①Evaluate JS (x+y²) dxdy over the Tregion R enclosed by the line y = x ^ 2 x = 2 y = 1 .

To evaluate the double integral \( \iint_R (x + y^2) \, dx \, dy \) over the region \( R \) enclosed by the curves \( y = x^2 \), \( x = 2 \), and \( y = 1 \), let's break down the problem step by step.

### 1. **Identify the Region \( R \):**
The region \( R \) is bounded by:
- The curve \( y = x^2 \) (a parabola).
- The vertical line \( x = 2 \).
- The horizontal line \( y = 1 \).

#### Intersection Points:
- At \( y = 1 \), substitute into \( y = x^2 \) to find \( x \). 
  \[
  1 = x^2 \quad \Rightarrow \quad x = \pm 1.
  \]
  So the parabola intersects the line \( y = 1 \) at the points \( (-1, 1) \) and \( (1, 1) \).

Thus, the region \( R \) is bounded by the parabola from \( x = -1 \) to \( x = 2 \), and the limits on \( y \) are determined by the parabola and the line \( y = 1 \).

### 2. **Set up the Double Integral:**

Since the region is simpler to describe in terms of \( y \), we can set up the integral in the form:
\[
\iint_R (x + y^2) \, dx \, dy = \int_{y=0}^{1} \int_{x=-\sqrt{y}}^{2} (x + y^2) \, dx \, dy.
\]

### 3. **Evaluate the Integral:**

#### Step 1: Integrate with respect to \( x \):
\[
\int_{-\sqrt{y}}^{2} (x + y^2) \, dx = \left[ \frac{x^2}{2} + y^2x \right]_{x=-\sqrt{y}}^{x=2}.
\]

Substitute the limits:

- When \( x = 2 \):
  \[
  \frac{2^2}{2} + y^2(2) = 2 + 2y^2.
  \]
- When \( x = -\sqrt{y} \):
  \[
  \frac{(-\sqrt{y})^2}{2} + y^2(-\sqrt{y}) = \frac{y}{2} - y^2\sqrt{y}.
  \]

So, the integral becomes:
\[
\left( 2 + 2y^2 \right) - \left( \frac{y}{2} - y^2\sqrt{y} \right) = 2 + 2y^2 - \frac{y}{2} + y^2\sqrt{y}.
\]

#### Step 2: Integrate with respect to \( y \):
Now, integrate the result with respect to \( y \) from 0 to 1:
\[
\int_0^1 \left( 2 + 2y^2 - \frac{y}{2} + y^2\sqrt{y} \right) \, dy.
\]

Let's evaluate each term separately:

1. \( \int_0^1 2 \, dy = 2 \times 1 = 2. \)
2. \( \int_0^1 2y^2 \, dy = \frac{2y^3}{3} \Big|_0^1 = \frac{2}{3}. \)
3. \( \int_0^1 -\frac{y}{2} \, dy = -\frac{1}{2} \times \frac{y^2}{2} \Big|_0^1 = -\frac{1}{4}. \)
4. \( \int_0^1 y^2\sqrt{y} \, dy = \int_0^1 y^{5/2} \, dy = \frac{y^{7/2}}{7/2} \Big|_0^1 = \frac{2}{7}. \)

#### Final Answer:
Adding up all the results:
\[
2 + \frac{2}{3} - \frac{1}{4} + \frac{2}{7}.
\]

To combine these fractions:
\[
= 2 + \frac{28}{42} - \frac{10.5

Learn how to evaluate the double integral \( \iint_R (x + y^2) \, dx \, dy \) over the region \( R \) enclosed by the curves \( y = x^2 \), \( x = 2 \), and \( y = 1 \). This problem involves advanced calculus concepts including double integrals, region integration, and limits of integration. Suitable for advanced college-level mathematics.

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