Find the smaller area enclosed between linex, if `y={x, if x >= 0 and -x, if x < 0` and curve `4x^2+9y^2=36`

Answer

To find the smaller area enclosed between the lines \(y = x\) (for \(x \geq 0\)) and \(y = -x\) (for \(x < 0\)), and the ellipse given by the equation \(4x^2 + 9y^2 = 36\), we will follow these steps:

### Step 1: Rewrite the equation of the ellipse
The equation of the ellipse is given by:
\[ 4x^2 + 9y^2 = 36 \]
Dividing the entire equation by 36, we get:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
This indicates that the ellipse is centered at the origin with semi-major axis \(a = 6\) (along the y-axis) and semi-minor axis \(b = 3\) (along the x-axis).

### Step 2: Find the points of intersection
To find the points of intersection of the lines and the ellipse, we substitute \(y = x\) into the ellipse equation:
\[ 4x^2 + 9(x^2) = 36 \]
This simplifies to:
\[ 4x^2 + 9x^2 = 36 \implies 13x^2 = 36 \implies x^2 = \frac{36}{13} \implies x = \pm \frac{6}{\sqrt{13}} \]
Thus, the points of intersection are:
\[ \left(\frac{6}{\sqrt{13}}, \frac{6}{\sqrt{13}}\right) \quad \text{and} \quad \left(-\frac{6}{\sqrt{13}}, -\frac{6}{\sqrt{13}}\right) \]

### Step 3: Set up the area integral
Since the area enclosed by the lines and the ellipse is symmetric about the y-axis, we can calculate the area in the first quadrant and then double it. The area \(A\) can be expressed as:
\[ A = 2 \int_{0}^{\frac{6}{\sqrt{13}}} \left( \text{upper curve} - \text{lower curve} \right) \, dx \]
The upper curve is the ellipse, and we need to express \(y\) in terms of \(x\):
From the ellipse equation:
\[ y = \frac{2}{3} \sqrt{36 - 4x^2} \]
Thus, the area becomes:
\[ A = 2 \int_{0}^{\frac{6}{\sqrt{13}}} \frac{2}{3} \sqrt{36 - 4x^2} \, dx \]

### Step 4: Evaluate the integral
To evaluate the integral, we can use the substitution \(u = 4x^2\), which gives \(du = 8x \, dx\) or \(dx = \frac{du}{8\sqrt{u/4}} = \frac{du}{4\sqrt{u}}\). The limits change accordingly:
- When \(x = 0\), \(u = 0\)
- When \(x = \frac{6}{\sqrt{13}}\), \(u = \frac{144}{13}\)

Now substituting into the integral:
\[ A = 2 \cdot \frac{2}{3} \cdot \frac{1}{4} \int_{0}^{\frac{144}{13}} \sqrt{36 - u} \cdot \frac{du}{\sqrt{u}} \]

### Step 5: Calculate the area of the triangle
The area of the triangle formed by the lines \(y = x\) and \(y = -x\) can be calculated as:
\[ \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{6}{\sqrt{13}} \times \frac{6}{\sqrt{13}} = \frac{18}{13} \]

### Step 6: Subtract the area of the triangle from the total area
Finally, we subtract the area of the triangle from the area under the curve to find the smaller area enclosed:
\[ \text{Smaller Area} = A - \text{Area of triangle} \]

### Conclusion
The final area can be expressed as:
\[ \text{Smaller Area} = 6 \sin^{-1}\left(\frac{2}{\sqrt{13}}\right) - \frac{18}{13} \]