Evaluate the double integral $\iint_S x \, dA$, where $S$ is the region between the curves $y = x$ and $y = x^3$. Note that the region $S$ has two parts.

1. Problem Description

Evaluate the double integral $\iint_S x \, dA$, where $S$ is the region between the curves $y = x$ and $y = x^3$. Note that the region $S$ has two parts.

2. Solution Steps

First, we need to find the points of intersection of the curves $y=x$ and $y=x^3$.

We set $x = x^3$, which gives $x^3 - x = 0$, so $x(x^2 - 1) = 0$, which means $x(x-1)(x+1) = 0$. The solutions are $x = -1$, $x = 0$, and $x = 1$. Thus, the intersection points are $(-1, -1)$, $(0, 0)$, and $(1, 1)$.

The region $S$ consists of two parts: $S_1$ where $-1 \le x \le 0$, and $S_2$ where $0 \le x \le 1$.

In $S_1$, $x^3 \ge x$, and in $S_2$, $x \ge x^3$.

Then the double integral can be written as the sum of two integrals:

$\iint_S x \, dA = \iint_{S_1} x \, dA + \iint_{S_2} x \, dA$

For $S_1$, we have

$\iint_{S_1} x \, dA = \int_{-1}^0 \int_{x}^{x^3} x \, dy \, dx = \int_{-1}^0 x(x^3 - x) \, dx = \int_{-1}^0 (x^4 - x^2) \, dx = \left[ \frac{x^5}{5} - \frac{x^3}{3} \right]_{-1}^0 = 0 - \left( \frac{(-1)^5}{5} - \frac{(-1)^3}{3} \right) = - \left( -\frac{1}{5} + \frac{1}{3} \right) = - \left( \frac{-3 + 5}{15} \right) = -\frac{2}{15}$.

For $S_2$, we have

$\iint_{S_2} x \, dA = \int_0^1 \int_{x^3}^{x} x \, dy \, dx = \int_0^1 x(x - x^3) \, dx = \int_0^1 (x^2 - x^4) \, dx = \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_0^1 = \frac{1}{3} - \frac{1}{5} - 0 = \frac{5 - 3}{15} = \frac{2}{15}$.

Therefore,

$\iint_S x \, dA = -\frac{2}{15} + \frac{2}{15} = 0$.

3. Final Answer

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