To find the mass of a prism with a variable density function, we integrate the density function over the volume of the prism. Let's go through the setup step-by-step.

Step 1: Define the Volume of the Prism

The prism is defined by the planes:

• $z = x$,
• $x = 1$,
• $y = 3$, and
• The coordinate planes $x = 0$, $y = 0$, $z = 0$.

So, the boundaries for each variable are:

• $0 \leq x \leq 1$,
• $0 \leq y \leq 3$, and
• $0 \leq z \leq x$ (since $z = x$ is the upper bound for $z$).

Step 2: Set Up the Integral for the Mass

The density function is given by: $\rho(x, y, z) = 3 + y$

The mass $M$ of the prism is calculated by integrating the density function over the volume $V$ of the prism: $M = \iiint_V \rho(x, y, z) \, dV$

Substitute the density function $\rho(x, y, z) = 3 + y$: $M = \iiint_V (3 + y) \, dV$

Step 3: Determine the Order of Integration

The integral bounds based on the constraints are:

$M = \int_0^1 \int_0^3 \int_0^x (3 + y) \, dz \, dy \, dx$

Step 4: Integrate with Respect to $z$

Since $3 + y$ does not depend on $z$, the integration with respect to $z$ is straightforward: $\int_0^x (3 + y) \, dz = (3 + y) \cdot z \Big|_0^x = (3 + y) \cdot x$

Thus, we have: $M = \int_0^1 \int_0^3 (3 + y) x \, dy \, dx$

Step 5: Integrate with Respect to $y$

Now, integrate with respect to $y$: $\int_0^3 (3 + y) x \, dy = x \int_0^3 (3 + y) \, dy$

Calculate the inner integral: $\int_0^3 (3 + y) \, dy = \left[ 3y + \frac{y^2}{2} \right]_0^3 = 3 \cdot 3 + \frac{3^2}{2} = 9 + \frac{9}{2} = \frac{27}{2}$

So we have: $M = \int_0^1 x \cdot \frac{27}{2} \, dx = \frac{27}{2} \int_0^1 x \, dx$

Step 6: Integrate with Respect to $x$

Now, integrate with respect to $x$: $\frac{27}{2} \int_0^1 x \, dx = \frac{27}{2} \cdot \frac{x^2}{2} \Big|_0^1 = \frac{27}{2} \cdot \frac{1}{2} = \frac{27}{4}$

Final Answer

The mass of the prism is: $M = \frac{27}{4}$

Would you like further details on any part of this solution?

Here are related questions to explore:

1. How would the solution change if the density function were $\rho(x, y, z) = x + y + z$?
2. How does the shape of the prism affect the limits of integration?
3. What if the density function depended only on $z$? How would that simplify the problem?
4. How is the setup different if the region were bounded by $x = 2$ instead of $x = 1$?
5. What is the impact of increasing $y$ on the mass given the density function?

Tip: Setting up the limits of integration correctly is crucial in solving mass integrals for regions with variable density.