Question

The area (in sq. units) of the region \((x, y) : 0 \leq y \leq 2|x| + 1, 0 \leq y \leq x^2 + 1, |x| \leq 3\) is:

• A. \(\frac{64}{3}\)
• B. \(\frac{17}{3}\)
• C. \(\frac{32}{3}\)
• D. \(\frac{80}{3}\)

Hint:

Always check the limits and symmetry in region integrations to simplify the work.

Answer 1

The Correct Option is A

Given: The region defined by: \[ 0 \leq y \leq 2|x| + 1 \] \[ 0 \leq y \leq x^2 + 1 \] \[ |x| \leq 3 \]

Step 1: Exploit Symmetry
Since the region is symmetric about the y-axis, we calculate the area for \(x \geq 0\) and multiply by 2.

Step 2: First Quadrant Inequalities
For \(x \geq 0\): \[ 0 \leq y \leq \min(2x + 1, x^2 + 1) \] \[ 0 \leq x \leq 3 \]

Step 3: Find Intersection Points
Set \(2x + 1 = x^2 + 1\): \[ x^2 - 2x = 0 \] \[ x = 0 \text{ or } x = 2 \]

Step 4: Determine Dominant Function
For \(0 \leq x \leq 2\): \(x^2 + 1 \leq 2x + 1\)
For \(2 \leq x \leq 3\): \(2x + 1 \leq x^2 + 1\)

Step 5: Calculate Area
First quadrant area: \[ \int_0^2 (x^2 + 1) dx + \int_2^3 (2x + 1) dx \] \[ = \left[ \frac{x^3}{3} + x \right]_0^2 + \left[ x^2 + x \right]_2^3 \] \[ = \left(\frac{8}{3} + 2\right) + (9 + 3) - (4 + 2) \] \[ = \frac{14}{3} + 6 = \frac{32}{3} \]

Step 6: Total Area
Multiply by 2 for symmetry: \[ 2 \times \frac{32}{3} = \frac{64}{3} \]

Verification:
Alternative calculation confirms: \[ 2\left( \frac{14}{3} + 6 \right) = \frac{64}{3} \]

Final Answer:
The area of the region is \(\dfrac{64}{3}\).