The area (in sq. units) of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle \(x^{2}+y^{2}=8\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\pi\)
The line y = x meets the circle where
\(2x^{2} = 8\), so \(x = 2\) (first quadrant)
The circle meets the x-axis at \(x = 2\sqrt{2}\). So the region is bounded above by the line from x = 0 to x = 2 and by the circle from x = 2 to \(x = 2\sqrt{2}\):
\( \displaystyle A = \int_{0}^{2} x\,dx + \int_{2}^{2\sqrt{2}}\sqrt{8-x^{2}}\,dx \)
The first part:
\( \displaystyle \int_{0}^{2} x\,dx = \left[\frac{x^{2}}{2}\right]_{0}^{2} = 2 \)
The second part, using \(\int\sqrt{a^{2}-x^{2}}\,dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\) with \(a^{2} = 8\):
\( \displaystyle \left[\frac{x}{2}\sqrt{8-x^{2}}+4\sin^{-1}\frac{x}{2\sqrt{2}}\right]_{2}^{2\sqrt{2}} \)
\( \displaystyle = \left(0+4\times\frac{\pi}{2}\right)-\left(1\times 2+4\times\frac{\pi}{4}\right) \)
\(= \pi - 2\)
Adding the two parts:
\(A = 2 + (\pi - 2) = \pi\)
Shortcut: The region is a sector of the circle of radius \(2\sqrt{2}\) between the angles 0 and \(\frac{\pi}{4}\), which is one eighth of the circle:
\( \displaystyle A = \frac{1}{8}\times\pi\times 8 = \pi \)