Triple Integrals in Cylindrical & Spherical Coordinates: Volume & Inertia

The center of mass is a weighted average of position. To find the x-coordinate (x-bar), we must calculate the "moment about the yz-plane" and divide it by the total mass of the object. The moment is calculated by integrating the position x multiplied by the density function p(x,y,z) over the volume. The total mass is calculated by integrating just the density function p(x,y,z) over the volume. Therefore, x-bar is the ratio of the moment integral (numerator) to the mass integral (denominator). Option D is incorrect because it assumes constant density (centroid), whereas the question specifies variable density.

The region is best described in cylindrical coordinates where x² + y² = r². The bounds become 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 2 (since z = 4 intersects the paraboloid at r² = 4, giving r = 2), and r² ≤ z ≤ 4. The volume integral is V = ∫₀2π  ∫₀² ∫{r²}⁴ r dz dr dθ. First integrate with respect to z: ∫{r²}⁴ dz = 4 - r². Multiply by r to get r(4 - r²) = 4r - r³. Next integrate with respect to r: ∫₀² (4r - r³) dr = 2r² - r⁴/4 evaluated from 0 to 2 gives (8 - 4) - (0 - 0) = 4. Finally integrate with respect to θ: ∫₀2π  4 dθ = 4θ evaluated from 0 to 2π gives 8π. The volume is 8π cubic units.

In cylindrical coordinates, x = r cos θ, y = r sin θ, z = z, and dV = r dr dθ dz. The cylinder x² + y² ≤ 9 is r ≤ 3, θ from 0 to 2π (full circle), z from 0 to 8. The Jacobian r must be included, and the function becomes f(r cos θ, r sin θ, z).

The total charge is found by integrating the density function over the volume. The charge Q = ∫₀¹ ∫₀¹ ∫₀¹ xyz dz dy dx. First integrate with respect to z: ∫₀¹ xyz dz = xy(z²/2) evaluated from 0 to 1 gives xy/2. Next integrate with respect to y: ∫₀¹ xy/2 dy = x(y²/4) evaluated from 0 to 1 gives x/4. Finally integrate with respect to x: ∫₀¹ x/4 dx = x²/8 evaluated from 0 to 1 gives 1/8. The total charge in the region is 1/8 coulombs (or appropriate units).

For a uniform distribution, probability equals volume of subregion divided by volume of total region. The total volume is Vₜotal = 2 × 2 × 2 = 8 cubic units. The subregion volume is Vₛub = 1 × 1 × 1 = 1 cubic unit. The probability P = Vₛub / Vₜotal = 1/8. Alternatively, by integration, the probability density function is f(x,y,z) = 1/8 over the total region, so P = ∫₀¹ ∫₀¹ ∫₀¹ (1/8) dz dy dx = (1/8) × 1 = 1/8.

The expected value (or mean) of a coordinate for a continuous probability distribution is calculated by multiplying that coordinate by the probability density function and integrating over the entire region. For the x-coordinate, this becomes E[x] = ∫∫∫_E x·f(x,y,z) dV. Option B is correct because it follows the definition of expected value for continuous distributions. Option A would give the expected value of x², option C gives the total probability (which equals 1), and option D gives the expected value of the sum of coordinates.

For a constant density, the center of mass coordinates are found by (x̄, ȳ, z̄) = (1/V)∫∫∫_E x dV, (1/V)∫∫∫_E y dV, (1/V)∫∫∫_E z dV. The volume V = 2 × 2 × 2 = 8. Due to symmetry, we know the answer must be at the midpoint (1,1,1), but let's verify. For x̄: ∫₀² ∫₀² ∫₀² x dz dy dx = ∫₀² ∫₀² (2x) dy dx = ∫₀² (4x) dx = 2x² evaluated from 0 to 2 gives 8. Then x̄ = (1/8) × 8 = 1. Similarly for ȳ and z̄, each gives 1. The center of mass is (1, 1, 1).

The moment of inertia about the z-axis is I_z = ∫∫∫_E (x² + y²)ρ dV. In cylindrical coordinates, x² + y² = r². The distance from the z-axis is r, so the integrand becomes r² × ρ dV. The volume element in cylindrical coordinates is dV = r dr dθ dz. Therefore, the integrand becomes ρ r² × r dr dθ dz = ρ r³ dr dθ dz. The region bounds are: z from 0 to 2, θ from 0 to 2π (full circle), and r from 0 to 2 (since x² + y² ≤ 4 gives r ≤ 2). The correct integral is ∫₀² ∫₀2π ∫₀² ρ r³ dr dθ dz.

This is the volume of a cone with height 5 and base radius 3. First ∫ dz from 0 to 5 - (5/3)r = 5 - (5/3)r. Then the integral becomes ∫₀2π dθ ∫0³ r (5 - (5/3)r) dr = 2π ∫0³ (5r - (5/3)r²) dr = 2π [ (5/2)r² - (5/9)r³ ]0³ = 2π [ (5/2)9 - (5/9)27 ] = 2π [ 45/2 - 15 ] = 2π [22.5 -15] = 2π *7.5 =15π.

The region between two concentric spheres is naturally described in spherical coordinates. The bounds are 1 ≤ ρ ≤ 2, 0 ≤ θ ≤ 2π, and 0 ≤ φ ≤ π. The volume integral is V = ∫₀2π  ∫₀π ∫₁² ρ² sin(φ) dρ dφ dθ. First integrate with respect to ρ: ∫₁² ρ² dρ = ρ³/3 evaluated from 1 to 2 gives (8/3 - 1/3) = 7/3. Next integrate with respect to φ: ∫₀π (7/3) sin(φ) dφ = -(7/3) cos(φ) evaluated from 0 to π gives -(7/3)(-1 - 1) = (14/3). Finally integrate with respect to θ: ∫₀2π  (14/3) dθ = (14/3)θ evaluated from 0 to 2π gives (28π)/3. The volume between the spheres is 28π/3 cubic units.

In the spherical coordinate system (ρ, θ, φ), there are three coordinates: ρ is the distance from the origin to the point, θ is the angle in the xy-plane measured from the positive x-axis (similar to polar coordinates), and φ is the angle measured from the positive z-axis down to the point. This is a standard convention in mathematics. The angle φ ranges from 0 to π, where φ = 0 points straight up along the positive z-axis, and φ = π points straight down along the negative z-axis. Option A describes θ, not φ.

To convert from rectangular (x, y, z) to spherical (ρ, θ, φ), we use the formulas: ρ = √(x² + y² + z²), θ = arctan(y/x), and φ = arccos(z/ρ). For the point (1, 1, √2): First compute ρ = √(1² + 1² + (√2)²) = √(1 + 1 + 2) = √4 = 2. Next compute θ = arctan(y/x) = arctan(1/1) = arctan(1) = π/4 (45 degrees). Finally compute φ = arccos(z/ρ) = arccos(√2/2) = arccos(1/√2) = π/4 (45 degrees). Therefore, the spherical coordinates are (2, π/4, π/4).

The center of mass coordinates are x̄ = (1/M) ∭ xδ dV, etc. First, find the mass M = ∭ xyz dV = ∫₀¹ ∫₀¹ ∫₀¹ xyz dz dy dx. This gives ∫₀¹ x dx ∫₀¹ y dy ∫₀¹ z dz = (1/2)(1/2)(1/2) = 1/8. For x̄: ∭ x · xyz dV = ∭ x²yz dV = ∫₀¹ x² dx ∫₀¹ y dy ∫₀¹ z dz = (⅓)(1/2)(1/2) = 1/12. So x̄ = (1/12)/(1/8) = (1/12) × 8 = 2/3.  Similarly, ȳ = 2/3 and z̄ = 2/3. The correct answer should be (2/3, 2/3, 2/3).

Spherical coordinates are most appropriate when the region of integration has spherical symmetry, meaning it is defined by equations involving x² + y² + z² (which becomes ρ²). Examples include spheres, spherical shells, portions of spheres, and cones (which are naturally described by constant φ). While cylindrical coordinates are better for cylindrical regions, and rectangular coordinates work well for box-like regions, spherical coordinates simplify integrals where the distance from the origin is a natural boundary. The presence of x² + y² + z² in the bounds or integrand is a strong indicator that spherical coordinates will simplify the calculation.