Problem 30 - 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 cone \(z=16-r\) and the plane \(z=0\)

Step-by-Step Solution

Answer

Based on the given solution, calculate the coordinates of the center of mass of the solid bounded by the cone \(z=16-r\) and the plane \(z=0\), assuming a constant density of 1. Answer: \((0,0,48)\)

Step 1: Sketch the region and choose a coordinate system

First, we want to sketch the solid bounded by the cone \(z=16-r\) and the plane \(z=0\). We notice that the cone intersects the \(xy\)-plane when \(z=0\), which implies \(r=16\). Therefore, the solid is contained within the cone with apex at \((0, 0, 16)\) and base radius of 16, and is above the \(xy\)-plane \(z=0\). Since the region is symmetric around the \(z\)-axis, it would be convenient to use cylindrical coordinates, with \(x = r\cos\theta\), \(y = r\sin\theta\), and \(z = z\).

Step 2: Find the mass of the solid

Let's compute the total mass of the solid, M: \(M = \iiint_R \rho dV = \iiint_R dV\), since the density is 1. Due to the cylindrical symmetry, we can integrate first over the \(\theta\) direction using \(\theta\) from \(0\) to \(2\pi\), then the \(r\) direction using \(r\) from \(0\) to \(16\), and lastly the \(z\) direction using \(z\) from \(0\) to \(16-r\). \(M = \int_{0}^{2\pi} \int_{0}^{16} \int_{0}^{16-r} rdz dr d\theta = \int_{0}^{2\pi} \int_{0}^{16} r(16-r) dr d\theta\). Now we integrate w.r.t \(r\): \(M = \int_{0}^{2\pi} \left[\frac{r^3}{3}-8r^2\right]_0^{16} d\theta= \int_{0}^{2\pi} \left(-\frac{4096}{3}\right) d\theta\). And finally we integrate w.r.t \(\theta\): \(M = \left[-\frac{4096}{3}\theta\right]_0^{2\pi} = -\frac{8192\pi}{3}\). Since this is the mass of the region, we know that it must be positive. So M=\(\frac{8192\pi}{3}\).

Step 3: Find the centroid coordinates

Due to the symmetry of the solid, we know that \(x_c = y_c = 0\). We just need to find \(z_c\): \(z_c = \frac{1}{M} \iiint_R z dV = \frac{1}{M} \int_{0}^{2\pi} \int_{0}^{16} \int_{0}^{16-r} zr dz dr d\theta = \frac{3}{8192\pi} \int_{0}^{2\pi} \int_{0}^{16} \int_{0}^{16-r} zr dz dr d\theta\). Now we integrate w.r.t \(z\): \(z_c = \frac{3}{8192\pi} \int_{0}^{2\pi} \int_{0}^{16} \left[\frac{z^2r}{2}\right]_0^{16-r} dr d\theta = \frac{3}{4096\pi} \int_{0}^{2\pi} \int_{0}^{16} (256r - r^3) dr d\theta\). Now we integrate w.r.t \(r\): \(z_c = \frac{3}{4096\pi} \int_{0}^{2\pi} \left[\frac{128r^3}{3} - \frac{r^5}{5}\right]_0^{16} d\theta = \frac{3}{4096\pi} \int_{0}^{2\pi} \left(\frac{65536}{5}\right) d\theta\). Finally, we integrate w.r.t \(\theta\): \(z_c=\frac{196608}{4096\pi}\left[\theta\right]_0^{2\pi}=\frac{196608}{4096}=\frac{48}{1}\). Hence, the centroid of the solid is located at \((x_c, y_c, z_c) = (0, 0, 48)\) on the \(z\)-axis.