To find the volume of the solid enclosed by the parabolic cylinder $y = x^2$ and the planes $z = 5 + y$ and $z = 6y$, we can set up an integral that calculates the volume of the region by subtracting two separate integrals, each representing a different volume.

Step 1: Setting up the volume integral

The volume can be found using the double integral $\text{Volume} = \int_a^b \int_c^d \big((6y) - (5 + y)\big) \, dy \, dx$ where:

• $6y$ is the upper bound in the $z$-direction,
• $5 + y$ is the lower bound in the $z$-direction.

Step 2: Determining the bounds

1. The region is bounded by $y = x^2$ in the $xy$-plane.
2. Since $y = x^2$, $y$ ranges from $0$ to $1$ (assuming the bounds for $y$ are determined by the intersection in the context).
3. Correspondingly, $x$ would range from $-1$ to $1$.

Thus, we can set:

• $a = -1$,
• $b = 1$,
• $c = 0$,
• $d = x^2$.

Step 3: Setting up the integral for the volume

Now the volume integral becomes: $\text{Volume} = \int_{-1}^1 \int_0^{x^2} \big((6y) - (5 + y)\big) \, dy \, dx.$

Step 4: Simplify the integrand

The integrand $(6y - (5 + y))$ simplifies to: $6y - 5 - y = 5y - 5.$ Thus, the integral becomes: $\text{Volume} = \int_{-1}^1 \int_0^{x^2} (5y - 5) \, dy \, dx.$

Step 5: Evaluate the inner integral

Now we integrate with respect to $y$: $\int_0^{x^2} (5y - 5) \, dy = \left[ \frac{5y^2}{2} - 5y \right]_0^{x^2} = \frac{5(x^2)^2}{2} - 5(x^2).$ Simplifying further, we get: $= \frac{5x^4}{2} - 5x^2.$

Step 6: Evaluate the outer integral

Now we integrate with respect to $x$: $\text{Volume} = \int_{-1}^1 \left( \frac{5x^4}{2} - 5x^2 \right) dx.$

Splitting this into two separate integrals: $\text{Volume} = \int_{-1}^1 \frac{5x^4}{2} \, dx - \int_{-1}^1 5x^2 \, dx.$

Step 7: Evaluate each integral

1. First integral: [ \int_{-1}^1 \frac{5x^4}{2} , dx = \frac{5}{2} \int_{-1}^1 x^4 , dx = \frac{5}{2} \cdot \frac{2}{5} = 0.