question_answer

The area of the region bounded by the curve \[x=2\,\cos \,\theta ,\,\,y=2\,\sin \theta ,\,x=0,\,x=2\] and the x-axis is (in sq unit)

A)  \[4\pi \]           

B)  \[2\pi \]

C)  \[\pi \]                 

D)  \[\frac{\pi }{2}\]

Correct Answer: D

Solution :

Given, curve is \[x=2\,\,\cos \,\,\theta \] \[y=2\,\,sin\,\,\theta \] From the given curve, it is clear that is a circle which is shown below. \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{4}=1\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}=4\] We are to determine the shaded area. Area \[OAB=\int_{x=0}^{2}{y\,dx}\] Area \[OAB=\int_{0}^{2}{\sqrt{1-\frac{{{x}^{2}}}{4}}}\,\,dx\] \[=\left[ \frac{x}{4}\sqrt{\frac{1-{{x}^{2}}}{4}}+\frac{1}{2}{{\sin }^{-1}}\frac{x}{2} \right]_{0}^{2}\] \[=\frac{1}{2}.\frac{\pi }{2}=\frac{\pi }{4}\] \[\therefore \] Area \[OBAC=2\times area\,OAB\] \[=2\times \frac{\pi }{4}=\frac{\pi }{2}\,\,sq\,units\]