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 $y = 1$, we'll follow these steps:

1. Understand the Region $R$

The region $R$ is bounded by:

• $y = x^2$ (a parabola)
• $x = 2$ (a vertical line)
• $y = 1$ (a horizontal line)

2. Find the Limits of Integration

To describe the region $R$, let's figure out the bounds for $x$ and $y$.

• For $y = x^2$, since $y = 1$ is the upper boundary, $y$ will range from the parabola $y = x^2$ up to the line $y = 1$.
• To find the corresponding $x$-values where these curves intersect, set $y = x^2 = 1$, which gives $x = \pm 1$. But, since the region $R$ is bounded by $x = 2$, and $x$ starts from the parabola to $x = 2$.

Thus, $x$ will range from $x = \sqrt{y}$ (where $y = x^2$) to $x = 2$. And $y$ will range from $y = 0$ to $y = 1$.

3. Set Up the Double Integral

We can now write the double integral:

$\iint_{R} (x + y^2) \, dx \, dy = \int_{y=0}^{1} \int_{x=\sqrt{y}}^{2} (x + y^2) \, dx \, dy$

4. Evaluate the Inner Integral

First, evaluate the integral with respect to $x$:

$\int_{\sqrt{y}}^{2} (x + y^2) \, dx$

This can be split into two integrals:

$\int_{\sqrt{y}}^{2} x \, dx + \int_{\sqrt{y}}^{2} y^2 \, dx$

Evaluate each:

1. $\int_{\sqrt{y}}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{\sqrt{y}}^{2} = \frac{2^2}{2} - \frac{(\sqrt{y})^2}{2} = \frac{4}{2} - \frac{y}{2} = 2 - \frac{y}{2}$

2. $\int_{\sqrt{y}}^{2} y^2 \, dx = y^2 \int_{\sqrt{y}}^{2} dx = y^2 [x]_{\sqrt{y}}^{2} = y^2(2 - \sqrt{y})$

So, the result of the inner integral is:

$\int_{\sqrt{y}}^{2} (x + y^2) \, dx = 2 - \frac{y}{2} + y^2(2 - \sqrt{y})$

5. Evaluate the Outer Integral

Now, integrate with respect to $y$:

$\int_{0}^{1} \left( 2 - \frac{y}{2} + y^2(2 - \sqrt{y}) \right) dy$

This integral can be split and evaluated term by term:

1. $\int_{0}^{1} 2 \, dy = 2[y]_{0}^{1} = 2(1 - 0) = 2$

2. $\int_{0}^{1} -\frac{y}{2} \, dy = -\frac{1}{2} \left[ \frac{y^2}{2} \right]_{0}^{1} = -\frac{1}{2} \cdot \frac{1}{2} = -\frac{1}{4}$

3. $\int_{0}^{1} 2y^2 \, dy = 2 \left[ \frac{y^3}{3} \right]_{0}^{1} = 2 \cdot \frac{1}{3} = \frac{2}{3}$

4. $\int_{0}^{1} -y^{5/2} \, dy = \left[ \frac{-2y^{7/2}}{7} \right]_{0}^{1} = \frac{-2}{7}$

6. Combine the Results

Summing these, the result is:

$\text{Integral} = 2 - \frac{1}{4} + \frac{2}{3} - \frac{2}{7} = \frac{168 - 21 + 56 - 12}{84} = \frac{191}{84}$

Thus, the value of the double integral is:

$\iint_{R} (x + y^2) \, dx \, dy = \frac{191}{84}$

Would you like more details on this, or have any other questions? Here are some related topics to explore:

1. How to identify regions of integration in double integrals.
2. Understanding the concept of changing the order of integration.
3. Application of Green's Theorem in evaluating double integrals.
4. Double integrals in polar coordinates.
5. Visualization of regions for double integrals in a 3D space.

Tip: When dealing with double integrals, always sketch the region first. It helps in determining the limits of integration accurately.