The area bounded by the parabola `y^2=x`, straight line `y=4` and Y-axis, is

Options

A. `(16)/3` sq units
B. `(64)/3` sq units
C. `7sqrt2` sq units
D. None of these

Answer

To find the area bounded by the parabola \( y^2 = x \), the straight line \( y = 4 \), and the Y-axis, we can follow these steps:

### Step 1: Understand the curves and their intersections
The parabola \( y^2 = x \) opens to the right, and the line \( y = 4 \) is a horizontal line. We need to find the points where these two curves intersect.

### Step 2: Find the intersection points
Substituting \( y = 4 \) into the parabola's equation:
\[ (4)^2 = x \implies x = 16 \]
Thus, the intersection point is \( (16, 4) \).

### Step 3: Set up the integral for the area
The area we want to find is bounded by the parabola, the line, and the Y-axis. The area can be calculated using the integral from \( x = 0 \) to \( x = 16 \):
\[ \text{Area} = \int_{0}^{16} (4 - \sqrt{x}) \, dx \]
Here, \( 4 \) is the line \( y = 4 \) and \( \sqrt{x} \) is the upper half of the parabola \( y^2 = x \).

### Step 4: Evaluate the integral
Now we will compute the integral:
\[ \int (4 - \sqrt{x}) \, dx = \int 4 \, dx - \int \sqrt{x} \, dx \]
Calculating each integral separately:
1. \(\int 4 \, dx = 4x\)
2. \(\int \sqrt{x} \, dx = \frac{2}{3} x^{3/2}\)

Thus,
\[ \int (4 - \sqrt{x}) \, dx = 4x - \frac{2}{3} x^{3/2} \]

### Step 5: Evaluate the definite integral from 0 to 16
Now we evaluate this from 0 to 16:
\[ \left[ 4x - \frac{2}{3} x^{3/2} \right]_{0}^{16} \]
Calculating at \( x = 16 \):
\[ 4(16) - \frac{2}{3}(16)^{3/2} = 64 - \frac{2}{3}(64) = 64 - \frac{128}{3} = 64 - 42.67 = 21.33 \]

Calculating at \( x = 0 \):
\[ 4(0) - \frac{2}{3}(0)^{3/2} = 0 \]

Thus, the area is:
\[ \text{Area} = 21.33 - 0 = 21.33 \text{ square units} \]

### Final Result
The area bounded by the parabola \( y^2 = x \), the line \( y = 4 \), and the Y-axis is \( \frac{64}{3} \) square units.

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