Question

Find the area bounded by circle $9x^2+16y^2=144$.

Answer 1

We have,

Given the curve is

$9x^2+16y^2=144$

$\frac{9x^2+16y^2}{144}=1$

$\frac{9x^2}{144}+\frac{16y^2}{144}=1$

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

$\frac{x^2}{4^2}+\frac{y^2}{3^2}=1$

Comparing that,

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

Then,

$a=\pm 4,b=\pm 3$

Area of ellipse $=AreaofABCD$

$=2\times \left[AreaofABC\right]$

$=2\times {\int }_{-4}^{4}ydx$

Finding $y$

We know that,

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

Now area of Ellipse is

$=2\times {\int }_{-4}^{4}ydx$

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

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

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

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

$y^2=\frac{9}{16}(16-x^2)$

$y=\pm \sqrt{\frac{9}{16}(16-x^2)}$

$y=\pm \frac{3}{4}\sqrt{(16-x^2)}$

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

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

$=\frac{3}{2}[\frac{4}{2}\sqrt{{\left(4\right)}^2-{\left(4\right)}^2}-\frac{(-4)}{2}\sqrt{{\left(4\right)}^2-{(-4)}^2}+\frac{16}{2}{\sin }^{-1}\left(\frac{4}{4}\right)-\frac{16}{2}{\sin }^{-1}\left(\frac{-4}{4}\right)]$

$=\frac{3}{2}[2\times 0+2\times 0+8{\sin }^{-1}\left(1\right)-8{\sin }^{-1}(-1)]$

$=\frac{3}{2}[8{\sin }^{-1}\left(1\right)+8{\sin }^{-1}\left(1\right)]$

$=\frac{3}{2}\times 16{\sin }^{-1}\left(1\right)$

$=\frac{3}{2}\times 16\times \frac{\pi }{2}$

$=12\pi$

Hence, this is the answer.