JEE Mains · Maths · STD 12 - 8. Application and integration
The area of the region, enclosed by the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}=2\) which is not common to the region bounded by the parabola \(y^{2}=x\) and the straight line \(\mathrm{y}=\mathrm{x},\) is
- A \(\frac{1}{3}(12 \pi-1)\)
- B \(\frac{1}{6}(12 \pi-1)\)
- C \(\frac{1}{6}(24 \pi-1)\)
- D \(\frac{1}{3}(6 \pi-1)\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{6}(12 \pi-1)\)
Step-by-step Solution
Detailed explanation
\(A=\int_{0}^{1}(\sqrt{x}-x) d x\) \(=\left[\frac{2}{3} x^{3 / 2}-\frac{x^{2}}{2}\right]_{0}^{1}=\frac{1}{6}\) Required Area : \(\pi r^{2}-\frac{1}{6}=\frac{1}{6}(12 \pi-1)\)
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