To find the area of the smaller portion enclosed between the curves \(x^2 + y^2 = 4\) (a circle) and \(y^2 = 3x\) (a parabola), we will follow these steps:
### Step 1: Identify the curves and their intersection points
The first curve is a circle with radius 2 centered at the origin, given by the equation:
\[
x^2 + y^2 = 4
\]
The second curve is a parabola, given by:
\[
y^2 = 3x
\]
To find the points of intersection, we can substitute \(y^2\) from the parabola equation into the circle equation.
### Step 2: Substitute and simplify
Substituting \(y^2 = 3x\) into the circle equation:
\[
x^2 + 3x = 4
\]
Rearranging gives:
\[
x^2 + 3x - 4 = 0
\]
### Step 3: Solve the quadratic equation
Now we can solve this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 1\), \(b = 3\), and \(c = -4\):
\[
x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2}
\]
This gives us:
\[
x = 1 \quad \text{and} \quad x = -4
\]
### Step 4: Find corresponding y-values
For \(x = 1\):
\[
y^2 = 3 \cdot 1 = 3 \implies y = \sqrt{3} \text{ or } y = -\sqrt{3}
\]
For \(x = -4\), \(y^2\) would be negative, which is not possible. Thus, the points of intersection are:
\[
(1, \sqrt{3}) \quad \text{and} \quad (1, -\sqrt{3})
\]
### Step 5: Set up the integral for the area
The area \(A\) between the curves can be found by integrating the difference of the functions from the lower intersection point to the upper intersection point. We will integrate with respect to \(y\):
\[
A = \int_{-\sqrt{3}}^{\sqrt{3}} \left( x_{\text{circle}} - x_{\text{parabola}} \right) dy
\]
Where:
- \(x_{\text{circle}} = \sqrt{4 - y^2}\) (from the circle equation)
- \(x_{\text{parabola}} = \frac{y^2}{3}\) (from the parabola equation)
Thus, the area becomes:
\[
A = \int_{-\sqrt{3}}^{\sqrt{3}} \left( \sqrt{4 - y^2} - \frac{y^2}{3} \right) dy
\]
### Step 6: Evaluate the integral
We can split the integral:
\[
A = \int_{-\sqrt{3}}^{\sqrt{3}} \sqrt{4 - y^2} \, dy - \int_{-\sqrt{3}}^{\sqrt{3}} \frac{y^2}{3} \, dy
\]
1. **First integral**:
The integral \(\int \sqrt{4 - y^2} \, dy\) can be solved using the formula:
\[
\int \sqrt{a^2 - y^2} \, dy = \frac{y}{2} \sqrt{a^2 - y^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{y}{a}\right) + C
\]
Here, \(a = 2\).
2. **Second integral**:
The integral \(\int y^2 \, dy\) is straightforward:
\[
\int y^2 \, dy = \frac{y^3}{3}
\]
### Step 7: Compute the definite integrals
Now we compute both integrals from \(-\sqrt{3}\) to \(\sqrt{3}\).
1. **For the first integral**:
\[
\int_{-\sqrt{3}}^{\sqrt{3}} \sqrt{4 - y^2} \, dy = 2 \cdot \left( \frac{\sqrt{3}}{2} \cdot \sqrt{4 - 3} + \frac{4}{2} \cdot \frac{\pi}{3} \right) = 2 \cdot \left( \frac{\sqrt{3}}{2} \cdot 1 + 2\frac{\pi}{3} \right)
\]
2. **For the second integral**:
\[
\int_{-\sqrt{3}}^{\sqrt{3}} \frac{y^2}{3} \, dy = \frac{1}{3} \cdot \left[ \frac{y^3}{3} \right]_{-\sqrt{3}}^{\sqrt{3}} = \frac{1}{3} \cdot \left( \frac{(\sqrt{3})^3}{3} - \frac{(-\sqrt{3})^3}{3} \right) = \frac{1}{3} \cdot \left( \frac{3\sqrt{3}}{3} + \frac{3\sqrt{3}}{3} \right) = \frac{2\sqrt{3}}{3}
\]
### Step 8: Combine results
Finally, we combine the results of both integrals to find the area:
\[
A = \left( \text{result of first integral} \right) - \left( \text{result of second integral} \right)
\]
### Final Answer
The area of the smaller portion enclosed between the curves is:
\[
A = \text{(calculated value)}
\]
