Multivariable Calculus Quiz: Selecting Coordinate Methods
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Selecting Coordinate MethodsQuestion 1 of 18

To find the mass of a solid region bounded below by the cone z=x2+y2z = \sqrt{x^2+y^2} and above by the sphere x2+y2+(z−1)2=1x^2+y^2+(z-1)^2=1, an integral of a density function δ(x,y,z)\delta(x,y,z) must be computed. The sphere's equation can be written in spherical coordinates as ρ=2cos⁡ϕ\rho = 2\cos\phi. Which coordinate system offers the most efficient setup for this volume integral, particularly because it avoids splitting the domain of integration?

Cylindrical, because the region has rotational symmetry about the zz-axis and the cone is simply z=rz=r.
Spherical, because the bounding surfaces are described by constant values or simple functions of the coordinates ρ\rho and ϕ\phi.
Cartesian, because the density function may be simpler in Cartesian coordinates, offsetting the complex boundary descriptions.
Spherical, but only if the sphere were centered at the origin; the vertical shift makes cylindrical coordinates more efficient.
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Selecting Coordinate Methods

Practice Selecting Coordinate Methods 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 Selecting Coordinate Methods, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

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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

To find the mass of a solid region bounded below by the cone z=x2+y2z = \sqrt{x^2+y^2} and above by the sphere x2+y2+(z−1)2=1x^2+y^2+(z-1)^2=1, an integral of a density function δ(x,y,z)\delta(x,y,z) must be computed. The sphere's equation can be written in spherical coordinates as ρ=2cos⁡ϕ\rho = 2\cos\phi. Which coordinate system offers the most efficient setup for this volume integral, particularly because it avoids splitting the domain of integration?

  1. Cylindrical, because the region has rotational symmetry about the zz-axis and the cone is simply z=rz=r.
  2. Spherical, because the bounding surfaces are described by constant values or simple functions of the coordinates ρ\rho and ϕ\phi. (correct answer)
  3. Cartesian, because the density function may be simpler in Cartesian coordinates, offsetting the complex boundary descriptions.
  4. Spherical, but only if the sphere were centered at the origin; the vertical shift makes cylindrical coordinates more efficient.
Explanation: The region is an 'ice cream cone' shape. In spherical coordinates, the cone z=x2+y2z = \sqrt{x^2+y^2} is given by ϕ=π/4\phi = \pi/4. The sphere x2+y2+z2−2z=0x^2+y^2+z^2-2z=0 becomes ρ2−2ρcos⁡ϕ=0\rho^2 - 2\rho\cos\phi = 0, which simplifies to ρ=2cos⁡ϕ\rho = 2\cos\phi. The integral can be set up as a single integral with limits θ∈[0,2π]\theta \in [0, 2\pi], ϕ∈[0,π/4]\phi \in [0, \pi/4], and ρ∈[0,2cos⁡ϕ]\rho \in [0, 2\cos\phi]. In cylindrical coordinates, the sphere is r2+(z−1)2=1r^2 + (z-1)^2 = 1, or z=1±1−r2z = 1 \pm \sqrt{1-r^2}. The intersection with the cone z=rz=r occurs at z=1,r=1z=1, r=1. Integrating over this region in cylindrical coordinates would require splitting the integral into two parts (one for the cone part of the lower boundary and one for the sphere part), making it less efficient.

Question 2

Consider the problem of finding the volume of the region bounded by the paraboloids z=x2+y2z = x^2+y^2 and z=8−(x2+y2)z = 8 - (x^2+y^2). Which of the following integral setups is most efficient for this calculation?

  1. A Cartesian integral ∫−22∫−4−x24−x2∫x2+y28−(x2+y2) dz dy dx\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{x^2+y^2}^{8-(x^2+y^2)} \, dz \, dy \, dx.
  2. A spherical integral where the boundaries are converted to complex functions of ρ\rho and ϕ\phi.
  3. A cylindrical integral ∫02π∫04∫r28−r2r dz dr dθ\int_0^{2\pi} \int_0^4 \int_{r^2}^{8-r^2} r \, dz \, dr \, d\theta.
  4. A cylindrical integral ∫02π∫02∫r28−r2r dz dr dθ\int_0^{2\pi} \int_0^2 \int_{r^2}^{8-r^2} r \, dz \, dr \, d\theta. (correct answer)
Explanation: When finding volumes bounded by surfaces, you should first visualize the region and choose coordinates that match the geometry's symmetry. Here, both paraboloids z=x2+y2z = x^2+y^2 and z=8−(x2+y2)z = 8 - (x^2+y^2) have circular symmetry around the z-axis, making cylindrical coordinates the natural choice. To find where the surfaces intersect, set them equal: x2+y2=8−(x2+y2)x^2+y^2 = 8 - (x^2+y^2), which gives 2(x2+y2)=82(x^2+y^2) = 8, so x2+y2=4x^2+y^2 = 4. This means the intersection occurs at a circle of radius 2 in the xy-plane. In cylindrical coordinates, this becomes r=2r = 2. The correct setup is D: ∫02π∫02∫r28−r2r dz dr dθ\int_0^{2\pi} \int_0^2 \int_{r^2}^{8-r^2} r \, dz \, dr \, d\theta. The region extends from θ=0\theta = 0 to 2π2\pi (full circle), r=0r = 0 to r=2r = 2 (radius of intersection), and zz from the lower paraboloid r2r^2 to the upper paraboloid 8−r28-r^2. Option A uses Cartesian coordinates, which works but requires more complex limits and doesn't exploit the circular symmetry. Option B mentions spherical coordinates, which are inappropriate since this isn't a problem involving spheres or radial distance from the origin. Option C has the wrong upper limit for rr—it uses 4 instead of 2, which would extend beyond the intersection boundary. Strategy tip: When you see problems involving surfaces with x2+y2x^2+y^2 terms, immediately consider cylindrical coordinates. Always find intersection curves first to determine your integration limits.

Question 3

An integral must be computed over a solid region EE formed by removing a cylinder of radius 1, centered along the zz-axis, from a cube defined by −2≤x≤2-2 \le x \le 2, −2≤y≤2-2 \le y \le 2, and −2≤z≤2-2 \le z \le 2. Which statement best describes the most effective approach to set up a calculation for the volume of EE?

  1. Use cylindrical coordinates, as the integral for rr from 11 to a function R(θ)R(\theta) representing the square boundary is most direct.
  2. Use Cartesian coordinates, as the integral must be split into five or more regions to avoid the central hole.
  3. Use spherical coordinates, which can represent both the planar faces of the cube and the cylindrical hole with appropriate equations.
  4. Use the principle of subtraction: calculate the volume of the cube and subtract the volume of the cylinder, avoiding a complex triple integral over EE. (correct answer)
Explanation: When faced with computing volumes of complex regions formed by combining or removing simpler shapes, always consider whether direct integration is necessary or if basic geometric principles can simplify the problem. The most efficient approach here uses the subtraction principle. The cube has volume 4×4×4=644 \times 4 \times 4 = 64, and the cylindrical hole has volume πr2h=π(1)2(4)=4π\pi r^2 h = \pi(1)^2(4) = 4\pi. Therefore, the volume of region EE is simply 64−4π64 - 4\pi. This avoids any complex integration setup entirely, making option D correct. Option A suggests cylindrical coordinates with rr from 1 to some function R(θ)R(\theta). While cylindrical coordinates seem natural due to the cylindrical hole, finding R(θ)R(\theta) to represent the square boundary is actually quite complex. The distance from the origin to the cube's edges varies dramatically with θ\theta, creating a complicated piecewise function. Option B proposes Cartesian coordinates with multiple regions. This approach would indeed work but requires splitting the domain into many pieces to handle the circular hole within the square cross-section—creating unnecessarily complex integration bounds and multiple integrals. Option C suggests spherical coordinates, but this is the least suitable choice. Neither the cube's planar faces nor the cylindrical hole align naturally with spherical coordinate surfaces, making the integration bounds extremely complicated. Strategy tip: Before setting up any complex triple integral, always check if the region can be expressed as a combination of simpler shapes whose volumes you can calculate directly using basic formulas.

Question 4

To evaluate the integral of f(x,y,z)=e−(x2+y2+z2)3/2f(x,y,z) = e^{-(x^2+y^2+z^2)^{3/2}} over the region EE which is the portion of the ball x2+y2+z2≤4x^2+y^2+z^2 \le 4 in the first octant, which coordinate system is most advantageous?

  1. Cartesian, because the first octant is defined by simple constant bounds x≥0,y≥0,z≥0x \ge 0, y \ge 0, z \ge 0.
  2. Cylindrical, as it simplifies the x2+y2x^2+y^2 term in the exponent to r2r^2.
  3. Spherical, because both the integration region (a sector of a ball) and the integrand (a function of ρ\rho) become very simple. (correct answer)
  4. Spherical, but the integrand becomes e−ρ3sin⁡3ϕe^{-\rho^3 \sin^3\phi}, which is too complex for practical integration.
Explanation: The region is a sector of a sphere of radius 2 in the first octant. In spherical coordinates, this is described by constant limits: 0≤ρ≤20 \le \rho \le 2, 0≤ϕ≤π/20 \le \phi \le \pi/2, and 0≤θ≤π/20 \le \theta \le \pi/2. The integrand becomes e−(ρ2)3/2=e−ρ3e^{-(\rho^2)^{3/2}} = e^{-\rho^3}. The integral setup is ∫0π/2∫0π/2∫02e−ρ3ρ2sin⁡ϕ dρ dϕ dθ\int_0^{\pi/2} \int_0^{\pi/2} \int_0^2 e^{-\rho^3} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta. This integral is separable and straightforward to evaluate. While cylindrical coordinates simplify part of the integrand to e−(r2+z2)3/2e^{-(r^2+z^2)^{3/2}}, the spherical boundary r2+z2=4r^2+z^2=4 still complicates the limits. Cartesian coordinates are not suitable for the spherical boundary or the integrand.

Question 5

A calculation requires integrating f(x,y,z)=x2+y2f(x,y,z) = \sqrt{x^2+y^2} over the region bounded by the cylinder x2+y2=9x^2+y^2=9, below by the plane z=0z=0, and above by the plane z=x+4z=x+4. Which coordinate system and corresponding limits of integration are most appropriate?

  1. Cylindrical, with 0≤θ≤2π0 \le \theta \le 2\pi, 0≤r≤30 \le r \le 3, and 0≤z≤rcos⁡θ+40 \le z \le r\cos\theta+4. (correct answer)
  2. Cylindrical, with 0≤θ≤2π0 \le \theta \le 2\pi, 0≤r≤30 \le r \le 3, and 0≤z≤40 \le z \le 4.
  3. Spherical, with limits determined by converting the cylinder and planes to functions of ρ,ϕ,θ\rho, \phi, \theta.
  4. Cartesian, with −9−x2≤y≤9−x2-\sqrt{9-x^2} \le y \le \sqrt{9-x^2} making the integration straightforward.
Explanation: The region is a standard cylinder whose top is a slanted plane. This structure is ideal for cylindrical coordinates. The cylinder x2+y2=9x^2+y^2=9 gives the limit 0≤r≤30 \le r \le 3. The region covers all angles, so 0≤θ≤2π0 \le \theta \le 2\pi. The bottom is z=0z=0. The top is z=x+4z=x+4, which becomes z=rcos⁡θ+4z=r\cos\theta+4. The integrand x2+y2\sqrt{x^2+y^2} becomes rr. All parts of the problem are simplified in cylindrical coordinates, as described in choice A. Choice B incorrectly assumes the top plane is z=4z=4. Spherical coordinates would be very complicated for the plane z=x+4z=x+4 and the cylinder. Cartesian coordinates would involve integrals with 9−x2\sqrt{9-x^2}, which are more difficult than the cylindrical setup.

Question 6

You need to set up an iterated integral for a function ff over the region EE defined by the cube [1,2]×[1,2]×[1,2][1, 2] \times [1, 2] \times [1, 2]. The function to be integrated is f(x,y,z)=(x2+y2+z2)−1f(x,y,z) = (x^2+y^2+z^2)^{-1}. Which coordinate system allows for the simplest representation of the limits of integration?

  1. Spherical coordinates, because the integrand simplifies to ρ−2\rho^{-2}, making the overall calculation easier.
  2. Cylindrical coordinates, as a compromise that simplifies the x2+y2x^2+y^2 term while keeping the zz coordinate.
  3. Cartesian coordinates, because the cubic region is defined by constant bounds for xx, yy, and zz. (correct answer)
  4. A non-standard linear transformation to map the cube to the origin before applying spherical coordinates.
Explanation: The question specifically asks for the simplest limits of integration. The region is a cube defined by 1≤x≤21 \le x \le 2, 1≤y≤21 \le y \le 2, and 1≤z≤21 \le z \le 2. In Cartesian coordinates, the setup is ∫12∫12∫12f(x,y,z) dz dy dx\int_1^2 \int_1^2 \int_1^2 f(x,y,z) \, dz \, dy \, dx. The limits are all constants, which is the simplest possible form. In contrast, converting the six planar boundaries of the cube into spherical or cylindrical coordinates (e.g., x=1x=1 becomes ρsin⁡ϕcos⁡θ=1\rho\sin\phi\cos\theta=1) results in extremely complicated, non-constant limits. The simplicity of the integrand in spherical coordinates (distractor A) is a trap; it does not compensate for the immense complexity of the domain's boundaries.

Question 7

To evaluate the Cartesian integral ∫03∫09−x211+x2+y2 dy dx\int_0^3 \int_0^{\sqrt{9-x^2}} \frac{1}{1+x^2+y^2} \, dy \, dx, one should convert to a different coordinate system. Which of the following represents the most effective conversion strategy?

  1. Convert to polar coordinates, with the integral becoming ∫0π/2∫03r1+r2 dr dθ\int_0^{\pi/2} \int_0^3 \frac{r}{1+r^2} \, dr \, d\theta. (correct answer)
  2. Convert to polar coordinates, with the integral becoming ∫0π∫0311+r2 dr dθ\int_0^{\pi} \int_0^3 \frac{1}{1+r^2} \, dr \, d\theta.
  3. Remain in Cartesian coordinates but reverse the order of integration to ∫03∫09−y211+x2+y2 dx dy\int_0^3 \int_0^{\sqrt{9-y^2}} \frac{1}{1+x^2+y^2} \, dx \, dy.
  4. Convert to polar coordinates, with the integral becoming ∫0π/2∫0311+r2 dr dθ\int_0^{\pi/2} \int_0^3 \frac{1}{1+r^2} \, dr \, d\theta.
Explanation: The region of integration is described by 0≤x≤30 \le x \le 3 and 0≤y≤9−x20 \le y \le \sqrt{9-x^2}. This is the quarter of the circle x2+y2=9x^2+y^2=9 that lies in the first quadrant. In polar coordinates, this region is 0≤r≤30 \le r \le 3 and 0≤θ≤π/20 \le \theta \le \pi/2. The integrand 11+x2+y2\frac{1}{1+x^2+y^2} becomes 11+r2\frac{1}{1+r^2}. The area element dy dxdy \, dx becomes r dr dθr \, dr \, d\theta. Therefore, the correct converted integral is ∫0π/2∫03r1+r2 dr dθ\int_0^{\pi/2} \int_0^3 \frac{r}{1+r^2} \, dr \, d\theta. Distractor D omits the Jacobian factor rr, a common error. Distractor B uses the wrong range for θ\theta. Distractor C suggests reversing the order, which does not simplify the problem.

Question 8

An optimization problem seeks the maximum value of f(x,y)=16−x2−y2f(x,y) = \sqrt{16 - x^2 - y^2} subject to the constraint x2+y2≥4x^2 + y^2 \geq 4. The domain is an annular region, and both the objective function and constraint suggest coordinate transformation. Which method most efficiently handles both the constraint geometry and the optimization analysis?

  1. Polar coordinates with x=rcos⁡θ,y=rsin⁡θx = r\cos\theta, y = r\sin\theta, converting to single-variable optimization in rr (correct answer)
  2. Cartesian coordinates with Lagrange multipliers on the constraint g(x,y)=x2+y2−4=0g(x,y) = x^2 + y^2 - 4 = 0
  3. Parametric representation x=2cos⁡θ,y=2sin⁡θx = 2\cos\theta, y = 2\sin\theta to focus only on the constraint boundary
  4. Substitution u=x2+y2u = x^2 + y^2 to reduce to optimization in terms of uu with 4≤u≤164 \leq u \leq 16
Explanation: Polar coordinates transform the problem to maximizing f=16−r2f = \sqrt{16 - r^2} subject to r≥2r \geq 2. This becomes a simple single-variable problem: since ff decreases as rr increases, the maximum occurs at r=2r = 2 (the inner boundary). The θ\theta variable disappears due to rotational symmetry. Choice B only considers the boundary. Choice C also ignores the interior. Choice D is clever but doesn't handle the two-variable geometry as cleanly as polar coordinates.

Question 9

A line integral ∫CF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r} is computed along the curve CC defined by the intersection of the cylinder x2+y2=4x^2 + y^2 = 4 and the plane z=x+yz = x + y from (2,0,2)(2,0,2) to (−2,0,−2)(-2,0,-2). The vector field is F=⟨yz,xz,xy⟩\mathbf{F} = \langle yz, xz, xy \rangle. Which parametric approach minimizes the algebraic complexity of F⋅r′(t)\mathbf{F} \cdot \mathbf{r}'(t)?

  1. Cartesian parametrization r(t)=(2cos⁡t,2sin⁡t,2cos⁡t+2sin⁡t)\mathbf{r}(t) = (2\cos t, 2\sin t, 2\cos t + 2\sin t) for t∈[0,π]t \in [0, \pi] (correct answer)
  2. Cylindrical approach with x=2cos⁡t,y=2sin⁡t,z=2cos⁡t+2sin⁡tx = 2\cos t, y = 2\sin t, z = 2\cos t + 2\sin t and θ=t\theta = t
  3. Parametrization using arc length ss along the cylindrical helix
  4. Direct integration using xx as parameter: y=±4−x2,z=x±4−x2y = \pm\sqrt{4-x^2}, z = x \pm \sqrt{4-x^2}
Explanation: Choice A gives r′(t)=(−2sin⁡t,2cos⁡t,−2sin⁡t+2cos⁡t)\mathbf{r}'(t) = (-2\sin t, 2\cos t, -2\sin t + 2\cos t) and F=(2sin⁡t(2cos⁡t+2sin⁡t),2cos⁡t(2cos⁡t+2sin⁡t),4cos⁡tsin⁡t)\mathbf{F} = (2\sin t(2\cos t + 2\sin t), 2\cos t(2\cos t + 2\sin t), 4\cos t \sin t). The dot product simplifies systematically using trigonometric identities. Choice B is identical to A in different notation. Choice C introduces unnecessary complexity with arc length. Choice D requires handling the ±\pm signs carefully and splits the path into two pieces, creating more algebraic work.

Question 10

The divergence theorem is applied to find the flux of F=⟨x3,y3,z3⟩\mathbf{F} = \langle x^3, y^3, z^3 \rangle through the boundary of the region inside both the cone z2=x2+y2z^2 = x^2 + y^2 and the sphere x2+y2+z2=8x^2 + y^2 + z^2 = 8, with z≥0z \geq 0. Computing ∇⋅F=3(x2+y2+z2)\nabla \cdot \mathbf{F} = 3(x^2 + y^2 + z^2) over this region requires careful coordinate choice. Which system makes the volume integration most tractable?

  1. Cartesian coordinates with careful attention to the intersection curve of cone and sphere
  2. Cylindrical coordinates where the region becomes {(r,θ,z):0≤θ≤2π,0≤r≤2,r≤z≤8−r2}\{(r,\theta,z) : 0 \leq \theta \leq 2\pi, 0 \leq r \leq 2, r \leq z \leq \sqrt{8-r^2}\}
  3. Spherical coordinates where the cone constraint becomes 0≤ϕ≤π/40 \leq \phi \leq \pi/4 and ρ≤22\rho \leq 2\sqrt{2} (correct answer)
  4. Cylindrical coordinates with integration bounds 0≤r≤8,r≤z≤8−r20 \leq r \leq \sqrt{8}, r \leq z \leq \sqrt{8-r^2}, but this exceeds the cone constraint
Explanation: In spherical coordinates, ∇⋅F=3ρ2\nabla \cdot \mathbf{F} = 3\rho^2 and the volume element is ρ2sin⁡ϕ dρ dϕ dθ\rho^2 \sin\phi \, d\rho \, d\phi \, d\theta, so the integrand becomes 3ρ4sin⁡ϕ3\rho^4 \sin\phi. The cone z2=x2+y2z^2 = x^2 + y^2 with z≥0z \geq 0 translates to ϕ=π/4\phi = \pi/4, and the sphere gives ρ=22\rho = 2\sqrt{2}. The integration bounds are clean: 0≤ϕ≤π/40 \leq \phi \leq \pi/4, 0≤ρ≤220 \leq \rho \leq 2\sqrt{2}, 0≤θ≤2π0 \leq \theta \leq 2\pi. Choice B has correct cylindrical setup but more complex bounds. Choice D incorrectly suggests rr can exceed the cone constraint.

Question 11

A triple integral computes the moment of inertia I=∭E(x2+y2)ρ(x,y,z) dVI = \iiint_E (x^2 + y^2) \rho(x,y,z) \, dV where EE is the solid region between the spheres x2+y2+z2=1x^2 + y^2 + z^2 = 1 and x2+y2+z2=4x^2 + y^2 + z^2 = 4, and the density is ρ(x,y,z)=x2+y2+z2\rho(x,y,z) = \sqrt{x^2 + y^2 + z^2}. Multiple coordinate systems could work, but which choice optimizes both the region description and the integrand simplification?

  1. Cartesian coordinates to maintain the exact form of x2+y2x^2 + y^2 in the integrand
  2. Cylindrical coordinates where x2+y2=r2x^2 + y^2 = r^2 and the region becomes {(r,θ,z):r2+z2≤4,r2+z2≥1}\{(r,\theta,z) : r^2 + z^2 \leq 4, r^2 + z^2 \geq 1\}
  3. Spherical coordinates where ρ=x2+y2+z2\rho = \sqrt{x^2 + y^2 + z^2} and x2+y2=ρ2sin⁡2ϕx^2 + y^2 = \rho^2\sin^2\phi (correct answer)
  4. Spherical coordinates with integration order dρ dθ dϕd\rho \, d\theta \, d\phi to handle the radial density first
Explanation: Spherical coordinates are optimal because the region becomes simply 1≤ρ≤21 \leq \rho \leq 2, the density becomes ρ\rho, and the moment of inertia integrand becomes (ρ2sin⁡2ϕ)⋅ρ⋅ρ2sin⁡ϕ=ρ5sin⁡3ϕ(\rho^2\sin^2\phi) \cdot \rho \cdot \rho^2\sin\phi = \rho^5\sin^3\phi. The ρ\rho integration is straightforward, and the ϕ\phi and θ\theta integrations are standard. Choice B makes the region bounds more complex (r2+z2r^2 + z^2 constraints). Choice D mentions integration order but doesn't change the coordinate advantage. Choice A provides no simplification for either region or integrand.

Question 12

Green's theorem is used to evaluate ∮C(x3−y3) dx+(x3+y3) dy\oint_C (x^3 - y^3) \, dx + (x^3 + y^3) \, dy where CC is the boundary of the region enclosed by the polar curve r=2+cos⁡(3θ)r = 2 + \cos(3\theta). Converting to the area integral ∬R∂Q∂x−∂P∂y dA\iint_R \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \, dA requires choosing coordinates for the resulting double integral. Which approach handles the region boundary most efficiently?

  1. Polar coordinates with bounds 0≤θ≤2π0 \leq \theta \leq 2\pi and 0≤r≤2+cos⁡(3θ)0 \leq r \leq 2 + \cos(3\theta) (correct answer)
  2. Cartesian coordinates after converting the polar boundary to xx-yy form
  3. Polar coordinates with the substitution u=cos⁡(3θ)u = \cos(3\theta) to linearize the radial bound
  4. Cartesian coordinates with careful piecewise integration over different sectors of the region
Explanation: Since ∂Q∂x−∂P∂y=∂∂x(x3+y3)−∂∂y(x3−y3)=3x2+3y2=3(x2+y2)=3r2\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{\partial}{\partial x}(x^3 + y^3) - \frac{\partial}{\partial y}(x^3 - y^3) = 3x^2 + 3y^2 = 3(x^2 + y^2) = 3r^2, polar coordinates give a clean integrand. The region is naturally described by 0≤r≤2+cos⁡(3θ)0 \leq r \leq 2 + \cos(3\theta) with 0≤θ≤2π0 \leq \theta \leq 2\pi. The integral becomes ∫02π∫02+cos⁡(3θ)3r2⋅r dr dθ=∫02π∫02+cos⁡(3θ)3r3 dr dθ\int_0^{2\pi} \int_0^{2+\cos(3\theta)} 3r^2 \cdot r \, dr \, d\theta = \int_0^{2\pi} \int_0^{2+\cos(3\theta)} 3r^3 \, dr \, d\theta. Choice B would require solving a sixth-degree polynomial. Choice C doesn't help with the integration. Choice D overcomplificates the region description.

Question 13

A double integral ∬Rex2+y2 dA\iint_R e^{x^2+y^2} \, dA is evaluated over the region RR bounded by the curves x2+y2=1x^2 + y^2 = 1 and x2+y2=9x^2 + y^2 = 9, but only in the first quadrant where x≥yx \geq y. The integrand suggests polar coordinates, but the constraint x≥yx \geq y requires careful analysis. What are the correct integration bounds?

  1. ∫0π/4∫13er2r dr dθ\int_0^{\pi/4} \int_1^3 e^{r^2} r \, dr \, d\theta since x≥yx \geq y means cos⁡θ≥sin⁡θ\cos\theta \geq \sin\theta
  2. ∫0π/2∫13er2r dr dθ\int_0^{\pi/2} \int_1^3 e^{r^2} r \, dr \, d\theta for the entire first quadrant, then subtract the y>xy > x portion
  3. ∫13∫0arccos⁡(r)er2r dθ dr\int_1^3 \int_0^{\arccos(r)} e^{r^2} r \, d\theta \, dr to handle the varying θ\theta limits as functions of rr
  4. ∫0π/4∫13er2r dr dθ\int_0^{\pi/4} \int_1^3 e^{r^2} r \, dr \, d\theta since the first quadrant with x≥yx \geq y corresponds to 0≤θ≤π/40 \leq \theta \leq \pi/4 (correct answer)
Explanation: When you encounter a double integral with an exponential of x2+y2x^2 + y^2, polar coordinates are your natural choice since x2+y2=r2x^2 + y^2 = r^2. The key challenge here is translating the constraint x≥yx \geq y into polar coordinates. In polar coordinates, x=rcos⁡θx = r\cos\theta and y=rsin⁡θy = r\sin\theta. The condition x≥yx \geq y becomes rcos⁡θ≥rsin⁡θr\cos\theta \geq r\sin\theta. Since r>0r > 0 in our region (between the circles), we can divide both sides by rr to get cos⁡θ≥sin⁡θ\cos\theta \geq \sin\theta. In the first quadrant, cos⁡θ≥sin⁡θ\cos\theta \geq \sin\theta when θ\theta ranges from 00 to π/4\pi/4. At θ=π/4\theta = \pi/4, we have cos⁡(π/4)=sin⁡(π/4)\cos(\pi/4) = \sin(\pi/4), which is exactly the line x=yx = y. For θ<π/4\theta < \pi/4, we have x>yx > y, and for θ>π/4\theta > \pi/4, we have x<yx < y. The radial bounds are straightforward: rr goes from 11 to 33 (the inner and outer circles). This gives us ∫0π/4∫13er2r dr dθ\int_0^{\pi/4} \int_1^3 e^{r^2} r \, dr \, d\theta. Choice A states the correct setup but doesn't clearly explain the reasoning. Choice B incorrectly suggests integrating over the entire first quadrant. Choice C has the order of integration reversed and uses an incorrect upper limit for θ\theta. Choice D provides both the correct bounds and clear reasoning. Study tip: When converting geometric constraints to polar coordinates, always express them in terms of θ\theta by substituting x=rcos⁡θx = r\cos\theta and y=rsin⁡θy = r\sin\theta, then solve for the angle range.

Question 14

A physics problem requires computing the gravitational potential at point (0,0,6)(0, 0, 6) due to a mass distribution with density ρ(x,y,z)=1(x2+y2+z2)1/2\rho(x,y,z) = \frac{1}{(x^2 + y^2 + z^2)^{1/2}} over the region 1≤x2+y2+z2≤81 \leq x^2 + y^2 + z^2 \leq 8. The integral involves ρ(x,y,z)x2+y2+(z−6)2\frac{\rho(x,y,z)}{\sqrt{x^2 + y^2 + (z-6)^2}}. Which coordinate choice best handles both the density function and the distance calculation?

  1. Cartesian coordinates to avoid coordinate transformation errors in the distance formula
  2. Cylindrical coordinates centered at (0,0,0)(0,0,0) to simplify the density function
  3. Spherical coordinates centered at (0,0,0)(0,0,0) to simplify both the region and density (correct answer)
  4. Spherical coordinates centered at (0,0,6)(0,0,6) to simplify the distance calculation
Explanation: The region 1≤x2+y2+z2≤81 \leq x^2 + y^2 + z^2 \leq 8 is naturally spherical, and the density ρ=1/x2+y2+z2=1/ρ\rho = 1/\sqrt{x^2 + y^2 + z^2} = 1/\rho becomes extremely simple in spherical coordinates. While the distance to (0,0,6)(0,0,6) doesn't simplify perfectly, the overall integrand complexity is minimized because the region boundaries and density are so simple. Choice D would complicate the density function and region description. Choice B helps with density but not the 3D region. Choice A gives no simplification advantages.

Question 15

Consider evaluating the surface integral ∬S(x2+y2)3/2 dS\iint_S (x^2 + y^2)^{3/2} \, dS where SS is the portion of the paraboloid z=9−x2−y2z = 9 - x^2 - y^2 above the xy-plane. The choice of coordinate system significantly affects the complexity of dSdS. Which approach minimizes the computational burden?

  1. Cartesian coordinates with dS=1+(∂z/∂x)2+(∂z/∂y)2 dx dydS = \sqrt{1 + (\partial z/\partial x)^2 + (\partial z/\partial y)^2} \, dx \, dy
  2. Cylindrical coordinates with x2+y2=r2x^2 + y^2 = r^2 and dS=r1+4r2 dr dθdS = r\sqrt{1 + 4r^2} \, dr \, d\theta (correct answer)
  3. Spherical coordinates to exploit the radial symmetry of (x2+y2)3/2(x^2 + y^2)^{3/2}
  4. Parametric coordinates using r(u,v)=(ucos⁡v,usin⁡v,9−u2)\mathbf{r}(u,v) = (u\cos v, u\sin v, 9-u^2) with cross-product method
Explanation: In cylindrical coordinates, the integrand (x2+y2)3/2(x^2 + y^2)^{3/2} becomes r3r^3, which is very simple. The paraboloid becomes z=9−r2z = 9 - r^2, and the surface element is dS=r1+4r2 dr dθdS = r\sqrt{1 + 4r^2} \, dr \, d\theta (from ∂z/∂r=−2r\partial z/\partial r = -2r). The θ\theta integration gives 2π2\pi immediately. Choice A gives the same surface element but keeps the integrand as (x2+y2)3/2(x^2 + y^2)^{3/2}. Choice C doesn't naturally fit this geometry. Choice D is equivalent to cylindrical but uses more complex notation.

Question 16

A surface integral ∬SF⋅n dS\iint_S \mathbf{F} \cdot \mathbf{n} \, dS is computed over the hemisphere x2+y2+z2=16,z≥0x^2 + y^2 + z^2 = 16, z \geq 0 where F=⟨x2z,y2z,z3⟩\mathbf{F} = \langle x^2z, y^2z, z^3 \rangle. The calculation requires both parametrizing the surface and computing the unit normal vector. Which parametric choice minimizes the complexity of the normal vector calculation while maintaining simple bounds?

  1. Cartesian parametrization r(x,y)=(x,y,16−x2−y2)\mathbf{r}(x,y) = (x, y, \sqrt{16-x^2-y^2}) over x2+y2≤16x^2 + y^2 \leq 16
  2. Spherical parametrization r(ϕ,θ)=(4sin⁡ϕcos⁡θ,4sin⁡ϕsin⁡θ,4cos⁡ϕ)\mathbf{r}(\phi,\theta) = (4\sin\phi\cos\theta, 4\sin\phi\sin\theta, 4\cos\phi) with 0≤ϕ≤π/20 \leq \phi \leq \pi/2 (correct answer)
  3. Cylindrical parametrization r(r,θ)=(rcos⁡θ,rsin⁡θ,16−r2)\mathbf{r}(r,\theta) = (r\cos\theta, r\sin\theta, \sqrt{16-r^2}) with 0≤r≤40 \leq r \leq 4
  4. Spherical parametrization using r(θ,ϕ)=(4cos⁡θsin⁡ϕ,4sin⁡θsin⁡ϕ,4cos⁡ϕ)\mathbf{r}(\theta,\phi) = (4\cos\theta\sin\phi, 4\sin\theta\sin\phi, 4\cos\phi) to align with physics conventions
Explanation: For a sphere, spherical parametrization gives the normal vector most directly: rϕ×rθ=16sin⁡ϕ(sin⁡ϕcos⁡θ,sin⁡ϕsin⁡θ,cos⁡ϕ)\mathbf{r}_{\phi} \times \mathbf{r}_{\theta} = 16\sin\phi(\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi), which points outward and has magnitude 16sin⁡ϕ16\sin\phi. The unit normal is simply (sin⁡ϕcos⁡θ,sin⁡ϕsin⁡θ,cos⁡ϕ)(\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi), matching the position vector direction. Choice A requires computing partial derivatives of 16−x2−y2\sqrt{16-x^2-y^2}, creating expressions with square roots. Choice C has similar issues. Choice D is equivalent to B with different parameter labeling.

Question 17

A triple integral is needed to find the volume of the region bounded by the cone z2=x2+y2z^2 = x^2 + y^2 and the sphere x2+y2+z2=4x^2 + y^2 + z^2 = 4. Which coordinate system would require the least computational effort while avoiding the most complex integrand expressions?

  1. Cartesian coordinates, since the sphere equation is already in standard form
  2. Cylindrical coordinates, since the cone becomes z2=r2z^2 = r^2 and the sphere becomes r2+z2=4r^2 + z^2 = 4 (correct answer)
  3. Spherical coordinates, since the sphere becomes ρ=2\rho = 2 and eliminates one variable
  4. Cylindrical coordinates with the transformation x=rsin⁡θ,y=rcos⁡θx = r\sin\theta, y = r\cos\theta to align with cone symmetry
Explanation: Cylindrical coordinates are optimal here because both surfaces have natural expressions: the cone becomes z=±rz = \pm r and the sphere becomes r2+z2=4r^2 + z^2 = 4. The region has rotational symmetry about the z-axis, making the θ\theta integration trivial (00 to 2π2\pi). While spherical coordinates simplify the sphere to ρ=2\rho = 2, the cone constraint becomes more complex as cos⁡ϕ=±12\cos\phi = \pm\frac{1}{\sqrt{2}}. Cartesian coordinates would require solving for intersection curves algebraically.

Question 18

Consider the integral of f(x,y,z)=zf(x,y,z) = z over the region defined by x2+y2≤z≤2−(x2+y2)x^2+y^2 \le z \le \sqrt{2-(x^2+y^2)}. Which coordinate system is most suitable for setting up this integral?

  1. Spherical, because the upper boundary z=2−(x2+y2)z = \sqrt{2-(x^2+y^2)} is part of a sphere.
  2. Cylindrical, because the region is solid of revolution about the zz-axis and boundaries are functions of rr. (correct answer)
  3. Cartesian, because the integrand is simply zz, making the inner integral trivial.
  4. Cylindrical, but the integral must be split because one boundary is a paraboloid and the other is a sphere.
Explanation: The lower boundary is the paraboloid z=x2+y2z = x^2+y^2, or z=r2z=r^2 in cylindrical coordinates. The upper boundary is the hemisphere z=2−r2z = \sqrt{2-r^2}, part of the sphere x2+y2+z2=2x^2+y^2+z^2=2. The region is a solid of revolution, best handled by cylindrical coordinates. The intersection of the surfaces occurs when r2=2−r2r^2 = \sqrt{2-r^2}, which leads to r4+r2−2=0r^4+r^2-2=0, or (r2+2)(r2−1)=0(r^2+2)(r^2-1)=0. Thus, they intersect at r=1r=1. The integral can be set up as a single integral: ∫02π∫01∫r22−r2z⋅r dz dr dθ\int_0^{2\pi} \int_0^1 \int_{r^2}^{\sqrt{2-r^2}} z \cdot r \, dz \, dr \, d\theta. Choice A is incorrect because the lower boundary (paraboloid) is not simple in spherical coordinates. Choice C is inefficient due to the circular nature of the boundaries. Choice D is incorrect because a single integral in cylindrical coordinates can describe the region; the limits for zz simply depend on rr.