Question

The area of the region bounded by the curve $9x^2+4y^2-36=0$ is

• A. $9\pi$
• B. $4\pi$
• C. $36\pi$
• D. $6\pi$

Answer 1

Answer: (D)

The given equation of curve is $9x^2+4y^2-36$
or it can written as $\frac{x^2}{4}+\frac{y^2}{9}=1$

Required area = Area of curve $ADCBA$
$=4$ Area of curve $OCBA$
$=4{\int }_0^{2}ydx$
$=4{\int }_0^{2}3\sqrt{1-\frac{x^2}{4}}dx=6{\int }_0^2\sqrt{4-x^2}dx$
$=4{\left[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}\right]}_0^2$
$=6\left[0+2{\sin }^{-1}1-(0+2(0))\right]$
$=6\cdot 2\cdot \frac{\pi }{2}=6\pi$