Q. 67 - Calculus

Question

Let a, b, and c be positive real numbers. In Exercises 65–68, let T be the tetrahedron with vertices (0, 0, 0), (a, 0, 0), (0, b, 0), and (0, 0,c).

Assume that the density at each point in T is proportional to the distance of the point from the xz-plane. Set up the integral expressions required to find the center of mass of T.

Step-by-Step Solution

Answer

The integral expression to find the center of a mass of T is $$\begin{gathered} \overline{x}=\frac{1}{M}{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{\int }_{z=0}^{c\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}-\frac{{\displaystyle y}}{{\displaystyle b}}\right)}kxy dxdydz \\ \overline{y}=\frac{1}{M}{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{\int }_{z=0}^{c\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}-\frac{{\displaystyle y}}{{\displaystyle b}}\right)}k{y}^2 dxdydz \\ \overline{z}=\frac{1}{M}{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{\int }_{z=0}^{c\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}-\frac{{\displaystyle y}}{{\displaystyle b}}\right)}kyz dxdydz \end{gathered}$$.

Step 1. Given Information.

The given vertices of the tetrahedron are $(0, 0, 0), (a, 0, 0), (0, b, 0), \mathrm{and} (0, 0,c).$

Step 2. Find the integral expression of the center of a mass of T.

It is given that the density at each point in T is proportional to the distance of the point from the xz-plane, so $\rho (x,y,z)=ky.$

To find the center of a mass of tetrahedron, let's first find the mass:

$$\begin{gathered} M=\iiint \rho (x,y,z)dxdydz \\ M={\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{\int }_{z=0}^{c\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}-\frac{{\displaystyle y}}{{\displaystyle b}}\right)}kydxdydz \end{gathered}$$

Now, the center of a mass of T is,

$$\begin{gathered} \overline{x}=\frac{M_{yz}}{M}=\frac{{\iiint }_{T}x\rho (x,y,z)dxdydz}{\iiint \rho (x,y,z)dxdydz} \\ \overline{x}=\frac{1}{M}{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{\int }_{z=0}^{c\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}-\frac{{\displaystyle y}}{{\displaystyle b}}\right)}kxy dxdydz \end{gathered}$$

Step 3. Solve.

By proceeding with the calculation further,

$$\begin{gathered} \overline{y}=\frac{M_{xz}}{M}=\frac{{\iiint }_{T}y\rho (x,y,z)dxdydz}{\iiint \rho (x,y,z)dxdydz} \\ \overline{y}=\frac{1}{M}{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{\int }_{z=0}^{c\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}-\frac{{\displaystyle y}}{{\displaystyle b}}\right)}k{y}^{2}dxdydz \end{gathered}$$

Now,

$$\begin{gathered} \overline{z}=\frac{M_{xy}}{M}=\frac{{\iiint }_{T}z\rho (x,y,z)dxdydz}{\iiint \rho (x,y,z)dxdydz} \\ \overline{z}=\frac{1}{M}{\int }_{x=0}^a{\int }_{y=0}^{b\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}\right)}{\int }_{z=0}^{c\left(1-\frac{{\displaystyle x}}{{\displaystyle a}}-\frac{{\displaystyle y}}{{\displaystyle b}}\right)}kyz dxdydz \end{gathered}$$