Question:
The area of the region \([(x,y): x^2 + y^2 \leq 1 \leq x + y]\) is
The area of the region \([(x,y): x^2 + y^2 \leq 1 \leq x + y]\) is
Show Hint
Sketch the region: quarter circle minus right triangle.
Updated On: Apr 23, 2026
- \(\frac{\pi}{5}\)
- \(\frac{\pi}{4}\)
- \(\frac{\pi^2}{3}\)
- \(\frac{\pi}{4} - \frac{1}{2}\)
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The Correct Option is D
Solution and Explanation
Step 1: Formula / Definition}
\[ \text{Area} = \text{Area of quarter circle} - \text{Area of triangle} \]
Step 2: Calculation / Simplification}
Circle: \(x^2 + y^2 = 1\). Line: \(x + y = 1\)
Intersection: \(x^2 + (1-x)^2 = 1 \Rightarrow 2x^2 - 2x = 0 \Rightarrow x = 0, 1\)
Points: \((0,1)\) and \((1,0)\)
Area of sector = \(\frac{\pi}{4}\)
Area of triangle = \(\frac{1}{2} \times 1 \times 1 = \frac{1}{2}\)
Area = \(\frac{\pi}{4} - \frac{1}{2}\)
Step 3: Final Answer
\[ \frac{\pi}{4} - \frac{1}{2} \]
\[ \text{Area} = \text{Area of quarter circle} - \text{Area of triangle} \]
Step 2: Calculation / Simplification}
Circle: \(x^2 + y^2 = 1\). Line: \(x + y = 1\)
Intersection: \(x^2 + (1-x)^2 = 1 \Rightarrow 2x^2 - 2x = 0 \Rightarrow x = 0, 1\)
Points: \((0,1)\) and \((1,0)\)
Area of sector = \(\frac{\pi}{4}\)
Area of triangle = \(\frac{1}{2} \times 1 \times 1 = \frac{1}{2}\)
Area = \(\frac{\pi}{4} - \frac{1}{2}\)
Step 3: Final Answer
\[ \frac{\pi}{4} - \frac{1}{2} \]
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