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Chapter 12: Problem 23

Find the mass and the indicated coordinates of the center of mass of the solid of given density bounded by the graphs of the equations. Find \(\bar{z}\) using \(\rho(x, y, z)=k x\) \(Q: z=4-x, z=0, y=0, y=4, x=0\)

Short Answer

Expert verified
The mass of the solid is \(32k\) and the z-coordinate of the center of mass of the solid is \(1\).

Step by step solution

01

Identify the Bounds of Integration

We first need to identify the bounds of integration, which are given by the equations \(z = 4 - x\), \(z = 0\), \(y = 0\), and \(y = 4\). These imply that \(x\) varies from 0 to 4, \(y\) varies from 0 to 4, and \(z\) from 0 to \(4 - x\). This gives us the limits of integration.
02

Compute Mass With Density and Volume

The mass of the solid can be calculated by integrating the solid's density over its volume. In this case, the density is \(\rho(x, y, z) = kx\). The volume element in Cartesian coordinates is \(dV = dx dy dz\). Hence the mass is given by the triple integral: \(M = \int \int \int \rho dV = \int_0^4 \int_0^4 \int_0^{4-x} kx dz dy dx\)
03

Evaluate the Integral to Find Mass

Evaluating the inner integral first will give: M = \(\int_0^4 \int_0^4 kx(4-x) dy dx\). Next, evaluating the middle integral will yield: M = \(\int_0^4 4kx(4-x) dx\). Finally, computing the outer integral, we find the total mass of the solid, \(M = 32k\).
04

Compute \(\bar{z}\)

The z-coordinate of the center of mass \(\bar{z}\) is the ratio of the moment about the z-axis to the total mass. The moment about the z-axis is given by the triple integral: \(zM = \int \int \int z \rho dV = \int_0^4 \int_0^4 \int_0^{4-x} zkx dz dy dx\). Compute this integral to find \(zM\).
05

Evaluate the Integral to Find \(zM\)

Evaluating the inner integral first, we have: \(zM = \int_0^4 \int_0^4 \tfrac{1}{2}kx(4-x)^2 dy dx\). Because \(dy\) is a constant in regards to this integral, compute the middle integral to yield: \(zM = \int_0^4 2kx(4-x)^2 dx\). Evaluating the outermost integral gives us \(zM = 32k\).
06

Calculate \(\bar{z}\)

The formula for \(\bar{z}\) is \(zM/M\). Substituting the values found in previous steps, we get \(\bar{z} = 32k/32k = 1\).

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Most popular questions from this chapter

In Exercises \(43-50\), sketch the region \(R\) whose area is given by the iterated integral. Then switch the order of integration and show that both orders yield the same area. $$ \int_{0}^{9} \int_{\sqrt{x}}^{3} d y d x $$

Find the mass and center of mass of the lamina bounded by the graphs of the equations for the given density or densities. (Hint: Some of the integrals are simpler in polar coordinates.) $$ y=\sin \frac{\pi x}{L}, y=0, x=0, x=L, \rho=k y $$

In Exercises \(37-42,\) sketch the region \(R\) of integration and switch the order of integration. $$ \int_{-1}^{1} \int_{x^{2}}^{1} f(x, y) d y d x $$

In Exercises \(37-42,\) sketch the region \(R\) of integration and switch the order of integration. $$ \int_{-\pi / 2}^{\pi / 2} \int_{0}^{\cos x} f(x, y) d y d x $$

In Exercises 1-4, evaluate the iterated integral. $$ \int_{0}^{4} \int_{0}^{\pi / 2} \int_{0}^{2} r \cos \theta d r d \theta d z $$
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