What is the area bounded by the curve `y=x^(2)` and the line y=16?

Options

A. `32//3`
B. `64//3`
C. `256//3`
D. `128/3`

Answer

To find the area bounded by the curve \( y = x^2 \) and the line \( y = 16 \), we will follow these steps:

### Step 1: Find the points of intersection
We need to find the points where the curve \( y = x^2 \) intersects the line \( y = 16 \). To do this, we set the two equations equal to each other:

\[ x^2 = 16 \]

Taking the square root of both sides, we find:

\[ x = 4 \quad \text{and} \quad x = -4 \]

### Step 2: Set up the integral for the area
The area bounded by the curve and the line can be calculated using the integral from \( -4 \) to \( 4 \). The area \( A \) is given by:

\[ A = \int_{-4}^{4} (16 - x^2) \, dx \]

Here, \( 16 \) is the upper function (the line) and \( x^2 \) is the lower function (the curve).

### Step 3: Calculate the integral
We can calculate the integral:

\[ A = \int_{-4}^{4} (16 - x^2) \, dx \]

This can be split into two separate integrals:

\[ A = \int_{-4}^{4} 16 \, dx - \int_{-4}^{4} x^2 \, dx \]

Calculating the first integral:

\[ \int_{-4}^{4} 16 \, dx = 16 \cdot (4 - (-4)) = 16 \cdot 8 = 128 \]

Now, calculating the second integral:

\[ \int_{-4}^{4} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-4}^{4} = \frac{4^3}{3} - \frac{(-4)^3}{3} = \frac{64}{3} - \frac{-64}{3} = \frac{64 + 64}{3} = \frac{128}{3} \]

### Step 4: Combine the results
Now, we can combine the results of our integrals:

\[ A = 128 - \frac{128}{3} \]

To combine these, we convert \( 128 \) into a fraction with a denominator of \( 3 \):

\[ 128 = \frac{384}{3} \]

So,

\[ A = \frac{384}{3} - \frac{128}{3} = \frac{384 - 128}{3} = \frac{256}{3} \]

### Final Answer
Thus, the area bounded by the curve \( y = x^2 \) and the line \( y = 16 \) is:

\[ \boxed{\frac{256}{3}} \]