evaluate double integral D xy^2dA where D is the region in the xy-plane bounded by the curves x=y^3 and y=x^2

To evaluate the double integral of \( xy^2 \) over the region \( D \) bounded by the curves \( x = y^3 \) and \( y = x^2 \), we need to set up the limits of integration carefully and perform the integration step by step.

### 1. **Understand the Region \( D \)**:
The curves are:
- \( x = y^3 \) (a cubic curve)
- \( y = x^2 \) (a parabolic curve)

We need to find the points of intersection to establish the boundaries of the region. To do this, solve for \( y \) in terms of \( x \), or \( x \) in terms of \( y \), by equating the two curves.

Set \( x = y^3 \) and \( y = x^2 \). Substituting \( y = x^2 \) into \( x = y^3 \), we get:
\[
x = (x^2)^3 = x^6
\]
\[
x - x^6 = 0
\]
\[
x(1 - x^5) = 0
\]
So, \( x = 0 \) or \( x^5 = 1 \). This gives \( x = 0 \) and \( x = 1 \) (since \( x^5 = 1 \) implies \( x = 1 \)).

At these points, the corresponding \( y \)-values are:
- For \( x = 0 \), \( y = 0^2 = 0 \).
- For \( x = 1 \), \( y = 1^2 = 1 \).

Thus, the region \( D \) is bounded between \( x = 0 \) and \( x = 1 \), and lies between the curves \( x = y^3 \) and \( y = x^2 \).

### 2. **Set up the Double Integral**:
We now express the double integral in terms of \( y \). The region is naturally bounded by:
- \( x = y^3 \) (lower boundary)
- \( x = 1 \) (upper boundary)

The limits for \( y \) go from 0 to 1. So the integral becomes:
\[
\int_0^1 \int_{y^3}^y xy^2 \, dx \, dy
\]

### 3. **Perform the Integration**:
#### Step 1: Integrate with respect to \( x \) (inner integral):
\[
\int_{y^3}^y xy^2 \, dx
\]
Since \( y^2 \) is constant with respect to \( x \), this becomes:
\[
y^2 \int_{y^3}^y x \, dx = y^2 \left[ \frac{x^2}{2} \right]_{y^3}^y = y^2 \left( \frac{y^2}{2} - \frac{(y^3)^2}{2} \right)
\]
\[
= y^2 \left( \frac{y^2}{2} - \frac{y^6}{2} \right) = \frac{y^4}{2} - \frac{y^8}{2}
\]

#### Step 2: Integrate with respect to \( y \) (outer integral):
Now, integrate from 0 to 1:
\[
\int_0^1 \left( \frac{y^4}{2} - \frac{y^8}{2} \right) \, dy
\]
This can be split into two separate integrals:
\[
\frac{1}{2} \int_0^1 y^4 \, dy - \frac{1}{2} \int_0^1 y^8 \, dy
\]
Each can be computed using the power rule \( \int y^n \, dy = \frac{y^{n+1}}{n+1} \):
\[
\frac{1}{2} \cdot \frac{y^5}{5} \Big|_0^1 = \frac{1}{2} \cdot \frac{1}{5} = \frac{1}{10}
\]
\[
\frac{1}{2} \cdot \frac{y^9}{9} \Big|_0^1 = \frac{1}{2} \cdot \frac{1}{9} = \frac{1}{18}
\]

Thus, the result of the integral is:
\[
\frac{1}{10} - \frac{1}{18} = \frac{9}{90} - \frac{5}{90} = \frac{4}{90} = \frac{2}{45}
\]

### Final Answer:
The value of the double integral is \( \boxed{\frac{2}{45}} \).

---

Would you like more details on any of the steps?

Here are 5 follow-up questions to expand on this:
1. How do you determine the points of intersection for more complex curves?
2. Why do we switch the order of integration for certain regions?
3. What are common techniques to handle regions that aren't simple rectangles?
4. How would you approach this problem if polar coordinates were needed?
5. Can you explain the geometric meaning of a double integral over a region?

**Tip**: Always visualize the region you're integrating over to ensure you're setting the limits of integration correctly.

Evaluate the double integral of xy^2 dA where D is the region in the xy-plane bounded by the curves x=y^3 and y=x^2.

This problem demonstrates how to evaluate a double integral of xy^2 over a region bounded by the curves x=y^3 and y=x^2. The step-by-step solution covers finding the limits of integration, performing the integration, and applying power rule techniques. Suitable for first-year college calculus students.

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