To find the area of the region enclosed by the curves \( y^2 \leq 3x \), \( x^2 + y^2 \leq 4 \), and \( y \geq 0 \), we will follow these steps:
### Step 1: Identify the curves
1. The equation \( y^2 = 3x \) represents a rightward-opening parabola.
2. The equation \( x^2 + y^2 = 4 \) represents a circle with a radius of 2 centered at the origin.
### Step 2: Sketch the curves
- Draw the parabola \( y^2 = 3x \) which opens to the right.
- Draw the circle \( x^2 + y^2 = 4 \) which intersects the axes at points (2,0), (0,2), (-2,0), and (0,-2).
- Since we are interested in the region where \( y \geq 0 \), we will only consider the upper half of the circle and the right half of the parabola.
### Step 3: Find the points of intersection
To find the area enclosed by these curves, we need to determine where they intersect. Set \( y^2 = 3x \) into the circle's equation:
\[
x^2 + y^2 = 4 \implies x^2 + 3x = 4
\]
Rearranging gives:
\[
x^2 + 3x - 4 = 0
\]
### Step 4: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 3, c = -4 \):
\[
x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2}
\]
Calculating the roots:
\[
x = 1 \quad \text{and} \quad x = -4
\]
Since \( x = -4 \) is outside the region of interest, we take \( x = 1 \).
### Step 5: Find the corresponding \( y \) value
Substituting \( x = 1 \) back into \( y^2 = 3x \):
\[
y^2 = 3 \cdot 1 = 3 \implies y = \sqrt{3}
\]
Thus, the points of intersection are \( (1, \sqrt{3}) \).
### Step 6: Set up the area integrals
The area \( A \) can be calculated as the sum of two integrals:
1. From \( x = 0 \) to \( x = 1 \) under the parabola.
2. From \( x = 1 \) to \( x = 2 \) under the circle.
The area is given by:
\[
A = \int_0^1 \sqrt{3x} \, dx + \int_1^2 \sqrt{4 - x^2} \, dx
\]
### Step 7: Calculate the first integral
For the first integral:
\[
\int_0^1 \sqrt{3x} \, dx = \sqrt{3} \int_0^1 x^{1/2} \, dx = \sqrt{3} \left[ \frac{x^{3/2}}{3/2} \right]_0^1 = \sqrt{3} \cdot \frac{2}{3} = \frac{2\sqrt{3}}{3}
\]
### Step 8: Calculate the second integral
For the second integral:
\[
\int_1^2 \sqrt{4 - x^2} \, dx
\]
Using the formula for the area under a semicircle:
\[
\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right)
\]
Here, \( a = 2 \):
Evaluating from 1 to 2:
\[
\left[ \frac{x}{2} \sqrt{4 - x^2} + 2 \sin^{-1} \left( \frac{x}{2} \right) \right]_1^2
\]
At \( x = 2 \):
\[
= 0 + 2 \cdot \frac{\pi}{2} = \pi
\]
At \( x = 1 \):
\[
= \frac{1}{2} \cdot \sqrt{3} + 2 \cdot \frac{\pi}{6} = \frac{\sqrt{3}}{2} + \frac{\pi}{3}
\]
Thus, the second integral evaluates to:
\[
\pi - \left( \frac{\sqrt{3}}{2} + \frac{\pi}{3} \right) = \pi - \frac{\pi}{3} - \frac{\sqrt{3}}{2} = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}
\]
### Step 9: Combine the areas
Combining both areas gives:
\[
A = \frac{2\sqrt{3}}{3} + \left( \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \right)
\]
### Step 10: Simplify the final area
Combining the terms:
\[
A = \frac{2\sqrt{3}}{3} - \frac{3\sqrt{3}}{6} + \frac{2\pi}{3} = \frac{4\sqrt{3}}{6} - \frac{3\sqrt{3}}{6} + \frac{2\pi}{3} = \frac{\sqrt{3}}{6} + \frac{2\pi}{3}
\]
### Final Answer
Thus, the area of the region enclosed by the given curves is:
\[
\frac{\sqrt{3}}{6} + \frac{2\pi}{3}
\]
