Draw the diagram to show the area enclosed by the curves :
`y ^(2) =16 x and x ^(2) = 16y.`
The straight line `x =4` divides the area into two parts. Find the area of the larger portion by integration.

Answer

To solve the problem, we need to find the area enclosed by the curves \( y^2 = 16x \) and \( x^2 = 16y \), and then determine the area of the larger portion divided by the line \( x = 4 \).

### Step 1: Find the points of intersection of the curves

1. The first curve is \( y^2 = 16x \).
2. The second curve is \( x^2 = 16y \).

To find the points of intersection, we can express \( y \) from the first equation and substitute it into the second equation.

From \( y^2 = 16x \), we have:
\[ y = 4\sqrt{x} \]

Substituting \( y \) into the second equation:
\[ x^2 = 16(4\sqrt{x}) \]
\[ x^2 = 64\sqrt{x} \]
Squaring both sides:
\[ x^4 = 4096x \]
Rearranging gives:
\[ x^4 - 4096x = 0 \]
Factoring out \( x \):
\[ x(x^3 - 4096) = 0 \]
This gives us \( x = 0 \) or \( x^3 = 4096 \). Solving for \( x \):
\[ x = 0 \quad \text{or} \quad x = 16 \]

Now substituting \( x = 16 \) back into \( y^2 = 16x \):
\[ y^2 = 16 \times 16 = 256 \quad \Rightarrow \quad y = 16 \quad \text{(taking positive root)} \]

Thus, the points of intersection are \( (0, 0) \) and \( (16, 16) \).

### Step 2: Set up the integral for the area

The area between the curves from \( x = 0 \) to \( x = 16 \) can be calculated using the formula:
\[ \text{Area} = \int_{x_1}^{x_2} (f(x) - g(x)) \, dx \]

Where:
- \( f(x) \) is the upper curve and \( g(x) \) is the lower curve.

From the equations:
- For \( y^2 = 16x \), \( y = 4\sqrt{x} \) (upper curve)
- For \( x^2 = 16y \), \( y = \frac{x^2}{16} \) (lower curve)

### Step 3: Calculate the area from \( x = 0 \) to \( x = 4 \)

The area from \( x = 0 \) to \( x = 4 \):
\[ \text{Area} = \int_{0}^{4} \left(4\sqrt{x} - \frac{x^2}{16}\right) \, dx \]

### Step 4: Calculate the area from \( x = 4 \) to \( x = 16 \)

The area from \( x = 4 \) to \( x = 16 \):
\[ \text{Area} = \int_{4}^{16} \left(4\sqrt{x} - \frac{x^2}{16}\right) \, dx \]

### Step 5: Solve the integral

1. **Integrate \( 4\sqrt{x} \)**:
\[ \int 4\sqrt{x} \, dx = \frac{4}{\frac{3}{2}} x^{\frac{3}{2}} = \frac{8}{3} x^{\frac{3}{2}} \]

2. **Integrate \( \frac{x^2}{16} \)**:
\[ \int \frac{x^2}{16} \, dx = \frac{1}{16} \cdot \frac{x^3}{3} = \frac{x^3}{48} \]

Putting it all together:
\[ \int \left(4\sqrt{x} - \frac{x^2}{16}\right) \, dx = \frac{8}{3} x^{\frac{3}{2}} - \frac{x^3}{48} \]

### Step 6: Evaluate the definite integral from \( 4 \) to \( 16 \)

Calculating:
\[ \left[\frac{8}{3} (16)^{\frac{3}{2}} - \frac{(16)^3}{48}\right] - \left[\frac{8}{3} (4)^{\frac{3}{2}} - \frac{(4)^3}{48}\right] \]

Calculating each term:
- For \( x = 16 \):
\[ \frac{8}{3} \cdot 64 - \frac{4096}{48} = \frac{512}{3} - \frac{256}{3} = \frac{256}{3} \]

- For \( x = 4 \):
\[ \frac{8}{3} \cdot 8 - \frac{64}{48} = \frac{64}{3} - \frac{4}{3} = \frac{60}{3} = 20 \]

Thus:
\[ \text{Area} = \frac{256}{3} - 20 = \frac{256}{3} - \frac{60}{3} = \frac{196}{3} \]

### Final Area of the Larger Portion

The area of the larger portion is:
\[ \text{Area} = \frac{196}{3} \text{ square units} \]