JEE Mains · Maths · STD 12 - 8. Application and integration
The area of the region in the first quadrant inside the circle \(x^2+y^2=8\) and outside the parabola \(\mathrm{y}^2=2 \mathrm{x}\) is equal to :
- A \(\frac{\pi}{2}-\frac{1}{3}\)
- B \(\pi-\frac{2}{3}\)
- C \(\frac{\pi}{2}-\frac{2}{3}\)
- D \(\pi-\frac{1}{3}\)
Answer & Solution
Correct Answer
(B) \(\pi-\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
Required area \(=\operatorname{Ar}\)( circle from \(0\) to \(2\))- \( \operatorname{ar}(\text { para from } 0 \text { to } 2) \) \( =\int_0^2 \sqrt{8-\mathrm{x}^2} \mathrm{dx}-\int_0^2 \sqrt{2 \mathrm{x}} \mathrm{dx} \)…
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