A) \[3\pi {{a}^{2}}\] sq. unit
B) \[\frac{3\pi {{a}^{2}}}{2}\] sq. unit
C) \[\frac{3\pi {{a}^{2}}}{4}\] sq. unit
D) \[\frac{6\pi {{a}^{2}}}{5}\] sq. unit
Correct Answer: B
Solution :
| [b] Let the equation of curve |
| \[{{y}^{2}}(2a-x)={{x}^{3}}...(i)\] |
| and equation of line \[x=2a...(ii)\] |
| The given curve is symmetrical about x-axis and passes through origin. |
| From (i) we have, \[{{y}^{2}}=\frac{{{x}^{3}}}{2a-x}\] |
| But \[\frac{{{x}^{3}}}{2a-x}<0\] for \[x>2a\] and \[x<0\] |
| So, curve does not lie in the portion \[x>2a\] and \[x<0\], therefore curve lies in \[0\le x\le 2a\]. |
| \[\therefore \] Area bounded by the curve and line |
| \[=\int\limits_{0}^{2a}{ydx=\int\limits_{0}^{2a}{\frac{{{x}^{3/2}}}{\sqrt{2a}-x}dx}}\] |
| Put \[x=2a{{\sin }^{2}}\theta \] and \[dx=4a\sin \theta \cos \theta d\theta \] |
| \[\therefore I=\int\limits_{0}^{\pi /2}{8{{a}^{2}}{{\sin }^{4}}\theta d\theta =8{{a}^{2}}\left[ \frac{3}{4}.\frac{1}{2}.\frac{\pi }{2} \right]}\] |
| \[=\frac{3\pi {{a}^{2}}}{2}\] sq. unit |
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