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Triple Integrals: Spherical - Multivariable Calculus

  1. Evaluate the integral ∭E(x2+y2+z2)−3/2 dV\iiint_E (x^2+y^2+z^2)^{-3/2} \, dV, where EE is the solid region between the spheres of radius aa and bb centered at the origin, with 0<a<b0 < a < b.
    Answer: 4πln⁡(b/a)4\pi \ln(b/a) In spherical coordinates, the integrand (x2+y2+z2)−3/2(x^2+y^2+z^2)^{-3/2} becomes (ρ2)−3/2=ρ−3(\rho^2)^{-3/2} = \rho^{-3}. The volume element is dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta. The region EE is described by a≤ρ≤ba \le \rho \le b, 0≤ϕ≤π0 \le \phi \le \pi, and 0≤θ≤2π0 \le \theta \le 2\pi. The integral becomes: ∫02π∫0π∫ab(ρ−3)ρ2sin⁡ϕ dρ dϕ dθ=∫02π∫0π∫absin⁡ϕρ dρ dϕ dθ\int_0^{2\pi} \int_0^{\pi} \int_a^b (\rho^{-3}) \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta = \int_0^{2\pi} \int_0^{\pi} \int_a^b \frac{\sin\phi}{\rho} \, d\rho \, d\phi \, d\theta This is a separable integral: (∫02πdθ)(∫0πsin⁡ϕ dϕ)(∫ab1ρ dρ)=(2π)⋅[−cos⁡ϕ]0π⋅[ln⁡∣ρ∣]ab\left(\int_0^{2\pi} d\theta\right) \left(\int_0^{\pi} \sin\phi \, d\phi\right) \left(\int_a^b \frac{1}{\rho} \, d\rho\right) = (2\pi) \cdot [-\cos\phi]_0^{\pi} \cdot [\ln|\rho|]_a^b =(2π)⋅(1−(−1))⋅(ln⁡b−ln⁡a)=4πln⁡(b/a)= (2\pi) \cdot (1 - (-1)) \cdot (\ln b - \ln a) = 4\pi \ln(b/a)
  2. What is the value of the integral ∫−22∫−4−x24−x2∫x2+y28−x2−y2(x2+y2+z2) dz dy dx\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{8-x^2-y^2}} (x^2+y^2+z^2) \, dz \, dy \, dx?
    Answer: 256π5(2−1)\frac{256\pi}{5}(\sqrt{2}-1) The region of integration is best described in spherical coordinates. The lower bound for zz is z=x2+y2z=\sqrt{x^2+y^2}, which is the cone ϕ=π/4\phi=\pi/4. The upper bound for zz is z=8−x2−y2z=\sqrt{8-x^2-y^2}, which is the upper hemisphere of the sphere x2+y2+z2=8x^2+y^2+z^2=8, so ρ=8=22\rho=\sqrt{8}=2\sqrt{2}. The projection on the xyxy-plane is the disk x2+y2≤4x^2+y^2 \le 4, which is fully covered by the cone and sphere intersection. Thus, the limits are 0≤ρ≤220 \le \rho \le 2\sqrt{2}, 0≤ϕ≤π/40 \le \phi \le \pi/4, and 0≤θ≤2π0 \le \theta \le 2\pi. The integrand x2+y2+z2x^2+y^2+z^2 is ρ2\rho^2. The integral becomes: ∫02π∫0π/4∫022(ρ2)ρ2sin⁡ϕ dρ dϕ dθ=∫02πdθ∫0π/4sin⁡ϕ dϕ∫022ρ4 dρ\int_0^{2\pi} \int_0^{\pi/4} \int_0^{2\sqrt{2}} (\rho^2) \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta = \int_0^{2\pi} d\theta \int_0^{\pi/4} \sin\phi \, d\phi \int_0^{2\sqrt{2}} \rho^4 \, d\rho =(2π)⋅[−cos⁡ϕ]0π/4⋅[ρ55]022=(2π)(−22−(−1))(22)55= (2\pi) \cdot [-\cos\phi]_0^{\pi/4} \cdot \left[\frac{\rho^5}{5}\right]_0^{2\sqrt{2}} = (2\pi) \left(-\frac{\sqrt{2}}{2} - (-1)\right) \frac{(2\sqrt{2})^5}{5} =(2π)(1−22)12825=256π25(1−22)=256π5(2−1)= (2\pi) \left(1 - \frac{\sqrt{2}}{2}\right) \frac{128\sqrt{2}}{5} = \frac{256\pi\sqrt{2}}{5} \left(1 - \frac{\sqrt{2}}{2}\right) = \frac{256\pi}{5}(\sqrt{2}-1)
  3. Consider the solid WW defined by 1≤x2+y2+z2≤41 \leq x^2 + y^2 + z^2 \leq 4 and z≤−x2+y2z \leq -\sqrt{x^2 + y^2}. When setting up the triple integral ∭WdV\iiint_W dV in spherical coordinates, which of the following gives the correct bounds?
    Answer: 1≤ρ≤21 \leq \rho \leq 2, 3π4≤ϕ≤π\frac{3\pi}{4} \leq \phi \leq \pi, 0≤θ≤2π0 \leq \theta \leq 2\pi When setting up triple integrals in spherical coordinates, you need to carefully analyze each constraint to determine the bounds for ρ\rho, ϕ\phi, and θ\theta. The constraint 1≤x2+y2+z2≤41 \leq x^2 + y^2 + z^2 \leq 4 translates directly to 1≤ρ2≤41 \leq \rho^2 \leq 4, giving us 1≤ρ≤21 \leq \rho \leq 2. Since there's no restriction on rotation around the z-axis, we have 0≤θ≤2π0 \leq \theta \leq 2\pi. The key challenge is finding the correct ϕ\phi bounds from the constraint z≤−x2+y2z \leq -\sqrt{x^2 + y^2}. In spherical coordinates, z=ρcos⁡ϕz = \rho\cos\phi and x2+y2=ρsin⁡ϕ\sqrt{x^2 + y^2} = \rho\sin\phi. Substituting these into z≤−x2+y2z \leq -\sqrt{x^2 + y^2} gives us ρcos⁡ϕ≤−ρsin⁡ϕ\rho\cos\phi \leq -\rho\sin\phi. Dividing by ρ\rho (which is positive), we get cos⁡ϕ≤−sin⁡ϕ\cos\phi \leq -\sin\phi, or cos⁡ϕ+sin⁡ϕ≤0\cos\phi + \sin\phi \leq 0. This simplifies to 2sin⁡(ϕ+π4)≤0\sqrt{2}\sin(\phi + \frac{\pi}{4}) \leq 0, which means sin⁡(ϕ+π4)≤0\sin(\phi + \frac{\pi}{4}) \leq 0. This occurs when ϕ+π4≥π\phi + \frac{\pi}{4} \geq \pi, so ϕ≥3π4\phi \geq \frac{3\pi}{4}. Combined with ϕ≤π\phi \leq \pi (since we're below the xy-plane), we get 3π4≤ϕ≤π\frac{3\pi}{4} \leq \phi \leq \pi. Answer D is correct. Answer A uses π2≤ϕ≤π\frac{\pi}{2} \leq \phi \leq \pi, which includes regions above the cone. Answer B has the wrong upper bound for ϕ\phi. Answer C uses π4≤ϕ≤π\frac{\pi}{4} \leq \phi \leq \pi, which includes far too much of the upper hemisphere. Remember: always convert inequality constraints systematically into spherical coordinates and solve the resulting inequalities carefully to avoid including unwanted regions.
  4. A solid is bounded by the cone z=3(x2+y2)z = \sqrt{3(x^2+y^2)} and the sphere x2+y2+z2=16x^2+y^2+z^2=16. The density at any point is inversely proportional to its distance from the origin. Find the total mass of the solid.
    Answer: 8π(2−3)8\pi(2-\sqrt{3}) The region is described by z≥3(x2+y2)z \ge \sqrt{3(x^2+y^2)} and ρ≤4\rho \le 4. The cone is ρcos⁡ϕ=3(ρsin⁡ϕ)2=3ρsin⁡ϕ\rho\cos\phi = \sqrt{3(\rho\sin\phi)^2} = \sqrt{3}\rho\sin\phi, which simplifies to cot⁡ϕ=3\cot\phi = \sqrt{3}, so ϕ=π/6\phi = \pi/6. The solid is defined by 0≤ρ≤40 \le \rho \le 4, 0≤ϕ≤π/60 \le \phi \le \pi/6, and 0≤θ≤2π0 \le \theta \le 2\pi. The density is δ=k/x2+y2+z2=k/ρ\delta = k/\sqrt{x^2+y^2+z^2} = k/\rho for some constant kk. Assuming k=1k=1, the mass M=∭Eδ dVM = \iiint_E \delta \, dV. M=∫02π∫0π/6∫04(1ρ)ρ2sin⁡ϕ dρ dϕ dθ=∫02π∫0π/6∫04ρsin⁡ϕ dρ dϕ dθM = \int_0^{2\pi} \int_0^{\pi/6} \int_0^4 \left(\frac{1}{\rho}\right) \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta = \int_0^{2\pi} \int_0^{\pi/6} \int_0^4 \rho \sin\phi \, d\rho \, d\phi \, d\theta =(∫02πdθ)(∫0π/6sin⁡ϕ dϕ)(∫04ρ dρ)=(2π)⋅[−cos⁡ϕ]0π/6⋅[ρ22]04= \left(\int_0^{2\pi} d\theta\right) \left(\int_0^{\pi/6} \sin\phi \, d\phi\right) \left(\int_0^4 \rho \, d\rho\right) = (2\pi) \cdot [-\cos\phi]_0^{\pi/6} \cdot \left[\frac{\rho^2}{2}\right]_0^4 =(2π)(−32−(−1))(8)=16π(1−32)=8π(2−3)= (2\pi) \left(-\frac{\sqrt{3}}{2} - (-1)\right) (8) = 16\pi \left(1 - \frac{\sqrt{3}}{2}\right) = 8\pi(2-\sqrt{3})
  5. Let EE be the solid region inside the sphere x2+y2+z2=4x^2+y^2+z^2=4 and outside the cylinder x2+y2=1x^2+y^2=1. Which of the following iterated integrals in spherical coordinates represents the volume of EE?
    Answer: ∫02π∫π/65π/6∫csc⁡ϕ2ρ2sin⁡ϕ dρ dϕ dθ\int_0^{2\pi} \int_{\pi/6}^{5\pi/6} \int_{\csc\phi}^2 \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta The region EE is described in spherical coordinates. The sphere x2+y2+z2=4x^2+y^2+z^2=4 is ρ=2\rho=2. The cylinder x2+y2=1x^2+y^2=1 is (ρsin⁡ϕ)2=1(\rho\sin\phi)^2=1, which gives ρsin⁡ϕ=1\rho\sin\phi=1 or ρ=csc⁡ϕ\rho=\csc\phi. Thus, for a given (ϕ,θ)(\phi, \theta), ρ\rho ranges from csc⁡ϕ\csc\phi to 22. The surfaces intersect when 2=csc⁡ϕ2=\csc\phi, which means sin⁡ϕ=1/2\sin\phi=1/2. This occurs at ϕ=π/6\phi=\pi/6 and ϕ=5π/6\phi=5\pi/6. The solid lies between these angles. The angle θ\theta ranges from 00 to 2π2\pi. The volume element in spherical coordinates is dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta. Therefore, the integral for the volume is ∫02π∫π/65π/6∫csc⁡ϕ2ρ2sin⁡ϕ dρ dϕ dθ\int_0^{2\pi} \int_{\pi/6}^{5\pi/6} \int_{\csc\phi}^2 \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.
  6. Consider the triple integral ∭Ez dV\iiint_E z \, dV where EE is bounded by the sphere ρ=4\rho = 4 and the half-cone ϕ=π3\phi = \frac{\pi}{3} (with z≥0z \geq 0). The value of this integral is:
    Answer: 24π24\pi The region is defined by 0≤ρ≤40 \leq \rho \leq 4, 0≤ϕ≤π30 \leq \phi \leq \frac{\pi}{3}, 0≤θ≤2π0 \leq \theta \leq 2\pi. In spherical coordinates, z=ρcos⁡ϕz = \rho\cos\phi. The integral becomes: ∭Ez dV=∫02π∫0π3∫04ρcos⁡ϕ⋅ρ2sin⁡ϕ dρ dϕ dθ=∫02π∫0π3∫04ρ3cos⁡ϕsin⁡ϕ dρ dϕ dθ\iiint_E z \, dV = \int_0^{2\pi} \int_0^{\frac{\pi}{3}} \int_0^4 \rho\cos\phi \cdot \rho^2\sin\phi \, d\rho \, d\phi \, d\theta = \int_0^{2\pi} \int_0^{\frac{\pi}{3}} \int_0^4 \rho^3\cos\phi\sin\phi \, d\rho \, d\phi \, d\theta. Evaluating each integral: ∫04ρ3dρ=444=64\int_0^4 \rho^3 d\rho = \frac{4^4}{4} = 64, ∫02πdθ=2π\int_0^{2\pi} d\theta = 2\pi, and ∫0π3cos⁡ϕsin⁡ϕ dϕ=∫0π312sin⁡(2ϕ) dϕ=12[−12cos⁡(2ϕ)]0π3=14[−cos⁡(2π3)+cos⁡(0)]=14[12+1]=38\int_0^{\frac{\pi}{3}} \cos\phi\sin\phi \, d\phi = \int_0^{\frac{\pi}{3}} \frac{1}{2}\sin(2\phi) \, d\phi = \frac{1}{2}[-\frac{1}{2}\cos(2\phi)]_0^{\frac{\pi}{3}} = \frac{1}{4}[-\cos(\frac{2\pi}{3}) + \cos(0)] = \frac{1}{4}[\frac{1}{2} + 1] = \frac{3}{8}. Therefore: ∭Ez dV=2π⋅64⋅38=24π\iiint_E z \, dV = 2\pi \cdot 64 \cdot \frac{3}{8} = 24\pi. Choices A, C, and D result from computational errors in the trigonometric integral or volume element.
  7. The Jacobian determinant for the transformation from spherical to Cartesian coordinates is ∣J∣=ρ2sin⁡ϕ|J| = \rho^2 \sin\phi. What is the average value of this Jacobian over the solid unit sphere centered at the origin?
    Answer: 3π20\frac{3\pi}{20} The average value of a function ff over a region EE is given by 1Volume(E)∭Ef dV\frac{1}{\text{Volume}(E)} \iiint_E f \, dV. Here, f(ρ,ϕ,θ)=ρ2sin⁡ϕf(\rho, \phi, \theta) = \rho^2 \sin\phi and EE is the unit sphere, whose volume is V=43π(1)3=4π3V = \frac{4}{3}\pi(1)^3 = \frac{4\pi}{3}. We need to compute the integral of the Jacobian over the sphere: ∭E(ρ2sin⁡ϕ) dV=∫02π∫0π∫01(ρ2sin⁡ϕ)(ρ2sin⁡ϕ) dρ dϕ dθ\iiint_E (\rho^2 \sin\phi) \, dV = \int_0^{2\pi} \int_0^{\pi} \int_0^1 (\rho^2 \sin\phi) (\rho^2 \sin\phi) \, d\rho \, d\phi \, d\theta =∫02π∫0π∫01ρ4sin⁡2ϕ dρ dϕ dθ= \int_0^{2\pi} \int_0^{\pi} \int_0^1 \rho^4 \sin^2\phi \, d\rho \, d\phi \, d\theta This separates into: (∫02πdθ)(∫0πsin⁡2ϕ dϕ)(∫01ρ4 dρ)=(2π)⋅(π2)⋅(15)=π25\left(\int_0^{2\pi} d\theta\right) \left(\int_0^{\pi} \sin^2\phi \, d\phi\right) \left(\int_0^1 \rho^4 \, d\rho\right) = (2\pi) \cdot \left(\frac{\pi}{2}\right) \cdot \left(\frac{1}{5}\right) = \frac{\pi^2}{5} The average value is this result divided by the volume: Average Value=π2/54π/3=π25⋅34π=3π20\text{Average Value} = \frac{\pi^2/5}{4\pi/3} = \frac{\pi^2}{5} \cdot \frac{3}{4\pi} = \frac{3\pi}{20}
  8. Evaluate ∭Ex2+y2+z2 dV\iiint_E \sqrt{x^2 + y^2 + z^2} \, 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 within the cone z≥3x2+y2z \geq \sqrt{3}\sqrt{x^2 + y^2}.
    Answer: 7π6\frac{7\pi}{6} In spherical coordinates: x2+y2+z2=ρ\sqrt{x^2 + y^2 + z^2} = \rho, the spheres give 1≤ρ≤21 \leq \rho \leq 2, and the cone z≥3x2+y2z \geq \sqrt{3}\sqrt{x^2 + y^2} becomes ρcos⁡ϕ≥3ρsin⁡ϕ\rho\cos\phi \geq \sqrt{3}\rho\sin\phi, so cot⁡ϕ≥3\cot\phi \geq \sqrt{3}, giving 0≤ϕ≤π60 \leq \phi \leq \frac{\pi}{6}. The integral becomes: ∫02π∫0π6∫12ρ⋅ρ2sin⁡ϕ dρ dϕ dθ=∫02π∫0π6sin⁡ϕ[ρ44]12dϕ dθ=∫02π∫0π615sin⁡ϕ4dϕ dθ=∫02π154[−cos⁡ϕ]0π6dθ=∫02π154(1−32)dθ=7π6\int_0^{2\pi} \int_0^{\frac{\pi}{6}} \int_1^2 \rho \cdot \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta = \int_0^{2\pi} \int_0^{\frac{\pi}{6}} \sin\phi \left[\frac{\rho^4}{4}\right]_1^2 d\phi \, d\theta = \int_0^{2\pi} \int_0^{\frac{\pi}{6}} \frac{15\sin\phi}{4} d\phi \, d\theta = \int_0^{2\pi} \frac{15}{4}\left[-\cos\phi\right]_0^{\frac{\pi}{6}} d\theta = \int_0^{2\pi} \frac{15}{4}\left(1 - \frac{\sqrt{3}}{2}\right) d\theta = \frac{7\pi}{6}. The other choices result from errors in the cone constraint or integration bounds.
  9. Consider the integral ∫02π∫0π4∫sec⁡ϕ2ρ2sin⁡ϕ dρ dϕ dθ\int_0^{2\pi} \int_0^{\frac{\pi}{4}} \int_{\sec\phi}^{2} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta. Which of the following best describes the region of integration?
    Answer: The region inside the sphere ρ=2\rho = 2 and above the plane z=1z = 1, for all θ\theta The lower limit ρ=sec⁡ϕ\rho = \sec\phi means ρ=1cos⁡ϕ\rho = \frac{1}{\cos\phi}, so ρcos⁡ϕ=1\rho\cos\phi = 1, which is z=1z = 1 in Cartesian coordinates. This represents the plane z=1z = 1. The upper limit ρ=2\rho = 2 is the sphere of radius 2. The range 0≤ϕ≤π40 \leq \phi \leq \frac{\pi}{4} with 0≤θ≤2π0 \leq \theta \leq 2\pi covers the region above the plane z=1z = 1 (since ϕ=π4\phi = \frac{\pi}{4} corresponds to z=ρcos⁡(π4)=ρ2z = \rho\cos(\frac{\pi}{4}) = \frac{\rho}{\sqrt{2}}, and smaller ϕ\phi values give larger zz). Choice A incorrectly identifies a cylinder. Choice B limits to first octant only. Choice D misinterprets the cone constraint.
  10. The volume of a solid is given by the iterated integral ∫0π/2∫0π/3∫0sec⁡ϕρ2sin⁡ϕ dρ dϕ dθ\int_0^{\pi/2} \int_0^{\pi/3} \int_0^{\sec\phi} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta. Which of the following best describes the solid?
    Answer: A cone with its vertex at the origin, capped by the plane z=1z=1, and lying in the first octant. Let's analyze the limits of integration. The limit for θ\theta is 0≤θ≤π/20 \le \theta \le \pi/2, which restricts the solid to the first octant. The limit for ϕ\phi is 0≤ϕ≤π/30 \le \phi \le \pi/3, which describes the region inside the cone opening from the positive zz-axis with an angle of π/3\pi/3. The limit for ρ\rho is 0≤ρ≤sec⁡ϕ0 \le \rho \le \sec\phi. The upper bound ρ=sec⁡ϕ\rho = \sec\phi can be rewritten as ρcos⁡ϕ=1\rho\cos\phi = 1, which is z=1z=1 in Cartesian coordinates. So, the solid is bounded by the cone ϕ=π/3\phi=\pi/3 and the plane z=1z=1, restricted to the first octant.
  11. A density function is given by ρ(x,y,z)=x2+y2+z2\rho(x,y,z) = x^2 + y^2 + z^2 in a solid hemisphere of radius aa in the upper half-space z≥0z \geq 0. Using spherical coordinates, the total mass is given by which integral?
    Answer: ∫02π∫0π2∫0aρ4sin⁡ϕ dρ dϕ dθ\int_0^{2\pi} \int_0^{\frac{\pi}{2}} \int_0^a \rho^4 \sin\phi \, d\rho \, d\phi \, d\theta The density function ρ(x,y,z)=x2+y2+z2=ρ2\rho(x,y,z) = x^2 + y^2 + z^2 = \rho^2 in spherical coordinates. The total mass is ∭Eρ(x,y,z) dV=∭Eρ2 dV\iiint_E \rho(x,y,z) \, dV = \iiint_E \rho^2 \, dV. For a hemisphere of radius aa in the upper half-space: 0≤ρ≤a0 \leq \rho \leq a, 0≤ϕ≤π20 \leq \phi \leq \frac{\pi}{2}, 0≤θ≤2π0 \leq \theta \leq 2\pi. The volume element is ρ2sin⁡ϕ dρ dϕ dθ\rho^2 \sin\phi \, d\rho \, d\phi \, d\theta. So the integral is ∫02π∫0π2∫0aρ2⋅ρ2sin⁡ϕ dρ dϕ dθ=∫02π∫0π2∫0aρ4sin⁡ϕ dρ dϕ dθ\int_0^{2\pi} \int_0^{\frac{\pi}{2}} \int_0^a \rho^2 \cdot \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta = \int_0^{2\pi} \int_0^{\frac{\pi}{2}} \int_0^a \rho^4 \sin\phi \, d\rho \, d\phi \, d\theta. Choice A shows the setup before simplification. Choice C uses wrong ϕ\phi limits (full sphere). Choice D uses wrong θ\theta limits.
  12. A solid is defined by the inequalities x2+y2+z2≤4x^2+y^2+z^2 \le 4, x≥0x \ge 0, y≥0y \ge 0, and z≤x2+y2z \le \sqrt{x^2+y^2}. Which of the following integrals represents the volume of this solid?
    Answer: ∫0π/2∫π/4π∫02ρ2sin⁡ϕ dρ dϕ dθ\int_0^{\pi/2} \int_{\pi/4}^{\pi} \int_0^2 \rho^2\sin\phi \, d\rho \, d\phi \, d\theta Let's convert the inequalities to spherical coordinates. x2+y2+z2≤4x^2+y^2+z^2 \le 4 means 0≤ρ≤20 \le \rho \le 2. The conditions x≥0x \ge 0 and y≥0y \ge 0 restrict the solid to the first quadrant in the xyxy-plane, so 0≤θ≤π/20 \le \theta \le \pi/2. The inequality z≤x2+y2z \le \sqrt{x^2+y^2} becomes ρcos⁡ϕ≤(ρsin⁡ϕ)2=ρsin⁡ϕ\rho\cos\phi \le \sqrt{(\rho\sin\phi)^2} = \rho\sin\phi. Dividing by ρ\rho (since ρ>0\rho > 0) and then by cos⁡ϕ\cos\phi gives 1≤tan⁡ϕ1 \le \tan\phi (assuming cos⁡ϕ>0\cos\phi > 0, i.e., 0≤ϕ<π/20 \le \phi < \pi/2) or cot⁡ϕ≤1\cot\phi \le 1. This implies ϕ≥π/4\phi \ge \pi/4. If cos⁡ϕ<0\cos\phi < 0 (i.e., ϕ>π/2\phi > \pi/2), the inequality ρcos⁡ϕ≤ρsin⁡ϕ\rho\cos\phi \le \rho\sin\phi is always true since the left side is negative and the right side is positive. Therefore, the condition is satisfied for all ϕ\phi from π/4\pi/4 to π\pi. The limits are 0≤ρ≤20 \le \rho \le 2, π/4≤ϕ≤π\pi/4 \le \phi \le \pi, and 0≤θ≤π/20 \le \theta \le \pi/2. The volume integral is ∫0π/2∫π/4π∫02ρ2sin⁡ϕ dρ dϕ dθ\int_0^{\pi/2} \int_{\pi/4}^{\pi} \int_0^2 \rho^2\sin\phi \, d\rho \, d\phi \, d\theta.
  13. Let EE be the solid region bounded by the spheres ρ=1\rho=1 and ρ=2\rho=2 and the cones ϕ=π/6\phi=\pi/6 and ϕ=π/3\phi=\pi/3. Which integral represents the moment of inertia IzI_z of EE about the zz-axis, assuming constant density δ=1\delta=1?
    Answer: ∫02π∫π/6π/3∫12ρ4sin⁡3ϕ dρ dϕ dθ\int_0^{2\pi} \int_{\pi/6}^{\pi/3} \int_1^2 \rho^4 \sin^3\phi \, d\rho \, d\phi \, d\theta The moment of inertia about the zz-axis is given by Iz=∭E(x2+y2) dVI_z = \iiint_E (x^2+y^2) \, dV. In spherical coordinates, the distance from the zz-axis is r=ρsin⁡ϕr = \rho\sin\phi, so x2+y2=r2=(ρsin⁡ϕ)2=ρ2sin⁡2ϕx^2+y^2 = r^2 = (\rho\sin\phi)^2 = \rho^2\sin^2\phi. The volume element is dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2\sin\phi \, d\rho \, d\phi \, d\theta. The integrand for IzI_z is thus (ρ2sin⁡2ϕ)(ρ2sin⁡ϕ)=ρ4sin⁡3ϕ(\rho^2\sin^2\phi)(\rho^2\sin\phi) = \rho^4\sin^3\phi. The limits of integration for the given region are 1≤ρ≤21 \le \rho \le 2, π/6≤ϕ≤π/3\pi/6 \le \phi \le \pi/3, and 0≤θ≤2π0 \le \theta \le 2\pi. Combining these gives the integral: ∫02π∫π/6π/3∫12ρ4sin⁡3ϕ dρ dϕ dθ\int_0^{2\pi} \int_{\pi/6}^{\pi/3} \int_1^2 \rho^4 \sin^3\phi \, d\rho \, d\phi \, d\theta
  14. The triple integral ∭Ef(x,y,z) dV\iiint_E f(x,y,z) \, dV is converted to spherical coordinates as ∫02π∫0π3∫04cos⁡ϕf(ρsin⁡ϕcos⁡θ,ρsin⁡ϕsin⁡θ,ρcos⁡ϕ)ρ2sin⁡ϕ dρ dϕ dθ\int_0^{2\pi} \int_0^{\frac{\pi}{3}} \int_0^{4\cos\phi} f(\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi) \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta. What is the geometric shape of region EE?
    Answer: The interior of a sphere of radius 2 centered at (0,0,2)(0, 0, 2), in the upper half-space z≥0z \geq 0 The upper limit ρ=4cos⁡ϕ\rho = 4\cos\phi represents a sphere. Converting back to Cartesian: ρ=4cos⁡ϕ\rho = 4\cos\phi becomes ρ2=4ρcos⁡ϕ\rho^2 = 4\rho\cos\phi, so x2+y2+z2=4zx^2 + y^2 + z^2 = 4z, which is x2+y2+(z−2)2=4x^2 + y^2 + (z-2)^2 = 4. This is a sphere of radius 2 centered at (0,0,2)(0,0,2). The limit 0≤ϕ≤π30 \leq \phi \leq \frac{\pi}{3} ensures we stay in the upper half-space. Choice A misinterprets the radius and center. Choice C has the wrong center location. Choice D has the wrong center coordinates.
  15. A solid occupies the region of a hemisphere defined by x2+y2+z2≤R2x^2+y^2+z^2 \le R^2 and z≥0z \ge 0. Assuming the solid has a constant density, what is the zz-coordinate of its center of mass?
    Answer: 3R8\frac{3R}{8} Let the density be δ=1\delta=1. The zz-coordinate of the center of mass is zˉ=Mxy/M\bar{z} = M_{xy}/M, where MM is the total mass (volume) and MxyM_{xy} is the moment about the xyxy-plane. The volume of the hemisphere is M=23πR3M = \frac{2}{3}\pi R^3. The moment MxyM_{xy} is ∭Ez dV\iiint_E z \, dV. In spherical coordinates, the region is 0≤ρ≤R0 \le \rho \le R, 0≤ϕ≤π/20 \le \phi \le \pi/2, 0≤θ≤2π0 \le \theta \le 2\pi. The integrand is z=ρcos⁡ϕz = \rho\cos\phi. So, Mxy=∫02π∫0π/2∫0R(ρcos⁡ϕ)ρ2sin⁡ϕ dρ dϕ dθM_{xy} = \int_0^{2\pi} \int_0^{\pi/2} \int_0^R (\rho\cos\phi) \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta =(∫02πdθ)(∫0π/2cos⁡ϕsin⁡ϕ dϕ)(∫0Rρ3 dρ)= \left(\int_0^{2\pi} d\theta\right) \left(\int_0^{\pi/2} \cos\phi\sin\phi \, d\phi\right) \left(\int_0^R \rho^3 \, d\rho\right) =(2π)⋅[sin⁡2ϕ2]0π/2⋅[ρ44]0R=(2π)(12)(R44)=πR44= (2\pi) \cdot \left[\frac{\sin^2\phi}{2}\right]_0^{\pi/2} \cdot \left[\frac{\rho^4}{4}\right]_0^R = (2\pi) \left(\frac{1}{2}\right) \left(\frac{R^4}{4}\right) = \frac{\pi R^4}{4} Therefore, zˉ=πR4/42πR3/3=πR44⋅32πR3=3R8\bar{z} = \frac{\pi R^4/4}{2\pi R^3/3} = \frac{\pi R^4}{4} \cdot \frac{3}{2\pi R^3} = \frac{3R}{8}.
  16. A solid has the shape of a spherical cap: the portion of the ball x2+y2+z2≤9x^2 + y^2 + z^2 \leq 9 with z≥32z \geq \frac{3}{2}. Using spherical coordinates, which integral correctly represents the volume of this solid?
    Answer: ∫02π∫0π3∫03ρ2sin⁡ϕ dρ dϕ dθ\int_0^{2\pi} \int_0^{\frac{\pi}{3}} \int_0^3 \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta When setting up triple integrals in spherical coordinates, you need to carefully determine the bounds for each variable based on the geometry of your region. In spherical coordinates, we have ρ\rho (distance from origin), ϕ\phi (angle from positive z-axis), and θ\theta (azimuthal angle). For this spherical cap problem, let's work through each bound systematically. The constraint z≥32z \geq \frac{3}{2} in spherical coordinates becomes ρcos⁡ϕ≥32\rho \cos\phi \geq \frac{3}{2}. Since we want the entire cap from this plane to the top of the sphere, ρ\rho ranges from 0 to 3 (the sphere's radius). To find the ϕ\phi bound, we determine where the plane z=32z = \frac{3}{2} intersects the sphere. At the boundary, ρ=3\rho = 3 and z=32z = \frac{3}{2}, so 3cos⁡ϕ=323\cos\phi = \frac{3}{2}, giving cos⁡ϕ=12\cos\phi = \frac{1}{2}, which means ϕ=π3\phi = \frac{\pi}{3}. Since ϕ\phi measures from the positive z-axis downward, our cap corresponds to 0≤ϕ≤π30 \leq \phi \leq \frac{\pi}{3}. The azimuthal angle θ\theta covers the full rotation: 0≤θ≤2π0 \leq \theta \leq 2\pi. Therefore, the correct integral is D: ∫02π∫0π3∫03ρ2sin⁡ϕ dρ dϕ dθ\int_0^{2\pi} \int_0^{\frac{\pi}{3}} \int_0^3 \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta. Looking at the wrong answers: A uses ϕ=π4\phi = \frac{\pi}{4} (corresponding to z=322z = \frac{3\sqrt{2}}{2}), B uses ϕ=π6\phi = \frac{\pi}{6} (corresponding to z=332z = \frac{3\sqrt{3}}{2}), and C restricts ρ\rho from 32\frac{3}{2} to 3, which incorrectly excludes the interior of the cap. Key strategy: Always convert geometric constraints to spherical coordinates first, then use the sphere's boundary conditions to find your integration limits.
  17. The moment of inertia about the zz-axis for a solid with density δ=1\delta = 1 is Iz=∭E(x2+y2) dVI_z = \iiint_E (x^2 + y^2) \, dV. If EE is the solid sphere x2+y2+z2≤R2x^2 + y^2 + z^2 \leq R^2, what is IzI_z in terms of RR?
    Answer: 8πR515\frac{8\pi R^5}{15} In spherical coordinates, x2+y2=ρ2sin⁡2ϕx^2 + y^2 = \rho^2\sin^2\phi. The integral becomes: Iz=∫02π∫0π∫0Rρ2sin⁡2ϕ⋅ρ2sin⁡ϕ dρ dϕ dθ=∫02π∫0π∫0Rρ4sin⁡3ϕ dρ dϕ dθI_z = \int_0^{2\pi} \int_0^{\pi} \int_0^R \rho^2\sin^2\phi \cdot \rho^2\sin\phi \, d\rho \, d\phi \, d\theta = \int_0^{2\pi} \int_0^{\pi} \int_0^R \rho^4\sin^3\phi \, d\rho \, d\phi \, d\theta. Evaluating: ∫0Rρ4dρ=R55\int_0^R \rho^4 d\rho = \frac{R^5}{5}, ∫02πdθ=2π\int_0^{2\pi} d\theta = 2\pi, and ∫0πsin⁡3ϕ dϕ=∫0πsin⁡ϕ(1−cos⁡2ϕ) dϕ=[−cos⁡ϕ+cos⁡3ϕ3]0π=2+23=43\int_0^{\pi} \sin^3\phi \, d\phi = \int_0^{\pi} \sin\phi(1-\cos^2\phi) \, d\phi = [-\cos\phi + \frac{\cos^3\phi}{3}]_0^{\pi} = 2 + \frac{2}{3} = \frac{4}{3}. Therefore: Iz=2π⋅R55⋅43=8πR515I_z = 2\pi \cdot \frac{R^5}{5} \cdot \frac{4}{3} = \frac{8\pi R^5}{15}. Choice B omits the sin⁡3ϕ\sin^3\phi integration. Choice C has an error in the sin⁡3ϕ\sin^3\phi integral. Choice D has errors in both the ρ\rho and ϕ\phi integrations.
  18. For which of the following combinations of a region EE and an integrand f(x,y,z)f(x,y,z) would the integral ∭Ef(x,y,z) dV\iiint_E f(x,y,z) \, dV be most efficiently evaluated using spherical coordinates?
    Answer: EE is the region inside the sphere x2+y2+z2=4x^2+y^2+z^2=4 and above the cone z=x2+y2z=\sqrt{x^2+y^2}, and f(x,y,z)=zx2+y2+z2f(x,y,z) = \frac{z}{x^2+y^2+z^2}. Spherical coordinates are most effective when the region of integration EE or the integrand f(x,y,z)f(x,y,z) has spherical symmetry (involving spheres or cones centered at the origin) and expressions of the form x2+y2+z2x^2+y^2+z^2. (A) The region is a cylinder and the integrand is z⋅rz \cdot r. This is ideal for cylindrical coordinates. (B) The region is a cube, which has planar sides. This is ideal for Cartesian coordinates. Describing a cube in spherical coordinates is extremely complicated. (C) The region is bounded by a sphere (ρ=2\rho=2) and a cone (ϕ=π/4\phi=\pi/4), and the integrand simplifies to ρcos⁡ϕρ2=cos⁡ϕρ\frac{\rho\cos\phi}{\rho^2} = \frac{\cos\phi}{\rho}. Both the region and the integrand are very simple in spherical coordinates. This is the best choice. (D) The region is a tetrahedron, which is bounded by planes. This is best handled with Cartesian coordinates.
  19. Evaluate the volume of the solid enclosed by the surface ρ=2cos⁡ϕ\rho = 2\cos\phi.
    Answer: 4π3\frac{4\pi}{3} The surface ρ=2cos⁡ϕ\rho = 2\cos\phi can be converted to Cartesian coordinates by multiplying by ρ\rho: ρ2=2ρcos⁡ϕ\rho^2 = 2\rho\cos\phi, which is x2+y2+z2=2zx^2+y^2+z^2 = 2z. Completing the square gives x2+y2+(z−1)2=1x^2+y^2+(z-1)^2 = 1. This is a sphere of radius r=1r=1 centered at (0,0,1)(0,0,1). The volume of a sphere is given by the formula V=43πr3V = \frac{4}{3}\pi r^3, so the volume is 43π(1)3=4π3\frac{4}{3}\pi (1)^3 = \frac{4\pi}{3}. Alternatively, one can compute the volume using a triple integral. The solid is defined by 0≤ρ≤2cos⁡ϕ0 \le \rho \le 2\cos\phi. For ρ\rho to be non-negative, cos⁡ϕ\cos\phi must be non-negative, so ϕ\phi ranges from 00 to π/2\pi/2. The integral is: V=∫02π∫0π/2∫02cos⁡ϕρ2sin⁡ϕ dρ dϕ dθ=2π∫0π/2[ρ33]02cos⁡ϕsin⁡ϕ dϕV = \int_0^{2\pi} \int_0^{\pi/2} \int_0^{2\cos\phi} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta = 2\pi \int_0^{\pi/2} \left[\frac{\rho^3}{3}\right]_0^{2\cos\phi} \sin\phi \, d\phi =2π∫0π/28cos⁡3ϕ3sin⁡ϕ dϕ=16π3[−cos⁡4ϕ4]0π/2=16π3(0−(−14))=4π3= 2\pi \int_0^{\pi/2} \frac{8\cos^3\phi}{3} \sin\phi \, d\phi = \frac{16\pi}{3} \left[-\frac{\cos^4\phi}{4}\right]_0^{\pi/2} = \frac{16\pi}{3} \left(0 - \left(-\frac{1}{4}\right)\right) = \frac{4\pi}{3}
  20. Consider the triple integral ∭Ez2 dV\iiint_E z^2 \, dV where EE is the region inside the sphere x2+y2+z2=9x^2 + y^2 + z^2 = 9 and above the cone z=x2+y2z = \sqrt{x^2 + y^2}. When converting to spherical coordinates, what are the correct limits of integration?
    Answer: 0≤ρ≤30 \leq \rho \leq 3, 0≤ϕ≤π40 \leq \phi \leq \frac{\pi}{4}, 0≤θ≤2π0 \leq \theta \leq 2\pi The sphere x2+y2+z2=9x^2 + y^2 + z^2 = 9 gives ρ=3\rho = 3. The cone z=x2+y2z = \sqrt{x^2 + y^2} becomes ρcos⁡ϕ=ρsin⁡ϕ\rho\cos\phi = \rho\sin\phi, which simplifies to cos⁡ϕ=sin⁡ϕ\cos\phi = \sin\phi, so ϕ=π4\phi = \frac{\pi}{4}. Since we want the region above the cone (larger zz), we need 0≤ϕ≤π40 \leq \phi \leq \frac{\pi}{4}. Choice B includes too much of the sphere (down to the xyxy-plane). Choice C incorrectly uses ϕ≥π4\phi \geq \frac{\pi}{4}, which is below the cone. Choice D uses an incorrect upper bound for ϕ\phi.

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