Question

Find the area of the region bounded by the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$.

• A. $6\pi$
• B. $12\pi$
• C. $18\pi$
• D. $24\pi$

Answer 1

The correct option is B $12\pi$

The given equation of the ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$.

Area bounded by ellipse = $4\times AreaofOAB$

Area of OAB = ${\int }_0^{4}ydx$

$=\frac{3}{4}{\int }_0^4\sqrt{16-x^2}dx$

$=\frac{3}{4}{[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}{\sin }^{-1}\frac{x}{4}]}_0^4$

$=\frac{3}{4}\times \frac{8\pi }{2}=3\pi$

Thus, required area = $4\times 3\pi =12\pi$ sq. units