The area of the smaller region in first quadrant bounded by \( x^{2}+y^{2}=4 \) and \( x=\displaystyle\frac{3}{2} \) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \pi-\displaystyle\frac{3\sqrt{7}}{8}-2\sin^{-1}\left(\displaystyle\frac{3}{4}\right) \) Sq.units
The line \( x=\frac{3}{2} \) cuts off the smaller region between \( x=\frac{3}{2} \) and \( x=2 \), under the arc \( y=\sqrt{4-x^{2}} \).
\( \displaystyle A=\int_{3/2}^{2}\sqrt{4-x^{2}}\,dx \)
By the standard integral, with a = 2,
\( \displaystyle \int\sqrt{a^{2}-x^{2}}\,dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a} \)
So
\( \displaystyle A=\left[\frac{x}{2}\sqrt{4-x^{2}}+2\sin^{-1}\frac{x}{2}\right]_{3/2}^{2} \)
At the upper limit the value is \( 0+2\cdot\frac{\pi}{2}=\pi \). At the lower limit, where \( \sqrt{4-\frac{9}{4}}=\frac{\sqrt{7}}{2} \), it is \( \frac{3\sqrt{7}}{8}+2\sin^{-1}\frac{3}{4} \). So
\( \displaystyle A=\pi-\frac{3\sqrt{7}}{8}-2\sin^{-1}\left(\frac{3}{4}\right) \)