The area of the region bounded by the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}={1}\) is

Calculation

The given equation of the ellipse,

\(\frac{x^2}{16} + \frac{y^2}{9} = 1\),

the standard form of ellipse is given by

\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

by comparing we get,

a = 4 and b = 3

It can be observed that the ellipse is symmetrical about x-axis and y-axis.

Area bounded by ellipse = 4 × Area of OAB

Area of OAB = \(\int_{0}^{4} y dx\) = \(\int_{0}^{4} y dx\)

= \(\int_{0}^{4} 3\sqrt{1 - \frac{x^2}{16}} dx\)

= \(\frac{3}{4} \int_{0}^{4} \sqrt{16 - x^2} dx\)

= \(\frac{3}{4} [\frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2} \sin^{-1}\frac{x}{4}]_{0}^{4}\)

= \(\frac{3}{4} [2\sqrt{16 - 16} + 8 \sin^{-1}(1) - 0 - 8 \sin^{-1}(0)]\)

= \(\frac{3}{4} [\frac{8\pi}{2}]\)

= \(\frac{3}{4} [4\pi]\) = 3π

Therefore, area bounded by the ellipse = 4 × 3π = 12π units

Hence option 1 is correct