Area of region bounded by the curve `y=(16-x^(2))/(4)` and `y=sec^(-1)[-sin^(2)x]` (where [x] denotes the greatest ingeger function) is

Options

A. `(1)/(3)(4-pi)^(3//2)`
B. `8(4-pi)^(3//2)`
C. `(8)/(3)(4-pi)^(3//2)`
D. `(8)/(3)(4-pi)^(1//2)`

Explanation

To find the area of the region bounded by the curves \( y = \frac{16 - x^2}{4} \) and \( y = \sec^{-1}[-\sin^2 x] \), we will follow these steps:

### Step 1: Identify the curves
The first curve is given by:
\[ y = \frac{16 - x^2}{4} \]
The second curve is:
\[ y = \sec^{-1}[-\sin^2 x] \]

### Step 2: Determine the range of \( y = \sec^{-1}[-\sin^2 x] \)
The function \( -\sin^2 x \) varies between -1 and 0. The secant inverse function is defined for values \( x \leq -1 \) or \( x \geq 1 \). Thus, we can evaluate \( y \) at these bounds:
- For \( -\sin^2 x = -1 \), we have \( y = \sec^{-1}(-1) = \pi \).
- For \( -\sin^2 x = 0 \), \( y \) is not defined.

### Step 3: Set the equations equal to find intersection points
To find the points of intersection, we set the two equations equal:
\[ \frac{16 - x^2}{4} = \pi \]
Multiplying through by 4:
\[ 16 - x^2 = 4\pi \]
Rearranging gives:
\[ x^2 = 16 - 4\pi \]
Taking the square root:
\[ x = \pm 2\sqrt{4 - \pi} \]

### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( x = -2\sqrt{4 - \pi} \) to \( x = 2\sqrt{4 - \pi} \) is given by:
\[ A = \int_{-2\sqrt{4 - \pi}}^{2\sqrt{4 - \pi}} \left( \frac{16 - x^2}{4} - \pi \right) \, dx \]

### Step 5: Simplify the integral
This can be simplified as:
\[ A = \int_{-2\sqrt{4 - \pi}}^{2\sqrt{4 - \pi}} \left( \frac{16 - x^2 - 4\pi}{4} \right) \, dx \]
\[ = \frac{1}{4} \int_{-2\sqrt{4 - \pi}}^{2\sqrt{4 - \pi}} (16 - x^2 - 4\pi) \, dx \]

### Step 6: Calculate the integral
Now we compute the integral:
\[ = \frac{1}{4} \left[ 16x - \frac{x^3}{3} - 4\pi x \right]_{-2\sqrt{4 - \pi}}^{2\sqrt{4 - \pi}} \]

### Step 7: Evaluate the definite integral
Calculating at the limits:
1. At \( x = 2\sqrt{4 - \pi} \):
\[ = 16(2\sqrt{4 - \pi}) - \frac{(2\sqrt{4 - \pi})^3}{3} - 4\pi(2\sqrt{4 - \pi}) \]
2. At \( x = -2\sqrt{4 - \pi} \):
\[ = 16(-2\sqrt{4 - \pi}) - \frac{(-2\sqrt{4 - \pi})^3}{3} - 4\pi(-2\sqrt{4 - \pi}) \]

Combining these results gives the total area.

### Final Result
After evaluating the integral and simplifying, we find:
\[ A = \frac{8}{3}(4 - \pi)^{3/2} \]