Problem 28 - Calculus Early Transcendentals

Question

Find the center of mass of the following solids, assuming a constant density of 1. Sketch the region and indicate the location of the centroid. Use symmetry when possible and choose a convenient coordinate system. The upper half of the ball \(x^{2}+y^{2}+z^{2} \leq 16(\text { for } z \geq 0)\)

Step-by-Step Solution

Answer

In summary, the centroid (center of mass) of the solid hemisphere given by the equation \(x^2 + y^2 + z^2 \leq 16\) and \(z \geq 0\) is at the coordinates (0, 0, 3/4).

Step 1: Identify the given region

The equation \(x^2 + y^2 + z^2 \leq 16\) represents a sphere with a radius of 4 centered at the origin. We are interested in the upper half of the sphere where \(z \geq 0\). This region is thus a hemispheric solid.

Step 2: Use symmetry

Since the hemisphere is symmetric about the origin, we have x̄ = 0 and ȳ = 0 due to symmetry. So, we just need to find the z̄ coordinate of the centroid.

Step 3: Choose coordinate system and volume element

We will use spherical coordinates, since it is convenient for a hemisphere. In spherical coordinates, we have: - \(x = r \sin(\theta) \cos(\phi)\) - \(y = r \sin(\theta) \sin(\phi)\) - \(z = r \cos(\theta)\) The volume element in spherical coordinates is given by \(dV = r^2 \sin(\theta) dr d\theta d\phi\). For the given hemisphere, we have: - Radius: \(0 \leq r \leq 4\) - Polar angle: \(0 \leq \theta \leq \frac{\pi}{2}\) - Azimuthal angle: \(0 \leq \phi \leq 2\pi\)

Step 4: Calculate the centroid's z-coordinate

To find the z-coordinate of the centroid, z̄, we will use the following formula: $$ z̄ = \frac{1}{V} \iiint_{\text{hemisphere}}{z \, dV} $$ Where V is the volume of the hemisphere. First, calculate the volume of the hemisphere using the volume element: $$ V = \iiint_{\text{hemisphere}}{dV} = \int_{0}^{4}{\int_{0}^{\frac{\pi}{2}}{\int_{0}^{2\pi}{r^2 \sin(\theta) dr d\theta d\phi}}} $$ Next, substitute \(z = r \cos(\theta)\) and evaluate the integral: $$ z̄ = \frac{1}{V} \iiint_{\text{hemisphere}}{r \cos(\theta) \, dV} = \frac{1}{V} \int_{0}^{4}{\int_{0}^{\frac{\pi}{2}}{\int_{0}^{2\pi}{r^3 \sin(\theta) \cos(\theta)dr d\theta d\phi}}} $$ Solve the integrals one by one: 1. \( \int_{0}^{4} r^3 dr = \frac{1}{4} r^4 \Big|_{0}^{4} = 64 \) 2. \( \int_{0}^{\frac{\pi}{2}} \sin(\theta) \cos(\theta) d\theta = \frac{1}{2} \sin^2(\theta) \Big|_{0}^{\frac{\pi}{2}} = \frac{1}{2} \) 3. \( \int_{0}^{2\pi} d\phi = \phi \Big|_{0}^{2\pi} = 2\pi \) Now, we multiply the results together: $$ z̄ = \frac{1}{V} \cdot 64 \cdot \frac{1}{2} \cdot 2\pi = \frac{64\pi}{V} $$ To find the volume of the hemisphere, observe that it is half the volume of a sphere with radius 4: $$ V = \frac{1}{2} \cdot \frac{4}{3}\pi(4)^3 = \frac{128}{3}\pi $$ Finally, we find the z-coordinate of the centroid: $$ z̄ = \frac{64\pi}{\frac{128}{3}\pi} = \frac{3}{4} $$

Step 5: Report the centroid

The centroid of the hemisphere is located at \((x̄, ȳ, z̄) = (0, 0, \frac{3}{4})\).