The area of the region in the first quadrant inside the circle \(x^2 + y^2 = 8\) and outside the parabola \(y^2 = 2x\) is equal to:
- \(\frac{\pi}{2} - \frac{1}{3}\)
- \(\pi - \frac{2}{3}\)
- \(\frac{\pi}{2} - \frac{2}{3}\)
- \(\pi - \frac{1}{3}\)
The Correct Option is B
Solution and Explanation
Required area = \( \text{Ar}( \text{circle from 0 to 2}) - \text{Ar}(\text{para from 0 to 2}) \)
\( = \int_0^2 \sqrt{8 - x^2} \, dx - \int_0^2 \sqrt{2x} \, dx \)
\( = \left[ \dfrac{x}{2} \sqrt{8 - x^2} + \dfrac{8}{2} \sin^{-1}\left(\dfrac{x}{2\sqrt{2}}\right) \right]_0^2 - \sqrt{2} \left[ \dfrac{x\sqrt{x}}{3/2} \right]_0^2 \)
\( = \dfrac{2}{2} \sqrt{8 - 4} + \dfrac{8}{2} \sin^{-1}\left(\dfrac{2}{2\sqrt{2}}\right) - \dfrac{2}{2\sqrt{2}} \left( 2\sqrt{2} - 0 \right) \)
\( \Rightarrow 2 + 4 \cdot \dfrac{\pi}{4} - \dfrac{8}{3} = \pi - \dfrac{2}{3} \)
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