Question

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

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

Answer 1

The correct option is D

$6\pi$ sq units

Explanation for the correct option:

Determining the area under the curve:

Given,

$$\begin{gathered} 9x^2+4y^2-36=0 \\ \text{dividing entire equation by}36 \\ \Rightarrow \frac{x^2}{4}+\frac{y^2}{9}=0 \\ \Rightarrow y=\pm 3\sqrt{1-\frac{x^2}{4}} \end{gathered}$$

Required Area$=$Area of total ellipse

$=$$4\times$ Area of part of the ellipse in $1^{st}$ quadrant(so +ve)

$$\begin{gathered} \text{∴ Area}=4{\int }_0^{2}ydx \\ =4{\int }_0^{2}3\sqrt{1-\frac{x^2}{4}}dx \\ =6{\int }_0^2\sqrt{4-x^2}dx \\ =6{\left[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}\right]}_0^2 \\ =6\left[0+2{\sin }^{-1}1-(0+0)\right] \\ =6\times 2\times \frac{\pi }{2} \\ =6\pi \end{gathered}$$
Hence, the correct answer is option (D)