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

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 \( V \).

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.

Discover how to calculate the volume of the region under the surface z = xy^2 and above the area bounded by the curves x = y^2 and x - 2y = 8. This step-by-step solution explores double integrals and the necessary mathematical concepts. Suitable for advanced high school students.

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