The area (in square units) of the region bounded by `x^(2)=8y,x=4` and x-axis is

Options

A. `(2)/(3)`
B. `(4)/(3)`
C. `(8)/(3)`
D. `(10)/(3)`

Answer

To find the area of the region bounded by the parabola \( x^2 = 8y \), the line \( x = 4 \), and the x-axis, we can follow these steps:

### Step 1: Understand the curves
The equation \( x^2 = 8y \) represents a parabola that opens upwards. The line \( x = 4 \) is a vertical line. The x-axis is represented by \( y = 0 \).

**Hint:** Sketch the curves to visualize the bounded area.

### Step 2: Express \( y \) in terms of \( x \)
From the equation of the parabola, we can express \( y \) as:
\[ y = \frac{x^2}{8} \]

**Hint:** Rearranging the equation helps in setting up the integral for area calculation.

### Step 3: Set up the integral
The area \( A \) under the curve from \( x = 0 \) to \( x = 4 \) can be calculated using the integral:
\[ A = \int_{0}^{4} \frac{x^2}{8} \, dx \]

**Hint:** The limits of integration correspond to the x-values where the region is bounded.

### Step 4: Integrate the function
Now, we integrate:
\[ A = \int_{0}^{4} \frac{x^2}{8} \, dx = \frac{1}{8} \int_{0}^{4} x^2 \, dx \]
The integral of \( x^2 \) is:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Thus:
\[ A = \frac{1}{8} \left[ \frac{x^3}{3} \right]_{0}^{4} \]

**Hint:** Remember to apply the fundamental theorem of calculus to evaluate the definite integral.

### Step 5: Evaluate the definite integral
Now, we evaluate the integral from 0 to 4:
\[ A = \frac{1}{8} \left( \frac{4^3}{3} - \frac{0^3}{3} \right) = \frac{1}{8} \left( \frac{64}{3} - 0 \right) = \frac{64}{24} \]

**Hint:** Simplifying fractions can help in finding the final area.

### Step 6: Simplify the result
Now, simplify \( \frac{64}{24} \):
\[ A = \frac{64 \div 8}{24 \div 8} = \frac{8}{3} \]

**Hint:** Always check if the fraction can be simplified for a cleaner answer.

### Final Answer
The area of the region bounded by the curves is:
\[ \boxed{\frac{8}{3}} \text{ square units} \]