To find the volume of the region under the surface $z = xy^2$ and above the area bounded by $x = y^2$ and $x - 2y = 8$, we can set up a double integral over the region in the $xy$-plane.

Step 1: Set Up the Region of Integration

We need to identify the region $R$ in the $xy$-plane bounded by the curves $x = y^2$ and $x - 2y = 8$.

1. Rewrite the line equation: Solve for $x$ in terms of $y$ in the equation $x - 2y = 8$: $x = 2y + 8$

2. Find the intersection points of $x = y^2$ and $x = 2y + 8$: Set $y^2 = 2y + 8$ and rearrange to get: $y^2 - 2y - 8 = 0$ This factors as: $(y - 4)(y + 2) = 0$ So, $y = 4$ and $y = -2$.

3. Determine the corresponding $x$-values for these $y$-values:

   • For $y = 4$: $x = 4^2 = 16$
   • For $y = -2$: $x = (-2)^2 = 4$

Thus, the region $R$ in the $xy$-plane is bounded by $x = y^2$ on the left and $x = 2y + 8$ on the right, with $y$ ranging from $-2$ to $4$.

Step 2: Set Up the Double Integral

The volume $V$ is given by: $V = \iint_R z \, dA = \iint_R xy^2 \, dA$ In terms of $x$ and $y$, this becomes: $V = \int_{y = -2}^{4} \int_{x = y^2}^{2y + 8} xy^2 \, dx \, dy$

Step 3: Integrate with Respect to $x$

Integrate $xy^2$ with respect to $x$: $\int_{y = -2}^{4} \int_{x = y^2}^{2y + 8} xy^2 \, dx \, dy = \int_{y = -2}^{4} \left[ \frac{y^2 x^2}{2} \right]_{x = y^2}^{x = 2y + 8} \, dy$

Substitute the limits for $x$: $= \int_{y = -2}^{4} \left( \frac{y^2 (2y + 8)^2}{2} - \frac{y^2 (y^2)^2}{2} \right) \, dy$ Simplify each term inside the integrand separately.

1. Expand $(2y + 8)^2$: $(2y + 8)^2 = 4y^2 + 32y + 64$ So, $\frac{y^2 (2y + 8)^2}{2} = \frac{y^2 (4y^2 + 32y + 64)}{2} = 2y^4 + 16y^3 + 32y^2$

2. Compute $\frac{y^2 (y^2)^2}{2}$: $\frac{y^2 (y^4)}{2} = \frac{y^6}{2}$

Putting it all together, we get: $V = \int_{y = -2}^{4} \left( 2y^4 + 16y^3 + 32y^2 - \frac{y^6}{2} \right) \, dy$

Step 4: Integrate with Respect to $y$

Now we integrate term-by-term: $V = \int_{y = -2}^{4} \left( 2y^4 + 16y^3 + 32y^2 - \frac{y^6}{2} \right) \, dy$

1. Integral of $2y^4$: $\int 2y^4 \, dy = \frac{2y^5}{5}$

2. Integral of $16y^3$: $\int 16y^3 \, dy = 4y^4$

3. Integral of $32y^2$: $\int 32y^2 \, dy = \frac{32y^3}{3}$

4. Integral of $-\frac{y^6}{2}$: $\int -\frac{y^6}{2} \, dy = -\frac{y^7}{14}$

After evaluating these integrals at $y = 4$ and $y = -2$ and combining the results, you will obtain the final volume $V$.