The area of the region {(x,y): x2 ≤ y ≤8-x2, y≤7} is
Step 1: Define the inequalities.
The region is bounded by the following inequalities: \[ y \geq x^2, \quad y \leq 8 - x^2, \quad y \leq 7. \] The points of intersection between the curves are: - \((-2, 4)\) and \((2, 4)\) where \(y = x^2\) intersects \(y = 8 - x^2\), - \((-1, 7)\) and \((1, 7)\) where \(y = 7\) intersects \(y = 8 - x^2\).
Step 2: Set up the integral.
We need to calculate the area between the curves using the following integral expression: \[ \text{Area} = 2 \left( 1.7 + \int_{-1}^{1} (8 - 2x^2) \, dx \right) - 2 \int_{0}^{1} x^2 \, dx. \] Step 3: Compute the first integral.
First, compute the integral for the region from \(x = -1\) to \(x = 1\) where the upper curve is \(8 - 2x^2\): \[ \int_{-1}^{1} (8 - 2x^2) \, dx = \left[ 8x - \frac{2x^3}{3} \right]_{-1}^{1}. \] Evaluating this: \[ = \left( 8(1) - \frac{2(1)^3}{3} \right) - \left( 8(-1) - \frac{2(-1)^3}{3} \right) \] \[ = \left( 8 - \frac{2}{3} \right) - \left( -8 + \frac{2}{3} \right) = 8 - \frac{2}{3} + 8 - \frac{2}{3} = 16 - \frac{4}{3}. \] Thus, the integral becomes: \[ \int_{-1}^{1} (8 - 2x^2) \, dx = 16 - \frac{4}{3} = \frac{44}{3}. \] Step 4: Compute the second integral.
Next, compute the integral for the region from \(x = 0\) to \(x = 1\) where the curve is \(x^2\): \[ \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}. \] Step 5: Combine the results.
Now combine the results from both integrals: \[ \text{Area} = 2 \left( 1.7 + \frac{44}{3} \right) - 2 \times \frac{1}{3}. \] Simplifying this expression: \[ = 2 \left( 7 + \frac{32}{3} - \frac{22}{3} \right) - \frac{2}{3} = 2 \left( 7 + \frac{10}{3} \right) - \frac{2}{3} = 2 \left( \frac{21}{3} + \frac{10}{3} \right) - \frac{2}{3} = 2 \times \frac{31}{3} - \frac{2}{3} \] \[ = \frac{62}{3} - \frac{2}{3} = \frac{60}{3} = 20. \] Final Answer: 20.