Multivariable Calculus Quiz: Choosing Parameterizations
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Choosing ParameterizationsQuestion 1 of 7

Consider the surface defined implicitly by x2+y2−z2=1x^2 + y^2 - z^2 = 1 for z≥0z \geq 0. You need to parameterize the portion of this surface where 1≤x2+y2≤41 \leq x^2 + y^2 \leq 4. Which approach provides the most efficient parameterization that avoids singularities?

Use cylindrical coordinates: r(r,θ)=(rcos⁡θ,rsin⁡θ,r2−1)\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, \sqrt{r^2-1}) with 1≤r≤21 \leq r \leq 2, 0≤θ≤2π0 \leq \theta \leq 2\pi
Use spherical coordinates: r(ρ,θ)=(ρcos⁡θ,ρsin⁡θ,ρ2−1)\mathbf{r}(\rho,\theta) = (\rho\cos\theta, \rho\sin\theta, \sqrt{\rho^2-1}) with 2≤ρ≤22\sqrt{2} \leq \rho \leq 2\sqrt{2}, 0≤θ≤2π0 \leq \theta \leq 2\pi
Use hyperbolic parameterization: r(u,v)=(cosh⁡ucos⁡v,cosh⁡usin⁡v,sinh⁡u)\mathbf{r}(u,v) = (\cosh u \cos v, \cosh u \sin v, \sinh u) with 0≤u≤ln⁡(2+3)0 \leq u \leq \ln(2+\sqrt{3}), 0≤v≤2π0 \leq v \leq 2\pi
Use Cartesian parameterization: r(x,y)=(x,y,x2+y2−1)\mathbf{r}(x,y) = (x, y, \sqrt{x^2+y^2-1}) with x2+y2∈[1,4]x^2 + y^2 \in [1,4]
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Choosing Parameterizations

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

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

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Question 1

Consider the surface defined implicitly by x2+y2−z2=1x^2 + y^2 - z^2 = 1 for z≥0z \geq 0. You need to parameterize the portion of this surface where 1≤x2+y2≤41 \leq x^2 + y^2 \leq 4. Which approach provides the most efficient parameterization that avoids singularities?

  1. Use cylindrical coordinates: r(r,θ)=(rcos⁡θ,rsin⁡θ,r2−1)\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, \sqrt{r^2-1}) with 1≤r≤21 \leq r \leq 2, 0≤θ≤2π0 \leq \theta \leq 2\pi (correct answer)
  2. Use spherical coordinates: r(ρ,θ)=(ρcos⁡θ,ρsin⁡θ,ρ2−1)\mathbf{r}(\rho,\theta) = (\rho\cos\theta, \rho\sin\theta, \sqrt{\rho^2-1}) with 2≤ρ≤22\sqrt{2} \leq \rho \leq 2\sqrt{2}, 0≤θ≤2π0 \leq \theta \leq 2\pi
  3. Use hyperbolic parameterization: r(u,v)=(cosh⁡ucos⁡v,cosh⁡usin⁡v,sinh⁡u)\mathbf{r}(u,v) = (\cosh u \cos v, \cosh u \sin v, \sinh u) with 0≤u≤ln⁡(2+3)0 \leq u \leq \ln(2+\sqrt{3}), 0≤v≤2π0 \leq v \leq 2\pi
  4. Use Cartesian parameterization: r(x,y)=(x,y,x2+y2−1)\mathbf{r}(x,y) = (x, y, \sqrt{x^2+y^2-1}) with x2+y2∈[1,4]x^2 + y^2 \in [1,4]
Explanation: This is a hyperboloid of one sheet. The constraint 1≤x2+y2≤41 \leq x^2 + y^2 \leq 4 naturally suggests cylindrical coordinates where r=x2+y2r = \sqrt{x^2+y^2}, so 1≤r≤21 \leq r \leq 2. From the surface equation x2+y2−z2=1x^2 + y^2 - z^2 = 1, we get z=r2−1z = \sqrt{r^2-1} (taking positive root since z≥0z \geq 0). This parameterization is well-defined and smooth throughout the domain. Choice B incorrectly uses ρ\rho in place of rr and has wrong bounds. Choice C uses hyperbolic functions which are unnecessarily complex for this region and the bounds are incorrect. Choice D is problematic because it doesn't clearly specify the domain and can lead to computational difficulties when integrating.

Question 2

The surface SS is defined by z=ln⁡(x2+y2)z = \ln(x^2 + y^2) for 1≤x2+y2≤e21 \leq x^2 + y^2 \leq e^2. For evaluating ∬Sf(x,y,z) dS\iint_S f(x,y,z)\,dS where ff depends significantly on the distance from the z-axis, which parameterization strategy provides the most computational advantage?

  1. Cartesian: r(x,y)=(x,y,ln⁡(x2+y2))\mathbf{r}(x,y) = (x, y, \ln(x^2 + y^2)) with domain 1≤x2+y2≤e21 \leq x^2 + y^2 \leq e^2
  2. Cylindrical: r(r,θ)=(rcos⁡θ,rsin⁡θ,ln⁡r2)\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, \ln r^2) with 1≤r≤e1 \leq r \leq e, 0≤θ≤2π0 \leq \theta \leq 2\pi
  3. Cylindrical: r(r,θ)=(rcos⁡θ,rsin⁡θ,2ln⁡r)\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, 2\ln r) with 1≤r≤e1 \leq r \leq e, 0≤θ≤2π0 \leq \theta \leq 2\pi (correct answer)
  4. Polar-like: r(ρ,ϕ)=(ρcos⁡ϕ,ρsin⁡ϕ,ln⁡ρ)\mathbf{r}(\rho,\phi) = (\sqrt{\rho}\cos\phi, \sqrt{\rho}\sin\phi, \ln\rho) with 1≤ρ≤e21 \leq \rho \leq e^2, 0≤ϕ≤2π0 \leq \phi \leq 2\pi
Explanation: Since ff depends significantly on the distance from the z-axis (which is x2+y2=r\sqrt{x^2 + y^2} = r), cylindrical coordinates are most appropriate. The surface becomes z=ln⁡(r2)=2ln⁡rz = \ln(r^2) = 2\ln r, so the parameterization is r(r,θ)=(rcos⁡θ,rsin⁡θ,2ln⁡r)\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, 2\ln r) with 1≤r≤e1 \leq r \leq e (since 1≤x2+y2≤e21 \leq x^2 + y^2 \leq e^2 means 1≤r2≤e21 \leq r^2 \leq e^2, so 1≤r≤e1 \leq r \leq e) and 0≤θ≤2π0 \leq \theta \leq 2\pi. This makes the dependence on distance from the z-axis explicit through the parameter rr. Choice A uses Cartesian coordinates which obscures the radial structure. Choice B incorrectly writes z=ln⁡r2z = \ln r^2 without simplifying to 2ln⁡r2\ln r. Choice D introduces an unnecessarily complex substitution that doesn't simplify the problem.

Question 3

Consider the curve formed by the intersection of the ellipsoid x24+y29+z216=1\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} = 1 and the plane x+2y−z=0x + 2y - z = 0. To parameterize this curve for computing work done by a conservative vector field, which approach most efficiently handles the constraint while maintaining parameter bounds that are easy to determine?

  1. Substitute z=x+2yz = x + 2y into the ellipsoid equation, solve the resulting equation in xx and yy for yy as a function of xx, then parameterize using xx as the parameter
  2. Use x=2cos⁡tx = 2\cos t, y=3sin⁡sy = 3\sin s, solve for zz from the plane constraint, then find a relationship between tt and ss from the ellipsoid equation
  3. Parameterize the plane as p(u,v)=(u,v,u+2v)\mathbf{p}(u,v) = (u, v, u + 2v) and substitute into the ellipsoid equation to find the constraint relating uu and vv, then use this constraint to eliminate one parameter (correct answer)
  4. Set x=2rcos⁡θx = 2r\cos\theta, y=3rsin⁡θy = 3r\sin\theta, z=2rcos⁡θ+6rsin⁡θz = 2r\cos\theta + 6r\sin\theta, then solve for rr in terms of θ\theta using the ellipsoid equation
Explanation: Method C is most systematic. Parameterizing the plane as (u,v,u+2v)(u, v, u + 2v) and substituting into the ellipsoid gives u24+v29+(u+2v)216=1\frac{u^2}{4} + \frac{v^2}{9} + \frac{(u + 2v)^2}{16} = 1. Expanding: u24+v29+u2+4uv+4v216=1\frac{u^2}{4} + \frac{v^2}{9} + \frac{u^2 + 4uv + 4v^2}{16} = 1, which simplifies to 5u216+uv4+25v2144=1\frac{5u^2}{16} + \frac{uv}{4} + \frac{25v^2}{144} = 1. This can be written in the form of an ellipse and parameterized trigonometrically. Choice A leads to a complicated quartic relationship that's difficult to solve explicitly. Choice B introduces two parameters unnecessarily and requires finding a complex relationship between them. Choice D assumes a specific radial form that may not match the actual intersection curve and could lead to rr being multi-valued as a function of θ\theta.

Question 4

A space curve is defined implicitly as the intersection of x2+z2=9x^2 + z^2 = 9 and y2+z2=9y^2 + z^2 = 9. For parameterizing this curve to compute its arc length, which single-parameter approach avoids issues with undefined derivatives at specific points?

  1. Use z=3cos⁡tz = 3\cos t as the parameter, then x=±3sin⁡tx = \pm 3\sin t and y=±3sin⁡ty = \pm 3\sin t, choosing signs to maintain continuity (correct answer)
  2. Parameterize as r(t)=(3cos⁡t,3cos⁡t,3sin⁡t)\mathbf{r}(t) = (3\cos t, 3\cos t, 3\sin t) for t∈[0,2π]t \in [0, 2\pi]
  3. Use x=3cos⁡tx = 3\cos t, then z=±3sin⁡tz = \pm 3\sin t and y=±9−z2=±3∣cos⁡t∣y = \pm\sqrt{9 - z^2} = \pm 3|\cos t|
  4. Parameterize as r(t)=(3sin⁡t,3sin⁡t,3cos⁡t)\mathbf{r}(t) = (3\sin t, 3\sin t, 3\cos t) for t∈[0,2π]t \in [0, 2\pi]
Explanation: The intersection of x2+z2=9x^2 + z^2 = 9 and y2+z2=9y^2 + z^2 = 9 gives us x2=y2x^2 = y^2, so x=±yx = \pm y. This means the curve consists of two components: one where x=yx = y and one where x=−yx = -y. Both lie on circles of radius 3. Using z=3cos⁡tz = 3\cos t as parameter gives x2=9−z2=9sin⁡2tx^2 = 9 - z^2 = 9\sin^2 t and y2=9−z2=9sin⁡2ty^2 = 9 - z^2 = 9\sin^2 t, so x=±3sin⁡tx = \pm 3\sin t and y=±3sin⁡ty = \pm 3\sin t. To avoid derivative issues, we choose signs consistently for each component. Choice B only captures one component (x=yx = y). Choice C introduces absolute value functions that create derivative discontinuities at t=π2,3π2t = \frac{\pi}{2}, \frac{3\pi}{2}. Choice D also only captures one component and has the same limitation as Choice B.

Question 5

A curve in 3D space is defined as the intersection of the surfaces x2+y2=4zx^2 + y^2 = 4z and x+y+z=6x + y + z = 6. To parameterize this curve efficiently for computing a line integral, which single-parameter approach is most appropriate?

  1. Set x=2cos⁡tx = 2\cos t, y=2sin⁡ty = 2\sin t, solve for zz from the constraint cos⁡t+sin⁡t+z=6\cos t + \sin t + z = 6
  2. Set z=tz = t, then x=2tcos⁡θx = 2\sqrt{t}\cos\theta, y=2tsin⁡θy = 2\sqrt{t}\sin\theta where θ\theta satisfies 2t(cos⁡θ+sin⁡θ)+t=62\sqrt{t}(\cos\theta + \sin\theta) + t = 6
  3. Use the constraint to write z=6−x−yz = 6 - x - y, substitute into x2+y2=4zx^2 + y^2 = 4z to get (x+2)2+(y+2)2=32(x+2)^2 + (y+2)^2 = 32, then parameterize this circle (correct answer)
  4. Set x=26−tcos⁡(t)x = 2\sqrt{6-t}\cos(\sqrt{t}), y=26−tsin⁡(t)y = 2\sqrt{6-t}\sin(\sqrt{t}), z=tz = t for appropriate bounds on tt
Explanation: Substituting z=6−x−yz = 6 - x - y into x2+y2=4zx^2 + y^2 = 4z gives x2+y2=4(6−x−y)=24−4x−4yx^2 + y^2 = 4(6-x-y) = 24 - 4x - 4y. Rearranging: x2+4x+y2+4y=24x^2 + 4x + y^2 + 4y = 24. Completing the square: (x+2)2+(y+2)2=24+4+4=32(x+2)^2 + (y+2)^2 = 24 + 4 + 4 = 32. This is a circle with center (−2,−2)(-2, -2) and radius 424\sqrt{2}, parameterized as x=−2+42cos⁡tx = -2 + 4\sqrt{2}\cos t, y=−2+42sin⁡ty = -2 + 4\sqrt{2}\sin t, z=6−x−y=10−42(cos⁡t+sin⁡t)z = 6 - x - y = 10 - 4\sqrt{2}(\cos t + \sin t). Choice A incorrectly assumes the projection is a circle of radius 2. Choice B introduces an unnecessary second parameter. Choice D has an artificial and complicated relationship between parameters that doesn't simplify the problem.

Question 6

Consider parameterizing the surface z=xyz = xy over the region where x2+y2≤4x^2 + y^2 \leq 4 and x≥0x \geq 0. For computing the flux of the vector field F=zk\mathbf{F} = z\mathbf{k} through this surface, which parameterization choice will result in the normal vector pointing generally upward (positive z-component)?

  1. r(r,θ)=(rcos⁡θ,rsin⁡θ,r2cos⁡θsin⁡θ)\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, r^2\cos\theta\sin\theta) with 0≤r≤20 \leq r \leq 2, −π2≤θ≤π2-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}
  2. r(x,y)=(x,y,xy)\mathbf{r}(x,y) = (x, y, xy) with x2+y2≤4x^2 + y^2 \leq 4, x≥0x \geq 0, computed as rx×ry\mathbf{r}_x \times \mathbf{r}_y (correct answer)
  3. r(u,v)=(u,v,uv)\mathbf{r}(u,v) = (u, v, uv) with u2+v2≤4u^2 + v^2 \leq 4, u≥0u \geq 0, computed as rv×ru\mathbf{r}_v \times \mathbf{r}_u
  4. r(r,θ)=(rcos⁡θ,rsin⁡θ,r2cos⁡θsin⁡θ)\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, r^2\cos\theta\sin\theta) with 0≤r≤20 \leq r \leq 2, π2≤θ≤3π2\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2}
Explanation: For a surface parameterized as r(u,v)\mathbf{r}(u,v), the normal vector is ru×rv\mathbf{r}_u \times \mathbf{r}_v. For r(x,y)=(x,y,xy)\mathbf{r}(x,y) = (x, y, xy), we have rx=(1,0,y)\mathbf{r}_x = (1, 0, y) and ry=(0,1,x)\mathbf{r}_y = (0, 1, x). Thus rx×ry=(−y,−x,1)\mathbf{r}_x \times \mathbf{r}_y = (-y, -x, 1), which has a positive z-component. Choice A uses the correct region but the cross product rr×rθ\mathbf{r}_r \times \mathbf{r}_{\theta} may not consistently point upward. Choice C computes rv×ru\mathbf{r}_v \times \mathbf{r}_u instead of ru×rv\mathbf{r}_u \times \mathbf{r}_v, which reverses the orientation and gives a downward normal. Choice D parameterizes the wrong half of the disk (x≤0x \leq 0 instead of x≥0x \geq 0).

Question 7

You need to parameterize the part of the paraboloid z=x2+y2z = x^2 + y^2 that lies between the cylinders x2+y2=1x^2 + y^2 = 1 and x2+y2=4x^2 + y^2 = 4, but only in the region where x≥∣y∣x \geq |y|. Which parameterization correctly captures this region and is suitable for surface integration?

  1. r(r,θ)=(rcos⁡θ,rsin⁡θ,r2)\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, r^2) with 1≤r≤21 \leq r \leq 2 and −π4≤θ≤π4-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}
  2. r(r,θ)=(rcos⁡θ,rsin⁡θ,r2)\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, r^2) with 1≤r≤21 \leq r \leq 2 and θ∈[−π4,π4]∪[3π4,5π4]\theta \in [-\frac{\pi}{4}, \frac{\pi}{4}] \cup [\frac{3\pi}{4}, \frac{5\pi}{4}]
  3. r(u,v)=(u,v,u2+v2)\mathbf{r}(u,v) = (u, v, u^2 + v^2) with 1≤u2+v2≤41 \leq u^2 + v^2 \leq 4 and u≥∣v∣u \geq |v|
  4. r(r,θ)=(rcos⁡θ,rsin⁡θ,r2)\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, r^2) with 1≤r≤21 \leq r \leq 2 and −π4≤θ≤π4-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} or 3π4≤θ≤5π4\frac{3\pi}{4} \leq \theta \leq \frac{5\pi}{4} (correct answer)
Explanation: The condition x≥∣y∣x \geq |y| means x≥yx \geq y when y≥0y \geq 0 and x≥−yx \geq -y when y<0y < 0. In polar coordinates, this corresponds to two angular regions: the first quadrant and parts of the second and fourth quadrants where ∣cos⁡θ∣≥∣sin⁡θ∣|\cos\theta| \geq |\sin\theta|. For θ∈[0,π4]\theta \in [0, \frac{\pi}{4}], we have cos⁡θ≥sin⁡θ≥0\cos\theta \geq \sin\theta \geq 0. For θ∈[−π4,0]\theta \in [-\frac{\pi}{4}, 0], we have cos⁡θ≥−sin⁡θ≥0\cos\theta \geq -\sin\theta \geq 0. By symmetry, we also need θ∈[3π4,5π4]\theta \in [\frac{3\pi}{4}, \frac{5\pi}{4}] to capture the region where x≤0x \leq 0 and x≥∣y∣x \geq |y|. Choice A misses the third quadrant portion. Choice B uses incorrect union notation that suggests simultaneous constraints rather than separate intervals. Choice C uses Cartesian parameterization which is less natural for the annular region and makes integration more complex.