question_answer

The area of the region bounded by tile curve \[9{{x}^{2}}+4{{y}^{2}}-36=0\] is :

A)  \[9\pi \]                              

B)  \[4\pi \]

C)  \[36\pi \]                                           

D)  \[6\pi \]

Correct Answer: D

Solution :

The given equation of curve is \[9{{x}^{2}}+4{{y}^{2}}-36\] or it can written as \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9}=1\] Required area = Area of curve ADCBA                 = 4 Area of curve OCBA                 \[=4\int_{0}^{3}{3\sqrt{1-\frac{{{x}^{2}}}{4}}\,dx=6}\,\int_{0}^{2}{\sqrt{4-{{x}^{2}}}\,dx}\]                 \[=6\,\left[ \frac{x}{2}\sqrt{4-{{x}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{x}{2} \right]_{0}^{2}\]                 \[=6\,\left[ 0+2{{\sin }^{-1}}1-(0+2\,(0)) \right]\]                 \[=6\,\,.\,\,2\,.\frac{\pi }{2}=6\,\pi \]