To evaluate the double integral $\iint_{R} (x^2 + y^2) \, dx \, dy$ over the region $R$ enclosed by the curve $y = x^2$, the line $x = 2$, and the line $y = 1$, we follow these steps:

Step 1: Define the Region $R$

The region $R$ is bounded by:

• The parabola $y = x^2$,
• The vertical line $x = 2$,
• The horizontal line $y = 1$.

Step 2: Set up the Integral

Since the region is easier to describe in terms of $x$ first, we can describe it as:

• $x$ varies from $x = 0$ (intersection of $y = x^2$ and $y = 0$) to $x = 2$.
• For a given $x$, $y$ varies from $y = x^2$ to $y = 1$.

The double integral becomes: $\iint_{R} (x^2 + y^2) \, dx \, dy = \int_{0}^{2} \int_{x^2}^{1} (x^2 + y^2) \, dy \, dx$

Step 3: Evaluate the Inner Integral

Evaluate the inner integral with respect to $y$: $\int_{x^2}^{1} (x^2 + y^2) \, dy$ This can be broken down as: $\int_{x^2}^{1} x^2 \, dy + \int_{x^2}^{1} y^2 \, dy$

For the first part: $\int_{x^2}^{1} x^2 \, dy = x^2 \cdot (1 - x^2)$

For the second part: $\int_{x^2}^{1} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{x^2}^{1} = \frac{1}{3} - \frac{(x^2)^3}{3} = \frac{1}{3} - \frac{x^6}{3}$

So, the inner integral becomes: $\int_{x^2}^{1} (x^2 + y^2) \, dy = x^2(1 - x^2) + \frac{1}{3} - \frac{x^6}{3}$

Simplifying: $= x^2 - x^4 + \frac{1}{3} - \frac{x^6}{3}$

Step 4: Evaluate the Outer Integral

Now, integrate with respect to $x$: $\int_{0}^{2} \left(x^2 - x^4 + \frac{1}{3} - \frac{x^6}{3}\right) dx$

Evaluate each term: $\int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{8}{3}$

$\int_{0}^{2} x^4 \, dx = \left[ \frac{x^5}{5} \right]_{0}^{2} = \frac{32}{5}$

$\int_{0}^{2} \frac{1}{3} \, dx = \frac{1}{3} \cdot \left[ x \right]_{0}^{2} = \frac{2}{3}$

$\int_{0}^{2} \frac{x^6}{3} \, dx = \frac{1}{3} \cdot \left[ \frac{x^7}{7} \right]_{0}^{2} = \frac{128}{21}$

Final Calculation:

Putting everything together: $\text{Integral} = \frac{8}{3} - \frac{32}{5} + \frac{2}{3} - \frac{128}{21}$

Let's simplify the expression to get the final result:

$\text{Integral} = \frac{8}{3} + \frac{2}{3} - \frac{32}{5} - \frac{128}{21} = \frac{10}{3} - \frac{32}{5} - \frac{128}{21}$

This expression can be further simplified to give the exact numeric value. Would you like the detailed computation or have any questions?

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Here are 5 related questions to deepen your understanding:

1. How does the choice of integration order affect the complexity of solving double integrals?
2. Can you describe another method to evaluate this integral using polar coordinates?
3. How would the integral change if the region $R$ was bounded by $y = x^3$ instead of $y = x^2$?
4. What are the general steps to sketch the region of integration for a double integral?
5. How would this problem differ if you had to evaluate a triple integral over a similar region?

Tip: When integrating over regions bounded by curves, always sketch the region first to determine the limits of integration accurately.