Find the area included between the curves `x^2=4y` and `y^2=4x`.

Options

A. `4//3`
B. `1//3`
C. `16//3`
D. `8//3`

Answer

To find the area included between the curves \( x^2 = 4y \) and \( y^2 = 4x \), we will follow these steps:

### Step 1: Identify the curves
The first curve is given by \( x^2 = 4y \), which can be rewritten as:
\[ y = \frac{x^2}{4} \]
The second curve is given by \( y^2 = 4x \), which can be rewritten as:
\[ y = 2\sqrt{x} \]
This shows that we have a parabola opening upwards and another parabola opening to the right.

### Step 2: Find the points of intersection
To find the points of intersection, we set the two equations equal to each other:
\[ \frac{x^2}{4} = 2\sqrt{x} \]
Multiplying both sides by 4 to eliminate the fraction:
\[ x^2 = 8\sqrt{x} \]
Now, we can rearrange this equation:
\[ x^2 - 8\sqrt{x} = 0 \]
Let \( u = \sqrt{x} \), then \( x = u^2 \). Substituting this in:
\[ u^4 - 8u = 0 \]
Factoring out \( u \):
\[ u(u^3 - 8) = 0 \]
This gives us:
\[ u = 0 \quad \text{or} \quad u^3 = 8 \Rightarrow u = 2 \]
Thus, \( \sqrt{x} = 0 \) gives \( x = 0 \) and \( \sqrt{x} = 2 \) gives \( x = 4 \). Therefore, the points of intersection are \( (0, 0) \) and \( (4, 4) \).

### Step 3: Set up the integral for the area
The area between the curves from \( x = 0 \) to \( x = 4 \) can be calculated using the integral:
\[ \text{Area} = \int_{0}^{4} \left( \text{upper function} - \text{lower function} \right) \, dx \]
Here, the upper function is \( y = 2\sqrt{x} \) and the lower function is \( y = \frac{x^2}{4} \). Thus, we have:
\[ \text{Area} = \int_{0}^{4} \left( 2\sqrt{x} - \frac{x^2}{4} \right) \, dx \]

### Step 4: Calculate the integral
Now we calculate the integral:
\[ \text{Area} = \int_{0}^{4} 2\sqrt{x} \, dx - \int_{0}^{4} \frac{x^2}{4} \, dx \]
Calculating the first integral:
\[ \int 2\sqrt{x} \, dx = \frac{2 \cdot 2}{3} x^{3/2} = \frac{4}{3} x^{3/2} \]
Evaluating from 0 to 4:
\[ \left[ \frac{4}{3} x^{3/2} \right]_{0}^{4} = \frac{4}{3} (4^{3/2}) - 0 = \frac{4}{3} \cdot 8 = \frac{32}{3} \]

Now for the second integral:
\[ \int \frac{x^2}{4} \, dx = \frac{1}{4} \cdot \frac{x^3}{3} = \frac{x^3}{12} \]
Evaluating from 0 to 4:
\[ \left[ \frac{x^3}{12} \right]_{0}^{4} = \frac{4^3}{12} - 0 = \frac{64}{12} = \frac{16}{3} \]

### Step 5: Combine the results
Now, we combine the results:
\[ \text{Area} = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} \]

Thus, the area included between the curves \( x^2 = 4y \) and \( y^2 = 4x \) is:
\[ \boxed{\frac{16}{3}} \]