Problem 32 - 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 solid bounded by the upper half \((z \geq 0)\) of the ellipsoid \(4 x^{2}+4 y^{2}+z^{2}=16\)

Step-by-Step Solution

Answer

Question: Find the centroid of the solid with constant density within the upper half of the ellipsoid \(4x^2 + 4y^2 + z^2 = 16\). Answer: The centroid of the given solid is at the point \((0, 0, 0)\).

Step 1: Identify spherical coordinate transformation

Convert the given equation from Cartesian coordinates to spherical coordinates. To do so, we will use the following transformations: \(r^2 = x^2 + y^2 + z^2\) \(x = r\sin(\phi)\cos(\theta)\) \(y = r\sin(\phi)\sin(\theta)\) \(z = r\cos(\phi)\) Now, we can rewrite the ellipsoid equation in spherical coordinates: \(4(r\sin(\phi)\cos(\theta))^2 + 4(r\sin(\phi)\sin(\theta))^2 + (r\cos(\phi))^2 = 16\)

Step 2: Simplify the equation in spherical coordinates

Simplify the ellipsoid equation in spherical coordinates and solve for r: \(4r^2\sin^2(\phi)\cos^2(\theta) + 4r^2\sin^2(\phi)\sin^2(\theta) + r^2\cos^2(\phi) = 16\) \(r^2(\sin^2(\phi)(\cos^2(\theta) + \sin^2(\theta)) + \cos^2(\phi)) = 16\) Since \(\cos^2(\theta) + \sin^2(\theta) = 1\), we have \(r^2(\sin^2(\phi) + \cos^2(\phi)) = 16\) \(r^2 = 16\) Now, we can identify the limits for integrating: \(r : 0 \to 4\) \(\theta : 0 \to 2\pi\) \(\phi : 0 \to \pi/2\)

Step 3: Find the volume of the region

For constant density, the center of mass is the centroid. To find the centroid, we need to find the average of the coordinates. We can use triple integrals to calculate the volume and the weighted coordinates of the region. First, calculate the volume of the region using the following integral: \(V = \int_{0}^{\pi/2} \int_{0}^{2\pi} \int_{0}^{4} r^2\sin(\phi)\cdot dr\,d\theta\,d\phi\)

Step 4: Evaluate the volume integral

Solve the integral for the volume: \(V = \int_{0}^{\pi/2} \int_{0}^{2\pi} \int_{0}^{4} r^2\sin(\phi)\, dr\,d\theta\,d\phi = \int_{0}^{\pi/2} \sin(\phi)\,d\phi \int_{0}^{2\pi} d\theta \int_{0}^{4} r^2\, dr = \int_{0}^{\pi/2} \sin(\phi)\,d\phi \int_{0}^{2\pi} d\theta\left[\frac{1}{3}r^3\right]_{0}^{4}\) \(V = \left[1 - \cos(\frac{\pi}{2})\right]\cdot2\pi\cdot\frac{64}{3}\) \(V = 2\pi\cdot\frac{64}{3}\)

Step 5: Find the centroid coordinates

Now, calculate the centroid coordinates using triple integrals with the weighted coordinates: \((\bar{x}, \bar{y}, \bar{z}) = \frac{1}{V}\cdot\left(\int_{0}^{\pi/2} \int_{0}^{2\pi} \int_{0}^{4} r^3\sin(\phi)\cos(\theta)\,dr\,d\theta\,d\phi,\int_{0}^{\pi/2} \int_{0}^{2\pi} \int_{0}^{4} r^3\sin(\phi)\sin(\theta)\,dr\,d\theta\,d\phi,\int_{0}^{\pi/2} \int_{0}^{2\pi} \int_{0}^{4} r^3\sin(\phi)\cos(\phi)\,dr\,d\theta\,d\phi\right)\) Due to the symmetry of the ellipsoid, we can see that the centroid will be at the origin which is \((0, 0, 0)\). To confirm that, evaluate the integrals: \(\bar{x} = \frac{1}{V}\cdot \int_{0}^{\pi/2} \int_{0}^{2\pi} \int_{0}^{4} r^3\sin(\phi)\cos(\theta)\,dr\,d\theta\,d\phi = \frac{1}{V}\cdot0\cdot0\) \(\bar{x} = 0\) Similarly, we can evaluate the integrals for \(\bar{y}\) and \(\bar{z}\), which would both also be zero due to the symmetry of the ellipsoid. Therefore, the centroid of the given solid is: \((\bar{x}, \bar{y}, \bar{z}) = (0, 0, 0)\).