Triple Integrals in Spherical - Calculus 3

1. A region $R$ is defined by $1 \leq x^2 + y^2 + z^2 \leq 4$ and $z \geq \sqrt{x^2 + y^2}$. When setting up the triple integral $\iiint_R \frac{1}{x^2 + y^2 + z^2} \, dV$ in spherical coordinates, which aspect requires the most careful attention?
   Answer: The constraint $z \geq \sqrt{x^2 + y^2}$ corresponds to $0 \leq \phi \leq \pi/4$, requiring careful trigonometric analysis The constraint $z \geq \sqrt{x^2 + y^2}$ translates to $\rho \cos \phi \geq \rho \sin \phi$, which gives $\cos \phi \geq \sin \phi$, or $\phi \leq \pi/4$. This is the key geometric insight needed. Choice A is incorrect because $\rho \geq 1$ prevents singularities. Choice C is wrong because $\frac{1}{\rho^2} \cdot \rho^2 \sin \phi = \sin \phi$, not a complete cancellation. Choice D misses that the geometric constraint is the primary challenge, not singularity concerns.
2. Consider the iterated integral $\int_0^{2\pi} \int_0^{\alpha} \int_0^{R} f(\rho) \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$ where $\alpha$ is a constant and $R > 0$. If this integral represents the total charge in a spherical cap (like the top portion of an orange), and the cap has height $h = \frac{R}{2}$, what is the value of $\alpha$?
   Answer: $\frac{\pi}{3}$ This question tests your understanding of spherical coordinates and how geometric constraints translate into integration limits. When you see a spherical coordinate integral with the form $\rho^2 \sin \phi$, you're working with volume elements where $\rho$ is the radial distance, $\phi$ is the angle from the positive z-axis (polar angle), and $\theta$ is the azimuthal angle. For a spherical cap of height $h = \frac{R}{2}$, you need to find where the cap's boundary intersects the sphere. The cap extends from the "north pole" ($\phi = 0$) down to where the sphere is cut by a plane. At the boundary, the z-coordinate is $z = R - h = R - \frac{R}{2} = \frac{R}{2}$. In spherical coordinates, $z = \rho \cos \phi$. At the sphere's surface, $\rho = R$, so the boundary condition becomes: $\frac{R}{2} = R \cos \alpha$, which gives $\cos \alpha = \frac{1}{2}$. Therefore, $\alpha = \frac{\pi}{3}$. Looking at the wrong answers: Choice A ($\frac{\pi}{6}$) would correspond to $\cos \phi = \frac{\sqrt{3}}{2}$, giving a much smaller cap with height $h = R(1-\frac{\sqrt{3}}{2}) \approx 0.134R$. Choice B ($\frac{\pi}{4}$) gives $\cos \phi = \frac{\sqrt{2}}{2}$, resulting in $h = R(1-\frac{\sqrt{2}}{2}) \approx 0.293R$. Choice D ($\frac{2\pi}{3}$) would give $\cos \phi = -\frac{1}{2}$, representing a cap that extends below the sphere's center. Remember: always convert geometric constraints to the coordinate system of your integral. Draw a diagram showing the relationship between the cap height and the polar angle $\phi$.
3. Consider the triple integral $\iiint_E z \, dV$ where $E$ is the region bounded by the sphere $x^2 + y^2 + z^2 = 4$ and above the cone $z = \sqrt{x^2 + y^2}$. When converting to spherical coordinates, which of the following correctly represents the limits of integration?
   Answer: $\int_0^{2\pi} \int_{\pi/4}^{\pi/2} \int_0^2 \rho \cos \phi \cdot \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$ In spherical coordinates, the cone $z = \sqrt{x^2 + y^2}$ becomes $\rho \cos \phi = \rho \sin \phi$, which gives $\phi = \pi/4$. Since we want the region above the cone (larger $z$ values), we need $\phi \in [\pi/4, \pi/2]$. The integrand $z$ becomes $\rho \cos \phi$, and the volume element is $\rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$. Choice A uses wrong $\phi$ limits. Choice B has incorrect integrand ($\rho^2 \cos \phi$ instead of $\rho \cos \phi$). Choice C has wrong $\phi$ limits and would give the region below the cone.
4. The moment of inertia about the $z$-axis of a solid sphere of radius $R$ with constant density $\delta_0$ can be computed using spherical coordinates as $I_z = \iiint_E \delta_0 (x^2 + y^2) \, dV$. After converting to spherical coordinates, the integral becomes $\delta_0 \int_0^{2\pi} \int_0^{\pi} \int_0^R \rho^4 \sin^3 \phi \, d\rho \, d\phi \, d\theta$. What is $I_z$?
   Answer: $\frac{8\pi \delta_0 R^5}{15}$ Evaluating the integral: $\int_0^R \rho^4 d\rho = \frac{R^5}{5}$, $\int_0^{2\pi} d\theta = 2\pi$, and $\int_0^{\pi} \sin^3 \phi \, d\phi$. For the $\phi$ integral, use $\sin^3 \phi = \sin \phi(1-\cos^2 \phi)$ with substitution $u = \cos \phi$: $\int_0^{\pi} \sin^3 \phi \, d\phi = \int_{1}^{-1} (1-u^2)(-du) = \int_{-1}^{1} (1-u^2) du = [u - \frac{u^3}{3}]_{-1}^{1} = (1-\frac{1}{3}) - (-1+\frac{1}{3}) = \frac{4}{3}$. Therefore: $I_z = \delta_0 \cdot 2\pi \cdot \frac{4}{3} \cdot \frac{R^5}{5} = \frac{8\pi \delta_0 R^5}{15}$. Choice B uses wrong bounds for $\sin^3 \phi$ integral. Choice C has computational error. Choice D omits the $\frac{1}{3}$ factor.
5. Consider evaluating $\iiint_E (x^2 + y^2) \, dV$ where $E$ is the solid bounded by $x^2 + y^2 + z^2 = 9$ and $x^2 + y^2 + z^2 = 1$, with $z \geq 0$. In spherical coordinates, this becomes $\int_0^{2\pi} \int_0^{\pi/2} \int_1^3 \rho^2 \sin^2 \phi \cdot \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$. What is the value of $\int_0^{\pi/2} \sin^3 \phi \, d\phi$?
   Answer: $\frac{2}{3}$ To evaluate $\int_0^{\pi/2} \sin^3 \phi \, d\phi$, use the identity $\sin^3 \phi = \sin \phi (1 - \cos^2 \phi)$. Let $u = \cos \phi$, so $du = -\sin \phi \, d\phi$. When $\phi = 0$, $u = 1$; when $\phi = \pi/2$, $u = 0$. The integral becomes $\int_1^0 (1-u^2)(-du) = \int_0^1 (1-u^2) du = [u - \frac{u^3}{3}]_0^1 = 1 - \frac{1}{3} = \frac{2}{3}$. Choice A forgets the $1$ term in the antiderivative. Choice C incorrectly uses $\int_0^{\pi/2} \sin \phi \, d\phi$. Choice D has a sign error in the substitution.

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