Area Under the Curve Question - None

Question

Consider the following regions in the plane:
$R_1 = \{(x, y) : 0 \leq x \leq 1 \text{ and } 0 \leq y \leq 1\}$
$R_2 = \{(x, y) : x^2 + y^2 \leq \frac{4}{3}\}$
The area of the region $R_1 \cap R_2$ can be expressed as $\frac{a\pi}{9} + \frac{b}{3}$, where $a$ and $b$ are integers. Find the value of $a + b$.

Accepted Answer

Given: $R_1$ is the unit square $[0,1] imes [0,1]$ and $R_2$ is a disk with radius $r = \sqrt{\frac{4}{3}} = \frac{2\sqrt{3}}{3}$

Step 1: The intersection $R_1 \cap R_2$ consists of the unit square minus the portion outside the disk.

Step 2: Using integration and geometric analysis, the area is computed as:

$A = \frac{3\pi}{9} + \frac{3}{3} = \frac{\pi}{3} + 1$

Step 3: Expressing in the form $\frac{a\pi}{9} + \frac{b}{3}$:

$\frac{3\pi}{9} + \frac{3}{3}$, so $a = 3$ and $b = 1$

∴ Answer: $a + b = 4$

Question:

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