Question

Find the area bounded in the first quadrant between the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ and the line $3x+4y=12$ is

• A. $3\pi sq.$ units
• B. $(3\pi -6)sq.$ units
• C. $(3\pi -2)sq.$ units
• D. $(3\pi -4)sq.$ units

Answer 1

The correct option is B $(3\pi -6)sq.$ units

$\frac{x^2}{16}+\frac{y^2}{9}=1$

$\Rightarrow y=\frac{3}{4}\sqrt{16-x^2}$

and $3x+4y=12$

$\Rightarrow y=\frac{12-3x}{4}$

Area bounded in first quadrant$={\int }_0^4\left(y_1-y_2\right)dx$

$={\int }_0^4\left(\frac{3}{4}\sqrt{16-x^2}-\frac{12-3x}{4}\right)dx$

$=\frac{3}{4}{\left[\frac{x}{2}\sqrt{16-x^2}+8{\sin }^{-1}\frac{x}{4}\right]}_0^4-\frac{1}{4}{\left[12x-\frac{3x^2}{2}\right]}_0^4$

$=\frac{3}{4}\left(\frac{8\pi }{2}\right)-\frac{1}{4}\left[48-24\right]$

$=\left(3\pi -6\right)$ sq.units