To find the mass $M$ and the center of mass $\bar{x}, \bar{y}, \bar{z}$ of the solid $E$ with density function $\rho(x, y, z) = y$, we need to compute these integrals over the region defined by the tetrahedron.

Step 1: Describe the Region $E$

The region $E$ is the tetrahedron bounded by the planes:

• $x = 0$,
• $y = 0$,
• $z = 0$,
• $x + y + z = 1$.

This region can be described as: $0 \leq x \leq 1, \quad 0 \leq y \leq 1 - x, \quad 0 \leq z \leq 1 - x - y.$

Step 2: Compute the Mass $M$

The mass of the solid is given by $M = \iiint_E \rho(x, y, z) \, dV = \iiint_E y \, dV.$

Using the bounds of $E$, we can set up the triple integral as follows: $M = \int_0^1 \int_0^{1 - x} \int_0^{1 - x - y} y \, dz \, dy \, dx.$

Evaluate the Integral

1. Integrate with respect to $z$: $\int_0^1 \int_0^{1 - x} \int_0^{1 - x - y} y \, dz \, dy \, dx = \int_0^1 \int_0^{1 - x} y \left[ z \right]_0^{1 - x - y} \, dy \, dx = \int_0^1 \int_0^{1 - x} y (1 - x - y) \, dy \, dx.$

2. Expand and integrate with respect to $y$: $= \int_0^1 \int_0^{1 - x} (y - xy - y^2) \, dy \, dx.$ Separating terms, we get: $= \int_0^1 \left( \int_0^{1 - x} y \, dy - x \int_0^{1 - x} y \, dy - \int_0^{1 - x} y^2 \, dy \right) dx.$

   Let's solve each part separately:

   • First term: $\int_0^{1 - x} y \, dy = \left[ \frac{y^2}{2} \right]_0^{1 - x} = \frac{(1 - x)^2}{2}$.
   • Second term: $x \int_0^{1 - x} y \, dy = x \cdot \frac{(1 - x)^2}{2}$.
   • Third term: $\int_0^{1 - x} y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^{1 - x} = \frac{(1 - x)^3}{3}$.

   Plugging these results back, we get: $M = \int_0^1 \left( \frac{(1 - x)^2}{2} - x \cdot \frac{(1 - x)^2}{2} - \frac{(1 - x)^3}{3} \right) dx.$

Combine and simplify, then integrate with respect to $x$ to find the value of $M$.

Step 3: Compute the Center of Mass $(\bar{x}, \bar{y}, \bar{z})$

The coordinates of the center of mass are given by: $\bar{x} = \frac{1}{M} \iiint_E x \rho(x, y, z) \, dV, \quad \bar{y} = \frac{1}{M} \iiint_E y \rho(x, y, z) \, dV, \quad \bar{z} = \frac{1}{M} \iiint_E z \rho(x, y, z) \, dV.$

Since $\rho(x, y, z) = y$, we substitute $y$ in each integral. Each of these integrals follows a similar setup as the mass calculation but with $x y$, $y^2$, and $z y$ in the integrands, respectively.

For example, to compute $\bar{x}$, we calculate: $\bar{x} = \frac{1}{M} \iiint_E x y \, dV.$

You would then follow similar integration steps for $\bar{y}$ and $\bar{z}$ using $\iiint_E y^2 \, dV$ and $\iiint_E z y \, dV$, respectively. The results yield the exact coordinates for the center of mass.

Would you like the step-by-step evaluation of each of these integrals for the center of mass?

Related Questions

1. How would the mass change if the density function was $\rho(x, y, z) = x + y + z$?
2. What is the volume of the tetrahedron using a different method?
3. How would the center of mass change if $\rho(x, y, z) = x + y$?
4. Can the same approach be applied for density functions in cylindrical coordinates?
5. What are the limits of integration if we change variables to spherical coordinates?

Tip:

In triple integrals, always determine the correct order and bounds of integration based on the region's geometry to simplify calculations.