To evaluate the double integral

$\iint_R (2x + 2y) \, dA$

where $R$ is the region bounded by $y = x^2$ and $y = x$, we need to follow these steps:

1. Determine the region of integration $R$: The region $R$ is bounded by the curves $y = x^2$ and $y = x$.

   To find the points of intersection, solve $x^2 = x$: $x^2 = x \implies x^2 - x = 0 \implies x(x - 1) = 0 \implies x = 0 \text{ or } x = 1.$ Thus, the points of intersection are $(0, 0)$ and $(1, 1)$.

2. Set up the double integral: We will integrate with respect to $y$ first, and then with respect to $x$. The limits for $x$ are from 0 to 1. For a fixed $x$, $y$ ranges from $y = x^2$ to $y = x$.

   Thus, the integral becomes: $\iint_R (2x + 2y) \, dA = \int_{0}^{1} \int_{x^2}^{x} (2x + 2y) \, dy \, dx.$

3. Evaluate the inner integral: Integrate $2x + 2y$ with respect to $y$: $\int_{x^2}^{x} (2x + 2y) \, dy.$

   $\int_{x^2}^{x} (2x + 2y) \, dy = \left[ 2xy + y^2 \right]_{y=x^2}^{y=x}.$

   Substitute the limits $y = x$ and $y = x^2$: [ \left[ 2x(x) + (x)^2 \right] - \left[ 2x(x^2) + (x^2)^2 \right] = \left[ 2x^2 + x^2 \right] - \left[ 2x^3 + x^4 \right] = (2x^2 + x^2) - (2x^3 + x^4) = 3x^2 - 2x^3 - x^4. ]

4. Evaluate the outer integral: Integrate $3x^2 - 2x^3 - x^4$ with respect to $x$: $\int_{0}^{1} (3x^2 - 2x^3 - x^4) \, dx.$

   Compute the antiderivative:

   = \left[ x^3 - \frac{1}{2} x^4 - \frac{1}{5} x^5 \right]_{0}^{1}.$$ Evaluate at the limits $$x = 1$$ and $$x = 0$$: $$\left(1^3 - \frac{1}{2} \cdot 1^4 - \frac{1}{5} \cdot 1^5 \right) - \left(0^3 - \frac{1}{2} \cdot 0^4 - \frac{1}{5} \cdot 0^5 \right) = 1 - \frac{1}{2} - \frac{1}{5} = 1 - 0.5 - 0.2 = 0.3.$$

Therefore, the value of the integral is:

$\boxed{0.3}$

Would you like further details or have any questions?

Related Questions:

1. How do you find the region of integration for more complex boundaries?
2. What is the significance of the order of integration in double integrals?
3. How can you change the order of integration in double integrals?
4. What is Fubini's Theorem and how is it used in evaluating double integrals?
5. How do you evaluate double integrals over non-rectangular regions?
6. What are some common applications of double integrals in physics?
7. How do you handle double integrals with polar coordinates?
8. What are the differences between iterated integrals and double integrals?

Tip:

When solving double integrals, sketching the region of integration can provide valuable insights into setting the correct limits for integration.