The are of the region `{(x,y):x^(2)leyle8-x^(2),yle7}` is

Options

A. `18`
B. `21`
C. `24`
D. `20`

Answer

To find the area of the region defined by the inequalities \( x^2 \leq y \leq 8 - x^2 \) and \( y \leq 7 \), we will follow these steps:

### Step 1: Identify the curves
We have two curves:
1. \( y = x^2 \) (a parabola opening upwards)
2. \( y = 8 - x^2 \) (a parabola opening downwards)

Additionally, we have the line \( y = 7 \).

### Step 2: Find the points of intersection
To find the points of intersection of the curves \( y = x^2 \) and \( y = 8 - x^2 \), we set them equal to each other:

\[ x^2 = 8 - x^2 \]

Solving this equation:

\[ 2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2 \]

So, the points of intersection are \( (2, 4) \) and \( (-2, 4) \).

Next, we find the intersection of \( y = 8 - x^2 \) and \( y = 7 \):

\[ 8 - x^2 = 7 \implies x^2 = 1 \implies x = \pm 1 \]

So, the points of intersection are \( (1, 7) \) and \( (-1, 7) \).

### Step 3: Determine the area bounded by the curves
The area we need to calculate is bounded by the curves from \( x = -2 \) to \( x = 1 \) and from \( x = -1 \) to \( x = 2 \).

1. From \( x = -2 \) to \( x = -1 \):
- The upper curve is \( y = 8 - x^2 \)
- The lower curve is \( y = x^2 \)

The area \( A_1 \) in this region is given by:

\[ A_1 = \int_{-2}^{-1} \left( (8 - x^2) - x^2 \right) \, dx = \int_{-2}^{-1} (8 - 2x^2) \, dx \]

2. From \( x = -1 \) to \( x = 1 \):
- The upper curve is \( y = 7 \)
- The lower curve is \( y = x^2 \)

The area \( A_2 \) in this region is given by:

\[ A_2 = \int_{-1}^{1} \left( 7 - x^2 \right) \, dx \]

3. From \( x = 1 \) to \( x = 2 \):
- The upper curve is \( y = 8 - x^2 \)
- The lower curve is \( y = 7 \)

The area \( A_3 \) in this region is given by:

\[ A_3 = \int_{1}^{2} \left( (8 - x^2) - 7 \right) \, dx = \int_{1}^{2} (1 - x^2) \, dx \]

### Step 4: Calculate the integrals
Now we calculate each area:

1. **For \( A_1 \)**:

\[ A_1 = \int_{-2}^{-1} (8 - 2x^2) \, dx = \left[ 8x - \frac{2}{3}x^3 \right]_{-2}^{-1} \]

Calculating this gives:

\[ = \left( 8(-1) - \frac{2}{3}(-1)^3 \right) - \left( 8(-2) - \frac{2}{3}(-2)^3 \right) \]
\[ = (-8 + \frac{2}{3}) - (-16 + \frac{16}{3}) = -8 + \frac{2}{3} + 16 - \frac{16}{3} = 8 - \frac{14}{3} = \frac{24 - 14}{3} = \frac{10}{3} \]

2. **For \( A_2 \)**:

\[ A_2 = \int_{-1}^{1} (7 - x^2) \, dx = \left[ 7x - \frac{1}{3}x^3 \right]_{-1}^{1} \]
Calculating this gives:

\[ = \left( 7(1) - \frac{1}{3}(1)^3 \right) - \left( 7(-1) - \frac{1}{3}(-1)^3 \right) \]
\[ = (7 - \frac{1}{3}) - (-7 + \frac{1}{3}) = 7 - \frac{1}{3} + 7 - \frac{1}{3} = 14 - \frac{2}{3} = \frac{42 - 2}{3} = \frac{40}{3} \]

3. **For \( A_3 \)**:

\[ A_3 = \int_{1}^{2} (1 - x^2) \, dx = \left[ x - \frac{1}{3}x^3 \right]_{1}^{2} \]
Calculating this gives:

\[ = \left( 2 - \frac{1}{3}(2)^3 \right) - \left( 1 - \frac{1}{3}(1)^3 \right) \]
\[ = \left( 2 - \frac{8}{3} \right) - \left( 1 - \frac{1}{3} \right) = \left( \frac{6}{3} - \frac{8}{3} \right) - \left( \frac{3}{3} - \frac{1}{3} \right) = -\frac{2}{3} - \frac{2}{3} = -\frac{4}{3} \]

### Step 5: Total Area
Now, the total area \( A \) is given by:

\[ A = A_1 + A_2 + A_3 = \frac{10}{3} + \frac{40}{3} + \left(-\frac{4}{3}\right) = \frac{10 + 40 - 4}{3} = \frac{46}{3} \]

Thus, the area of the region is:

\[ \boxed{\frac{46}{3}} \]