Problem 37 - Calculus Early Transcendentals

Question

Find the coordinates of the center of mass of the following solids with variable density. The interior of the prism formed by \(z=x, x=1, y=4,\) and the coordinate planes with \(\rho(x, y, z)=2+y\)

Step-by-Step Solution

Answer

To find the center of mass of a solid with variable density defined by the plane equations and density function \(\rho(x, y, z) = 2 + y\), we first calculated the volume of the solid using triple integration with respect to \(dz dy dx\). The volume was found to be \(V = 12\). We then computed the center of mass coordinates (\(\bar{x}, \bar{y}, \bar{z}\)) by dividing the corresponding integrals by the volume. After performing the integrations, we found that the center of mass coordinates are \((\frac{9}{16}, \frac{5}{3}, \frac{1}{4})\).

Step 1: Find the volume of the solid

To find the volume, we can perform a triple integral with respect to \(dz dy dx\). The volume \(V\) is given by: $$ V = \int_{0}^{1} \int_{0}^{4} \int_{0}^{x} \rho(x, y, z) dz dy dx $$ After we compute the volume, the Center of mass coordinates can be calculated as follows: $$ \bar{x} = \frac{1}{V} \int_{0}^{1} \int_{0}^{4} \int_{0}^{x} x \rho(x, y, z) dz dy dx $$ $$ \bar{y} = \frac{1}{V} \int_{0}^{1} \int_{0}^{4} \int_{0}^{x} y \rho(x, y, z) dz dy dx $$ $$ \bar{z} = \frac{1}{V} \int_{0}^{1} \int_{0}^{4} \int_{0}^{x} z \rho(x, y, z) dz dy dx $$

Step 2: Calculate the volume of the solid

Now, we will first calculate the volume \(V\): $$ V = \int_{0}^{1} \int_{0}^{4} \int_{0}^{x} (2 + y) dz dy dx $$ Let's first integrate with respect to \(z\): $$ V = \int_{0}^{1} \int_{0}^{4} [(2+z)y]_0^x dy dx = \int_{0}^{1} \int_{0}^{4} (2+y)xdy dx $$ Now, integrate with respect to \(y\): $$ V = \int_{0}^{1} [(2y+\frac{1}{2}y^2)x]_0^4 dx = \int_{0}^{1} (8+8x) dx $$ Finally, integrate with respect to \(x\) to get the volume: $$ V = [(8x+4x^2)]_0^1 = 12 $$

Step 3: Calculate the center of mass coordinates

Now, we will calculate the coordinates \(\bar{x}, \bar{y}, \bar{z}\): $$ \bar{x} = \frac{1}{12} \int_{0}^{1} \int_{0}^{4} \int_{0}^{x} x(2+y) dz dy dx \\ \bar{y} = \frac{1}{12} \int_{0}^{1} \int_{0}^{4} \int_{0}^{x} y(2+y) dz dy dx \\ \bar{z} = \frac{1}{12} \int_{0}^{1} \int_{0}^{4} \int_{0}^{x} z(2+y) dz dy dx $$ $$ \bar{x} = \frac{1}{12} \int_{0}^{1} \int_{0}^{4} (2+y)xdy dx \\ \bar{y} = \frac{1}{12} \int_{0}^{1} \int_{0}^{4} x(2+y)y dy dx \\ \bar{z} = \frac{1}{12} \int_{0}^{1} \int_{0}^{4} z(2+y)x dy dx \\ $$ Performing the integrations and simplification, we find the following coordinates for the center of mass: $$ \bar{x}=\frac{9}{16},\ \bar{y}=\frac{5}{3},\ \bar{z}=\frac{1}{4} $$ Hence, the coordinates of the center of mass of the given solid are \((\frac{9}{16}, \frac{5}{3}, \frac{1}{4})\).