Triple Integrals & Volume - Calculus 3

1. Consider the triple integral $\iiint_E (x^2 + y^2) \, dV$ where $E$ is the region inside the cylinder $x^2 + y^2 = 2y$ and between the planes $z = 0$ and $z = 3$. After converting to cylindrical coordinates, this integral becomes:
   Answer: $\int_0^{\pi} \int_0^{2\sin\theta} \int_0^3 r^3 \, dz \, dr \, d\theta$ The cylinder x² + y² = 2y can be rewritten as x² + (y-1)² = 1, which is a circle of radius 1 centered at (0,1). In cylindrical coordinates, x² + y² = 2y becomes r² = 2r sin θ, so r = 2 sin θ. Since the circle is entirely in the upper half-plane (y ≥ 0), we need sin θ ≥ 0, which gives 0 ≤ θ ≤ π. The integrand x² + y² = r² in cylindrical coordinates, and with the volume element r dr dθ dz, we get r³. Choice A uses the wrong θ range (full circle instead of semicircle). Choice C uses an incorrect θ range. Choice D incorrectly uses cos θ instead of sin θ.
2. The volume of the region enclosed by the surfaces $x^2 + y^2 = z^2$ and $x^2 + y^2 + z^2 = 2z$ is:
   Answer: $\frac{\pi}{3}$ When you encounter a volume problem with two intersecting surfaces, you need to identify the region of intersection and set up an appropriate coordinate system. Here we have a cone $x^2 + y^2 = z^2$ and a sphere $x^2 + y^2 + z^2 = 2z$. First, rewrite the sphere equation by completing the square: $x^2 + y^2 + z^2 - 2z = 0$ becomes $x^2 + y^2 + (z-1)^2 = 1$. This is a sphere centered at $(0,0,1)$ with radius 1. The cone $x^2 + y^2 = z^2$ opens upward and downward from the origin. To find where they intersect, substitute the cone equation into the sphere: $z^2 + z^2 - 2z = 0$, which gives $2z^2 - 2z = 0$, so $z = 0$ or $z = 1$. The enclosed region lies between $z = 0$ and $z = 1$. Using cylindrical coordinates where $x^2 + y^2 = r^2$, the cone becomes $r = z$ and the sphere becomes $r^2 + z^2 = 2z$. For a fixed $z$, $r$ ranges from $0$ to $\min(z, \sqrt{2z - z^2})$. Since $z \leq \sqrt{2z - z^2}$ for $0 \leq z \leq 1$, we have $r$ from $0$ to $z$. The volume integral is: $V = \int_0^{2\pi} \int_0^1 \int_0^z r \, dr \, dz \, d\theta = 2\pi \int_0^1 \frac{z^2}{2} dz = \pi \int_0^1 z^2 dz = \pi \cdot \frac{1}{3} = \frac{\pi}{3}$. Answer A ($\frac{4\pi}{3}$) likely comes from calculating the full sphere volume. Answer B ($\frac{2\pi}{3}$) might result from doubling the correct answer. Answer D ($\frac{\pi}{6}$) could come from halving the correct result. Always sketch the surfaces first to visualize the enclosed region before setting up your integral.
3. The volume of the region $E$ defined by $x^2 + y^2 + z^2 \leq 4$, $x^2 + y^2 \leq z^2$, and $z \geq 0$ is equal to:
   Answer: $\frac{8\pi}{3}(2 - \sqrt{2})$ The region E is the intersection of a sphere of radius 2, a cone x² + y² = z², and the upper half-space z ≥ 0. In spherical coordinates, the cone becomes φ = π/4 (since tan φ = √(x² + y²)/z = 1). The region extends from φ = 0 to φ = π/4, with ρ from 0 to 2. The volume integral is ∫₀²π ∫₀π/⁴ ∫₀² ρ² sin φ dρ dφ dθ = 2π · (8/3) · ∫₀π/⁴ sin φ dφ = (16π/3)[-cos φ]₀π/⁴ = (16π/3)[1 - 1/√2] = (8π/3)(2 - √2). Choice B has an incorrect factor of 2. Choice C incorrectly computes the cosine integral. Choice D has wrong signs and factors.
4. Consider the solid $T$ defined by $x^2 + y^2 + z^2 \leq 9$, $z \geq \sqrt{x^2 + y^2}$, and $z \geq 0$. Using spherical coordinates to find the volume of $T$, the correct integral setup requires:
   Answer: Integration with $0 \leq \phi \leq \pi/4$ and $0 \leq \rho \leq 3$ When setting up triple integrals in spherical coordinates, you need to carefully analyze each constraint to determine the correct bounds. The key is translating each condition into spherical coordinate language where $x = \rho\sin\phi\cos\theta$, $y = \rho\sin\phi\sin\theta$, and $z = \rho\cos\phi$. Let's examine each constraint systematically. The condition $x^2 + y^2 + z^2 \leq 9$ becomes $\rho^2 \leq 9$, so $0 \leq \rho \leq 3$. The constraint $z \geq 0$ means $\rho\cos\phi \geq 0$, which gives us $0 \leq \phi \leq \pi/2$ since $\rho > 0$. The critical constraint is $z \geq \sqrt{x^2 + y^2}$. In spherical coordinates, this becomes $\rho\cos\phi \geq \sqrt{\rho^2\sin^2\phi\cos^2\theta + \rho^2\sin^2\phi\sin^2\theta} = \rho\sin\phi$. Dividing by $\rho$ gives $\cos\phi \geq \sin\phi$, which means $\tan\phi \leq 1$, so $\phi \leq \pi/4$. Combining with $z \geq 0$, we get $0 \leq \phi \leq \pi/4$, making choice C correct. Choice A uses $\pi/4 \leq \phi \leq \pi/2$, which represents the region where $z \leq \sqrt{x^2 + y^2}$ - the opposite of what we want. Choice B incorrectly uses $\phi \leq \pi/3$, which would allow points where the cone constraint is violated. Choice D attempts to modify $\rho$ bounds unnecessarily when the sphere already provides the correct radial limit. Remember: when you see cone-like constraints involving $z$ and $\sqrt{x^2 + y^2}$, convert to spherical coordinates and solve the resulting trigonometric inequality to find your $\phi$ bounds.
5. The volume of the solid bounded by the surfaces $z = \sqrt{x^2 + y^2}$, $z = \sqrt{8 - x^2 - y^2}$, and $z \geq 0$ can be computed as a triple integral. What is this volume?
   Answer: $\frac{16\pi}{3}$ When finding volumes bounded by multiple surfaces, you need to identify the region of integration and set up appropriate bounds. This problem involves two surfaces that intersect to create a bounded solid. The surfaces are $z = \sqrt{x^2 + y^2}$ (a cone opening upward) and $z = \sqrt{8 - x^2 - y^2}$ (the upper half of a sphere with radius $2\sqrt{2}$). To find where they intersect, set them equal: $\sqrt{x^2 + y^2} = \sqrt{8 - x^2 - y^2}$. Squaring both sides gives $x^2 + y^2 = 8 - x^2 - y^2$, so $2(x^2 + y^2) = 8$, meaning $x^2 + y^2 = 4$. The surfaces intersect along a circle of radius 2. Using cylindrical coordinates where $x = r\cos\theta$, $y = r\sin\theta$, the volume integral becomes: $V = \int_0^{2\pi} \int_0^2 \int_r^{\sqrt{8-r^2}} r \, dz \, dr \, d\theta$ Evaluating: $V = \int_0^{2\pi} \int_0^2 r(\sqrt{8-r^2} - r) \, dr \, d\theta = 2\pi \int_0^2 (r\sqrt{8-r^2} - r^2) \, dr$ Using substitution $u = 8-r^2$ for the first term and direct integration for the second: $V = 2\pi\left[\frac{8}{3} - \frac{8}{3}\right] = 2\pi \cdot \frac{8}{3} = \frac{16\pi}{3}$ Answer D is correct. Answer A ($\frac{4\pi}{3}$) likely comes from incorrect bounds or missing the factor of $2\pi$. Answer B ($\frac{8\pi}{3}$) suggests forgetting the $2\pi$ from the $\theta$ integration. Answer C ($\frac{32\pi}{3}$) indicates an error in the radial integration, possibly doubling somewhere. Remember: always visualize the solid first, then choose coordinates that simplify the bounds—cylindrical coordinates are ideal when circular symmetry appears.
6. A solid $W$ is bounded by the surfaces $z = x^2 + y^2$, $z = 4$, and $x^2 + y^2 = 1$. To find the volume of $W$ using a triple integral in cylindrical coordinates, which of the following represents the correct setup?
   Answer: $\int_0^{2\pi} \int_1^2 \int_{r^2}^4 r \, dz \, dr \, d\theta$ The solid W is bounded above by z = 4, below by the paraboloid z = x² + y², and laterally by the cylinder x² + y² = 1. In cylindrical coordinates, the paraboloid becomes z = r² and the cylinder becomes r = 1. Since the region extends from the cylinder r = 1 outward to where the paraboloid meets the plane z = 4 (which occurs at r = 2), we have 1 ≤ r ≤ 2. For each fixed r in this range, z varies from r² to 4. Choice A incorrectly starts r at 0 instead of 1. Choice C uses incorrect z-limits. Choice D incorrectly limits r from 0 to 1, missing the outer portion of the solid.

Start this quiz