What is the area of the shared region in the figure above that is bounded by the -axis and the curve with the equation $$ y = x \sqrt{1 - x^2} $$?

To find the area of the region bounded by the $ x $-axis and the curve $ y = x \sqrt{1 - x^2} $, we need to set up and evaluate the appropriate definite integral.

### Step 1: Identify the interval
The curve $ y = x \sqrt{1 - x^2} $ is defined for values of $ x $ where $ 1 - x^2 \geq 0 $. This inequality holds for $ -1 \leq x \leq 1 $. Therefore, the relevant interval for $ x $ is $ [-1, 1] $.

### Step 2: Set up the integral
The area of the region bounded by the curve and the $ x $-axis is given by the integral of $ y $ with respect to $ x $ over the interval $ [-1, 1] $:

\[
\text{Area} = \int_{-1}^{1} x \sqrt{1 - x^2} \, dx
\]

### Step 3: Use symmetry to simplify the calculation
Notice that the function $ y = x \sqrt{1 - x^2} $ is odd (i.e., $ f(-x) = -f(x) $). Therefore, the area calculation can be simplified by considering only the interval $ [0, 1] $ and then doubling the result:

\[
\text{Area} = 2 \int_0^1 x \sqrt{1 - x^2} \, dx
\]

### Step 4: Evaluate the integral
To evaluate the integral, use the substitution $ u = 1 - x^2 $, which implies $ du = -2x \, dx $ or $ dx = -\frac{du}{2x} $. When $ x = 0 $, $ u = 1 $. When $ x = 1 $, $ u = 0 $.

Substituting into the integral:

\[
\int_0^1 x \sqrt{1 - x^2} \, dx = \int_1^0 \sqrt{u} \left(-\frac{du}{2}\right) = \frac{1}{2} \int_0^1 \sqrt{u} \, du
\]

Now, evaluate the integral:

\[
\frac{1}{2} \int_0^1 u^{1/2} \, du = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_0^1 = \frac{1}{3}
\]

### Step 5: Compute the area using the result from the integral
Since the area over the interval $ [0, 1] $ is $ \frac{1}{3} $, the total area, considering both sides of the $ y $-axis, is:

\[
\text{Total Area} = 2 \times \frac{1}{3} = \frac{2}{3}
\]

Therefore, the area of the shared region is $\boxed{\frac{2}{3}}$.

Would you like to go over any specific part of the solution or have additional questions?

### Related Questions:
1. How would you find the area under the curve $ y = x^2 \sqrt{1 - x^2} $ between $ x = 0 $ and $ x = 1 $?
2. What is the integral of $ \sqrt{1 - x^2} $ over the interval $[-1, 1]$?
3. How can you determine if a function is odd or even, and why is this property useful when calculating areas?
4. Can you find the area of the region enclosed by the curve $ y = \sqrt{1 - x^2} $ and the $ x $-axis?
5. How does the substitution method work, and when is it helpful in solving integrals?

### Tip:
Always check for symmetry in functions before setting up an integral. Symmetry can simplify calculations and reduce the amount of work needed.

Learn how to calculate the area of the region bounded by the curve y = x * sqrt(1 - x^2) and the x-axis using definite integrals and symmetry in functions. This problem involves integral calculus and is suitable for advanced high school students or college-level calculus courses.

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