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Chapter 13: Problem 25

Find the mass of the following objects with the given density functions. The solid cone \(D=\\{(r, \theta, z): 0 \leq z \leq 6-r, 0 \leq r \leq 6\\}\) with density \(\rho(r, \theta, z)=7-z\)

Short Answer

Expert verified
Answer: The mass of the solid cone is \(930\pi\) units.

Step by step solution

01

Set up the triple integral with the given density function

We will first set up the integral to find the mass of the cone: $$ Mass = \iiint_\mathcal{D} \rho(r, \theta, z) \, dv $$ Here, \(dv\) represents the volume differential in polar coordinates, which is given as $$ dv = r\,dz\,dr\,d\theta. $$
02

Determine the limits of integration

To find the limits of integration, let's analyze the cone's constraints: - \(0 \leq z \leq 6-r\): This tells us that z goes from 0 to \(6-r\). - \(0 \leq r \leq 6\): This tells us that r goes from 0 to 6. - Since the cone is closed and makes a full circle, the limits for \(\theta\) will be \(0\) to \(2\pi\). Now, we can combine these limits with our integral: $$ Mass = \int_{0}^{2\pi} \int_{0}^{6} \int_{0}^{6-r} \rho(r, \theta, z) \, r\,dz\,dr\,d\theta $$
03

Substitute the density function into the integral

We are given the density function \(\rho(r, \theta, z) = 7-z\). Substituting this into the integral, we get: $$ Mass = \int_{0}^{2\pi} \int_{0}^{6} \int_{0}^{6-r} (7-z) \, r\,dz\,dr\,d\theta $$
04

Integrate with respect to z

First, we will integrate with respect to z: $$ Mass = \int_{0}^{2\pi} \int_{0}^{6} \left[\int_{0}^{6-r} (7z - z^2)r\,dz\right] \, dr\,d\theta $$ Integrating, we get: $$ Mass = \int_{0}^{2\pi} \int_{0}^{6} (42r - 12r^2 + r^3) \, dr\,d\theta $$
05

Integrate with respect to r

Now, we will integrate with respect to r: $$ Mass = \int_{0}^{2\pi} \left[\int_{0}^{6} (42r - 12r^2 + r^3) \, dr\right] \, d\theta $$ Integrating, we get: $$ Mass = \int_{0}^{2\pi} (756 - 372 + 81) \, d\theta $$ Simplifying, we get $$ Mass = \int_{0}^{2\pi} 465\, d\theta $$
06

Integrate with respect to 胃

Finally, we will integrate with respect to 胃: $$ Mass = \int_{0}^{2\pi} 465\, d\theta $$ Integrating and simplifying, we get: $$ Mass = 930\pi $$ Thus, the mass of the solid cone is \(930\pi\) units.

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