∫01​∫x^2,2-x^2 (xy))​dydx reverse order and solve

To reverse the order of integration and solve the double integral:

\[
I = \int_0^1 \int_{x^2}^{2-x^2} xy \, dy \, dx
\]

### Step 1: Identify the region of integration
We first need to understand the limits of the integral. The inner integral is with respect to \(y\), and it ranges from \(y = x^2\) to \(y = 2 - x^2\). The outer integral is with respect to \(x\), and it ranges from \(x = 0\) to \(x = 1\).

So the region of integration \(R\) is defined by:
\[
x^2 \leq y \leq 2 - x^2 \quad \text{and} \quad 0 \leq x \leq 1
\]

### Step 2: Reverse the order of integration
Now we want to switch the order of integration, i.e., integrate with respect to \(x\) first, then with respect to \(y\).

From the bounds, we can express the region as:
- For a fixed \(y\), the values of \(x\) range from \(x = 0\) to \(x = \sqrt{y}\) (since \(y = x^2\)) and from \(x = \sqrt{2 - y}\) to \(x = 1\) (since \(y = 2 - x^2\)).

Thus, the new integral becomes:
\[
I = \int_0^1 \int_0^{\sqrt{y}} xy \, dx \, dy + \int_0^1 \int_{\sqrt{2-y}}^1 xy \, dx \, dy
\]

### Step 3: Solve the integral

#### First part: \(\int_0^1 \int_0^{\sqrt{y}} xy \, dx \, dy\)
- Integrate with respect to \(x\):
\[
\int_0^{\sqrt{y}} xy \, dx = \frac{y}{2} x^2 \Bigg|_0^{\sqrt{y}} = \frac{y}{2} \cdot y = \frac{y^2}{2}
\]
- Now, integrate with respect to \(y\):
\[
\int_0^1 \frac{y^2}{2} \, dy = \frac{1}{2} \cdot \frac{y^3}{3} \Bigg|_0^1 = \frac{1}{6}
\]

#### Second part: \(\int_0^1 \int_{\sqrt{2-y}}^1 xy \, dx \, dy\)
- Integrate with respect to \(x\):
\[
\int_{\sqrt{2-y}}^1 xy \, dx = \frac{y}{2} \left[ x^2 \right]_{\sqrt{2-y}}^1 = \frac{y}{2} \left( 1 - (2 - y) \right) = \frac{y}{2} \cdot y = \frac{y^2}{2}
\]
- Now, integrate with respect to \(y\):
\[
\int_0^1 \frac{y^2}{2} \, dy = \frac{1}{6}
\]

### Step 4: Final result
Adding both parts together:
\[
I = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}
\]

Thus, the value of the double integral is \( \frac{1}{3} \).

Would you like a detailed explanation on any step? 

Here are 5 related questions:
1. How do you determine the region of integration from given bounds?
2. What techniques can be used to reverse the order of integration for non-rectangular regions?
3. How can substitution help simplify double integrals?
4. What are common applications of double integrals in physics or engineering?
5. How does changing the order of integration affect the limits?

**Tip:** Always sketch the region of integration when reversing the order of integration for clarity.

Learn how to reverse the order of integration for the double integral ∫01​∫x^2,2-x^2 (xy))​ dydx. This detailed step-by-step solution includes identifying the region of integration, reversing the bounds, and calculating the final result. Ideal for students studying undergraduate calculus.

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