Question 10

Let SS be the portion of the sphere x2+y2+z2=4x^2+y^2+z^2=4 that lies above the xyxy-plane and within the cylinder x2+y2=3x^2+y^2=3. If SS is parameterized as a graph z=f(x,y)z=f(x,y), its surface area element is dS=g(x,y) dx dydS = g(x,y) \,dx \,dy. What is g(x,y)g(x,y)?

  1. g(x,y)=24−x2−y2g(x,y) = \frac{2}{\sqrt{4-x^2-y^2}} (correct answer)
  2. g(x,y)=44−x2−y2g(x,y) = \frac{4}{\sqrt{4-x^2-y^2}}
  3. g(x,y)=4−x2−y22g(x,y) = \frac{\sqrt{4-x^2-y^2}}{2}
  4. g(x,y)=2zg(x,y) = \frac{2}{z}
Explanation: The surface is the graph of z=f(x,y)=4−x2−y2z = f(x,y) = \sqrt{4-x^2-y^2}. The surface area element is given by dS=1+(∂z∂x)2+(∂z∂y)2 dx dydS = \sqrt{1 + (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2} \,dx \,dy. We compute the partial derivatives: ∂z∂x=−2x24−x2−y2=−x4−x2−y2\frac{\partial z}{\partial x} = \frac{-2x}{2\sqrt{4-x^2-y^2}} = \frac{-x}{\sqrt{4-x^2-y^2}} and ∂z∂y=−y4−x2−y2\frac{\partial z}{\partial y} = \frac{-y}{\sqrt{4-x^2-y^2}}. Squaring and adding them gives (∂z∂x)2+(∂z∂y)2=x24−x2−y2+y24−x2−y2=x2+y24−x2−y2(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 = \frac{x^2}{4-x^2-y^2} + \frac{y^2}{4-x^2-y^2} = \frac{x^2+y^2}{4-x^2-y^2}. Then, 1+(∂z∂x)2+(∂z∂y)2=1+x2+y24−x2−y2=(4−x2−y2)+(x2+y2)4−x2−y2=44−x2−y21 + (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 = 1 + \frac{x^2+y^2}{4-x^2-y^2} = \frac{(4-x^2-y^2) + (x^2+y^2)}{4-x^2-y^2} = \frac{4}{4-x^2-y^2}. Taking the square root, we get g(x,y)=44−x2−y2=24−x2−y2g(x,y) = \sqrt{\frac{4}{4-x^2-y^2}} = \frac{2}{\sqrt{4-x^2-y^2}}. While g(x,y)=2/zg(x,y)=2/z is technically correct, standard practice requires expressing the function in terms of the parameters xx and yy.

Question 11

The surface area element for a surface parameterized by r⃗(u,v)\vec{r}(u,v) is found to be dS=1+4u2+4v2 du dvdS = \sqrt{1+4u^2+4v^2} \,du \,dv. Which of the following is most likely the surface being described, using uu and vv as parameters for the xx and yy coordinates, respectively?

  1. The cone z=x2+y2z = \sqrt{x^2+y^2}
  2. The paraboloid z=x2+y2z = x^2+y^2 (correct answer)
  3. The plane z=2x+2yz = 2x+2y
  4. The hemisphere z=1−x2−y2z = \sqrt{1-x^2-y^2}
Explanation: We can test each surface by parameterizing it as r⃗(u,v)=⟨u,v,f(u,v)⟩\vec{r}(u,v) = \langle u, v, f(u,v) \rangle and calculating its surface area element factor 1+(fu)2+(fv)2\sqrt{1 + (f_u)^2 + (f_v)^2}. For choice B, the surface is z=f(x,y)=x2+y2z=f(x,y)=x^2+y^2. We use parameters u=x,v=yu=x, v=y, so f(u,v)=u2+v2f(u,v)=u^2+v^2. The partial derivatives are fu=2uf_u = 2u and fv=2vf_v = 2v. The magnitude of the normal vector is 1+(2u)2+(2v)2=1+4u2+4v2\sqrt{1 + (2u)^2 + (2v)^2} = \sqrt{1+4u^2+4v^2}. This matches the given expression. For the other choices: A) Cone z=u2+v2z=\sqrt{u^2+v^2} gives 1+(uu2+v2)2+(vu2+v2)2=2\sqrt{1 + (\frac{u}{\sqrt{u^2+v^2}})^2 + (\frac{v}{\sqrt{u^2+v^2}})^2} = \sqrt{2}. C) Plane z=2u+2vz=2u+2v gives 1+22+22=9=3\sqrt{1+2^2+2^2}=\sqrt{9}=3. D) Hemisphere z=1−u2−v2z=\sqrt{1-u^2-v^2} gives a more complex expression 11−u2−v2\frac{1}{\sqrt{1-u^2-v^2}}.

Question 12

The graph of z=x2−y2z = x^2 - y^2 over the square region [−1,1]×[−1,1][-1,1] \times [-1,1] forms a surface SS. If we use the standard parametrization r(x,y)=(x,y,x2−y2)\mathbf{r}(x,y) = (x, y, x^2 - y^2), what is the surface area element dSdS in terms of xx and yy?

  1. 1+4x2−4y2 dx dy\sqrt{1 + 4x^2 - 4y^2} \, dx \, dy
  2. 1+2x2+2y2 dx dy\sqrt{1 + 2x^2 + 2y^2} \, dx \, dy
  3. 1+4x2+4y2 dx dy\sqrt{1 + 4x^2 + 4y^2} \, dx \, dy (correct answer)
  4. 1+x2+y2 dx dy\sqrt{1 + x^2 + y^2} \, dx \, dy
Explanation: When you encounter a surface area problem in multivariable calculus, you need to find the surface area element dSdS using the formula dS=∥rx×ry∥ dx dydS = \|\mathbf{r}_x \times \mathbf{r}_y\| \, dx \, dy, where rx\mathbf{r}_x and ry\mathbf{r}_y are the partial derivatives of the parametrization vector. Given r(x,y)=(x,y,x2−y2)\mathbf{r}(x,y) = (x, y, x^2 - y^2), let's find the partial derivatives:
  • rx=(1,0,2x)\mathbf{r}_x = (1, 0, 2x)
  • ry=(0,1,−2y)\mathbf{r}_y = (0, 1, -2y)
Now compute the cross product rx×ry\mathbf{r}_x \times \mathbf{r}_y: rx×ry=∣ijk102x01−2y∣=(−2x,2y,1)\mathbf{r}_x \times \mathbf{r}_y = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 2x \\ 0 & 1 & -2y \end{vmatrix} = (-2x, 2y, 1) The magnitude is: ∥rx×ry∥=(−2x)2+(2y)2+12=4x2+4y2+1=1+4x2+4y2\|\mathbf{r}_x \times \mathbf{r}_y\| = \sqrt{(-2x)^2 + (2y)^2 + 1^2} = \sqrt{4x^2 + 4y^2 + 1} = \sqrt{1 + 4x^2 + 4y^2} Therefore, dS=1+4x2+4y2 dx dydS = \sqrt{1 + 4x^2 + 4y^2} \, dx \, dy, which is answer choice C. Answer A incorrectly subtracts the yy term, likely from confusing the original function x2−y2x^2 - y^2 with the cross product calculation. Answer B uses coefficients of 2 instead of 4, suggesting an error in squaring the partial derivatives 2x2x and −2y-2y. Answer D omits the factor of 4 entirely, possibly from forgetting to square the partial derivatives. Remember: when computing surface area elements, always square each component of the cross product before adding them under the square root. The coefficients matter!

Question 13

A hemisphere of radius 3 is parametrized by r(ϕ,θ)=(3sin⁡ϕcos⁡θ,3sin⁡ϕsin⁡θ,3cos⁡ϕ)\mathbf{r}(\phi, \theta) = (3\sin\phi\cos\theta, 3\sin\phi\sin\theta, 3\cos\phi) where 0≤ϕ≤π20 \leq \phi \leq \frac{\pi}{2} and 0≤θ≤2π0 \leq \theta \leq 2\pi. Which expression gives the correct magnitude ∣∣rϕ×rθ∣∣||\mathbf{r}_{\phi} \times \mathbf{r}_{\theta}|| for the surface area element?

  1. 3sin⁡ϕ3\sin\phi
  2. 9sin⁡ϕ9\sin\phi (correct answer)
  3. 9sin⁡2ϕ9\sin^2\phi
  4. 9cos⁡ϕ9\cos\phi
Explanation: When you encounter a parametric surface and need to find the surface area element, you're looking for the magnitude of the cross product of the partial derivative vectors. This cross product gives you a vector normal to the surface, and its magnitude tells you how much the surface area is "stretched" at each point. To find ∣∣rϕ×rθ∣∣||\mathbf{r}_{\phi} \times \mathbf{r}_{\theta}||, you need to compute the partial derivatives first: rϕ=(3cos⁡ϕcos⁡θ,3cos⁡ϕsin⁡θ,−3sin⁡ϕ)\mathbf{r}_{\phi} = (3\cos\phi\cos\theta, 3\cos\phi\sin\theta, -3\sin\phi) rθ=(−3sin⁡ϕsin⁡θ,3sin⁡ϕcos⁡θ,0)\mathbf{r}_{\theta} = (-3\sin\phi\sin\theta, 3\sin\phi\cos\theta, 0) Computing their cross product: rϕ×rθ=(9sin⁡2ϕcos⁡θ,9sin⁡2ϕsin⁡θ,9sin⁡ϕcos⁡ϕ)\mathbf{r}_{\phi} \times \mathbf{r}_{\theta} = (9\sin^2\phi\cos\theta, 9\sin^2\phi\sin\theta, 9\sin\phi\cos\phi) The magnitude is: ∣∣rϕ×rθ∣∣=81sin⁡4ϕcos⁡2θ+81sin⁡4ϕsin⁡2θ+81sin⁡2ϕcos⁡2ϕ||\mathbf{r}_{\phi} \times \mathbf{r}_{\theta}|| = \sqrt{81\sin^4\phi\cos^2\theta + 81\sin^4\phi\sin^2\theta + 81\sin^2\phi\cos^2\phi} Factoring out 81sin⁡2ϕ81\sin^2\phi: =9sin⁡ϕsin⁡2ϕ(cos⁡2θ+sin⁡2θ)+cos⁡2ϕ=9sin⁡ϕsin⁡2ϕ+cos⁡2ϕ=9sin⁡ϕ= 9\sin\phi\sqrt{\sin^2\phi(\cos^2\theta + \sin^2\theta) + \cos^2\phi} = 9\sin\phi\sqrt{\sin^2\phi + \cos^2\phi} = 9\sin\phi This confirms answer (B) is correct. (A) 3sin⁡ϕ3\sin\phi misses the factor of 3 that comes from squaring the radius. (C) 9sin⁡2ϕ9\sin^2\phi appears if you forget to take the square root when computing the magnitude. (D) 9cos⁡ϕ9\cos\phi would result from confusing ϕ\phi and θ\theta or misremembering spherical coordinate identities. Study tip: For spherical coordinates, the surface area element always involves sin⁡ϕ\sin\phi (never cos⁡ϕ\cos\phi), and remember that the radius gets squared in the cross product magnitude.

Question 14

Consider two different parametrizations of the same surface: r1(u,v)=(u,v,uv)\mathbf{r}_1(u,v) = (u, v, uv) and r2(s,t)=(s2,t2,s2t2)\mathbf{r}_2(s,t) = (s^2, \frac{t}{2}, \frac{s^2 t}{2}) where the domains are chosen so that both cover the same portion of the surface. If ∣∣r1u×r1v∣∣=1+u2+v2||\mathbf{r}_{1u} \times \mathbf{r}_{1v}|| = \sqrt{1 + u^2 + v^2} at a point (u0,v0)(u_0, v_0), and the corresponding point in the second parametrization is (s0,t0)(s_0, t_0) where s02=u0s_0^2 = u_0 and t02=v0\frac{t_0}{2} = v_0, what is ∣∣r2s×r2t∣∣||\mathbf{r}_{2s} \times \mathbf{r}_{2t}|| at (s0,t0)(s_0, t_0)?

  1. s021+s04+t024\frac{s_0}{2}\sqrt{1 + s_0^4 + \frac{t_0^2}{4}}
  2. s01+s04+t024s_0\sqrt{1 + s_0^4 + \frac{t_0^2}{4}} (correct answer)
  3. 2s01+s04+t0242s_0\sqrt{1 + s_0^4 + \frac{t_0^2}{4}}
  4. s04+4s04+t02s_0\sqrt{4 + 4s_0^4 + t_0^2}
Explanation: When you encounter surface parametrizations, the key insight is understanding how different coordinate systems relate through the Jacobian of the transformation. The magnitude of the cross product of partial derivatives (which gives surface area elements) transforms according to this Jacobian determinant. First, let's establish the coordinate transformation. From the given relationships: u=s2u = s^2, v=t2v = \frac{t}{2}, and w=uv=s2⋅t2=s2t2w = uv = s^2 \cdot \frac{t}{2} = \frac{s^2t}{2} The Jacobian matrix of this transformation is: The determinant is ∣J∣=2s⋅12=s|J| = 2s \cdot \frac{1}{2} = s. For surface parametrizations, the cross product magnitudes are related by: ∣∣r2s×r2t∣∣=∣J∣⋅∣∣r1u×r1v∣∣||\mathbf{r}_{2s} \times \mathbf{r}_{2t}|| = |J| \cdot ||\mathbf{r}_{1u} \times \mathbf{r}_{1v}|| At the corresponding points, u0=s02u_0 = s_0^2 and v0=t02v_0 = \frac{t_0}{2}, so: ∣∣r2s×r2t∣∣=s0⋅1+u02+v02=s01+s04+t024||\mathbf{r}_{2s} \times \mathbf{r}_{2t}|| = s_0 \cdot \sqrt{1 + u_0^2 + v_0^2} = s_0\sqrt{1 + s_0^4 + \frac{t_0^2}{4}} This matches answer B. A incorrectly uses s02\frac{s_0}{2} instead of s0s_0 as the Jacobian factor. C uses 2s02s_0, likely from confusing which transformation direction to use. D incorrectly manipulates the expression under the square root, possibly from algebraic errors in the substitution. Study tip: Always compute the Jacobian determinant carefully when transforming between parametrizations—the cross product magnitude scales by this factor.

Question 15

A surface SS is parametrized by r(u,v)=(ucos⁡v,usin⁡v,u2)\mathbf{r}(u,v) = (u\cos v, u\sin v, u^2) where 0≤u≤20 \leq u \leq 2 and 0≤v≤π0 \leq v \leq \pi. If the surface area element is dS=∣∣ru×rv∣∣ du dvdS = ||\mathbf{r}_u \times \mathbf{r}_v|| \, du \, dv, which expression correctly represents ∣∣ru×rv∣∣||\mathbf{r}_u \times \mathbf{r}_v||?

  1. u1+4u2u\sqrt{1 + 4u^2} (correct answer)
  2. u1+2u2u\sqrt{1 + 2u^2}
  3. u2+4u4\sqrt{u^2 + 4u^4}
  4. 1+4u2\sqrt{1 + 4u^2}
Explanation: First find the partial derivatives: ru=(cos⁡v,sin⁡v,2u)\mathbf{r}_u = (\cos v, \sin v, 2u) and rv=(−usin⁡v,ucos⁡v,0)\mathbf{r}_v = (-u\sin v, u\cos v, 0). Then compute the cross product: ru×rv=(−2u2cos⁡v,−2u2sin⁡v,u)\mathbf{r}_u \times \mathbf{r}_v = (-2u^2\cos v, -2u^2\sin v, u). The magnitude is ∣∣ru×rv∣∣=4u4cos⁡2v+4u4sin⁡2v+u2=4u4+u2=u4u2+1=u1+4u2||\mathbf{r}_u \times \mathbf{r}_v|| = \sqrt{4u^4\cos^2 v + 4u^4\sin^2 v + u^2} = \sqrt{4u^4 + u^2} = u\sqrt{4u^2 + 1} = u\sqrt{1 + 4u^2}. Choice B incorrectly uses 2u22u^2 instead of 4u24u^2. Choice C factors out uu incorrectly. Choice D omits the factor of uu entirely.

Question 16

Consider the portion of the cylinder x2+y2=4x^2 + y^2 = 4 between the planes z=0z = 0 and z=3z = 3. If parametrized as 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, what is the surface area of this cylindrical surface?

  1. 6π6\pi
  2. 24π24\pi
  3. 12π12\pi (correct answer)
  4. 36π36\pi
Explanation: When you encounter surface area problems involving parametric surfaces, you need to use the formula Surface Area=∬D∥ru×rv∥ du dv\text{Surface Area} = \iint_D \|\mathbf{r}_u \times \mathbf{r}_v\| \, du \, dv, where ru\mathbf{r}_u and rv\mathbf{r}_v are partial derivatives with respect to the parameters. For the parametrization r(θ,z)=(2cos⁡θ,2sin⁡θ,z)\mathbf{r}(\theta, z) = (2\cos\theta, 2\sin\theta, z), let's find the partial derivatives:
  • rθ=(−2sin⁡θ,2cos⁡θ,0)\mathbf{r}_\theta = (-2\sin\theta, 2\cos\theta, 0)
  • rz=(0,0,1)\mathbf{r}_z = (0, 0, 1)
The cross product is: rθ×rz=∣ijk−2sin⁡θ2cos⁡θ0001∣=(2cos⁡θ,2sin⁡θ,0)\mathbf{r}_\theta \times \mathbf{r}_z = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2\sin\theta & 2\cos\theta & 0 \\ 0 & 0 & 1 \end{vmatrix} = (2\cos\theta, 2\sin\theta, 0) The magnitude is ∥rθ×rz∥=4cos⁡2θ+4sin⁡2θ=4=2\|\mathbf{r}_\theta \times \mathbf{r}_z\| = \sqrt{4\cos^2\theta + 4\sin^2\theta} = \sqrt{4} = 2. Therefore: Surface Area=∫02π∫032 dz dθ=∫02π6 dθ=12π\text{Surface Area} = \int_0^{2\pi} \int_0^3 2 \, dz \, d\theta = \int_0^{2\pi} 6 \, d\theta = 12\pi Answer choice A (6π6\pi) results from forgetting the factor of 2 in the radius when computing the cross product magnitude. Answer choice B (24π24\pi) comes from incorrectly squaring the magnitude or doubling the final result. Answer choice D (36π36\pi) likely results from using the wrong bounds or miscalculating the cross product entirely. Remember: for cylindrical surfaces, the cross product magnitude equals the radius of the cylinder. Always double-check your partial derivatives and cross product calculations—they're the most common sources of error in parametric surface area problems.

Question 17

Consider the surface defined by z=xyz = xy over the region D:x2+y2≤4D: x^2 + y^2 \leq 4. If this surface is parametrized using r(x,y)=(x,y,xy)\mathbf{r}(x,y) = (x, y, xy), what is the surface area element dSdS?

  1. 1+x2+y2 dx dy\sqrt{1 + x^2 + y^2} \, dx \, dy (correct answer)
  2. 1+x2+y2+x2y2 dx dy\sqrt{1 + x^2 + y^2 + x^2y^2} \, dx \, dy
  3. 1+2xy+x2y2 dx dy\sqrt{1 + 2xy + x^2y^2} \, dx \, dy
  4. 1+x4+y4 dx dy\sqrt{1 + x^4 + y^4} \, dx \, dy
Explanation: For a surface z=f(x,y)z = f(x,y), the surface area element is dS=1+fx2+fy2 dx dydS = \sqrt{1 + f_x^2 + f_y^2} \, dx \, dy. Here f(x,y)=xyf(x,y) = xy, so fx=yf_x = y and fy=xf_y = x. Therefore dS=1+y2+x2 dx dy=1+x2+y2 dx dydS = \sqrt{1 + y^2 + x^2} \, dx \, dy = \sqrt{1 + x^2 + y^2} \, dx \, dy. Choice B incorrectly includes an extra x2y2x^2y^2 term. Choice C uses the product (fx)(fy)=xy(f_x)(f_y) = xy instead of fx2+fy2f_x^2 + f_y^2. Choice D incorrectly uses fx4+fy4f_x^4 + f_y^4 instead of fx2+fy2f_x^2 + f_y^2.