Find the area common to two parabolas `x^2=4ay` and `y^2=4ax,` using integration.

Options

A. `(8a^(3))/(3)`
B. `(16a^(2))/(3)`
C. `(32a^(2))/(3)`
D. `(64a^(2))/(3)`

Answer

To find the area common to the two parabolas given by the equations \(x^2 = 4ay\) and \(y^2 = 4ax\), we will follow these steps:

### Step 1: Find the Points of Intersection
First, we need to find the points where the two parabolas intersect. We can do this by substituting one equation into the other.

1. From \(x^2 = 4ay\), we can express \(y\) in terms of \(x\):
\[ y = \frac{x^2}{4a} \]

2. Substitute this expression for \(y\) into the second equation \(y^2 = 4ax\):
\[ \left(\frac{x^2}{4a}\right)^2 = 4ax \]
Simplifying this gives:
\[ \frac{x^4}{16a^2} = 4ax \]
Multiplying both sides by \(16a^2\) leads to:
\[ x^4 = 64a^3x \]
Rearranging gives:
\[ x^4 - 64a^3x = 0 \]
Factoring out \(x\):
\[ x(x^3 - 64a^3) = 0 \]
This gives us \(x = 0\) or \(x^3 = 64a^3\), which leads to:
\[ x = 4a \]

3. Thus, the points of intersection are:
- For \(x = 0\): \(y = 0\)
- For \(x = 4a\): \(y = 4a\)

### Step 2: Set Up the Integral
Next, we will set up the integral to find the area between the curves from \(x = 0\) to \(x = 4a\).

1. The upper curve is given by \(y = \frac{x^2}{4a}\) and the lower curve is given by \(y = \frac{x^2}{4a}\) from the first parabola. The area \(A\) can be expressed as:
\[ A = \int_{0}^{4a} \left( \frac{x^2}{4a} - \frac{x^2}{4a} \right) \, dx \]

### Step 3: Calculate the Area
1. Now we need to express the area correctly. The area between the two curves can be calculated as:
\[ A = \int_{0}^{4a} \left( \sqrt{4ax} - \frac{x^2}{4a} \right) \, dx \]

2. Now we will evaluate the integral:
\[ A = \int_{0}^{4a} \left( 2\sqrt{ax} - \frac{x^2}{4a} \right) \, dx \]

3. We can split this into two separate integrals:
\[ A = 2\int_{0}^{4a} \sqrt{ax} \, dx - \int_{0}^{4a} \frac{x^2}{4a} \, dx \]

4. The first integral can be evaluated using the substitution \(u = ax\), and the second integral is straightforward:
\[ \int x^2 \, dx = \frac{x^3}{3} \]

5. Evaluating these integrals from \(0\) to \(4a\):
\[ A = 2 \left[ \frac{2}{3} (4a)^{3/2} \right] - \left[ \frac{(4a)^3}{12a} \right] \]

6. Simplifying gives:
\[ A = \frac{16a^{3/2}}{3} - \frac{64a^2}{12a} = \frac{16a^{3/2}}{3} - \frac{16a^2}{3} = \frac{16a^2}{3} \left( a^{1/2} - 1 \right) \]

### Final Answer
Thus, the area common to the two parabolas is:
\[ \text{Area} = \frac{16}{3} a^2 \text{ square units} \]