Question:
The area of the region described by \( A = \{(x,y) : x^2 + y^2 \le 1 \text{ and } y^2 \le 1 - x \} \) is
The area of the region described by \( A = \{(x,y) : x^2 + y^2 \le 1 \text{ and } y^2 \le 1 - x \} \) is
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Split region when two curves dominate in different intervals.
Updated On: Apr 22, 2026
- \( \frac{\pi}{2} + \frac{4}{3} \)
- \( \frac{\pi}{2} - \frac{1}{2} \)
- \( \frac{\pi}{4} + \frac{2}{3} \)
- \( \frac{\pi}{2} + \frac{2}{3} \)
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The Correct Option is D
Solution and Explanation
Concept:
Region is intersection of:
\[
x^2 + y^2 \le 1 \quad (\text{circle})
\]
\[
y^2 \le 1 - x \quad (\text{parabola})
\]
Step 1: Find limits.
Intersection gives: \[ x^2 + (1-x) = 1 \Rightarrow x(x-1)=0 \Rightarrow x=0,1 \]
Step 2: Area expression.
\[ \text{Area} = \int_0^1 2\sqrt{1-x}\,dx + \int_{-1}^0 2\sqrt{1-x^2}\,dx \]
Step 3: Evaluate.
\[ \int_0^1 2\sqrt{1-x}\,dx = \frac{4}{3} \] \[ \int_{-1}^0 2\sqrt{1-x^2}\,dx = \frac{\pi}{2} \]
Step 4: Total area.
\[ = \frac{\pi}{2} + \frac{2}{3} \]
Step 1: Find limits.
Intersection gives: \[ x^2 + (1-x) = 1 \Rightarrow x(x-1)=0 \Rightarrow x=0,1 \]
Step 2: Area expression.
\[ \text{Area} = \int_0^1 2\sqrt{1-x}\,dx + \int_{-1}^0 2\sqrt{1-x^2}\,dx \]
Step 3: Evaluate.
\[ \int_0^1 2\sqrt{1-x}\,dx = \frac{4}{3} \] \[ \int_{-1}^0 2\sqrt{1-x^2}\,dx = \frac{\pi}{2} \]
Step 4: Total area.
\[ = \frac{\pi}{2} + \frac{2}{3} \]
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