Question

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 \(\bar{y}\) using \(\rho(x, y, z)=k\) \(Q: \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1(a, b, c>0), x=0, y=0, z=0\)

Short answer

The mass \( M \) of the solid is \( \frac{abc}{6}k \) and the y-coordinate of the center of mass \( \bar{y} \) is \( \frac{b}{4} \).

Step-by-step solution

Identify the Limits of Integration

First, we need to identify the bounds of the integration for the variables x, y and z. The given inequalities suggest that \( 0 \leq x \leq a \), \( 0 \leq y \leq b(1 - \frac{x}{a}) \), and \( 0 \leq z \leq c[1 - \frac{x}{a} - \frac{y}{b}] \)

Find the Total Mass

We are given that the density of the solid is \( k \). The mass M of the solid is therefore defined by the triple integral of the density. This can be obtained by integrating over the whole volume V: \( M = \int\int\int k \,dx\,dy\,dz = \int_0^a \int_0^{b(1 - \frac{x}{a})} \int_0^{c[1 - \frac{x}{a} - \frac{y}{b}]} k \,dz\,dy\,dx \). The mass is then computed to be \( M = \frac{abc}{6}k \).

Find the Intermediate Quantity

Now, we need to compute the numerator of the \( \bar{y} \) by integrating \( y \rho dV \) over the volume V. That is \( \int\int\int ky \,dx\,dy\,dz \). Here, we must integrate \( y \) with respect to \( z \), then \( x \), and then \( y \). After computing this integration, we get the result as \( I = \frac{abc}{24}k \).

Find the Center of Mass

Finally, the y-coordinate of the center of mass \( \bar{y} \) of the solid is found using the formula \( \bar{y} = \frac{1}{M} \int y \rho dV \). Substituting \( I \) and \( M \) into this formula, we get \( \bar{y} = \frac{I}{M} = \frac{abc}{24}k / \frac{abc}{6}k = \frac{b}{4} \).