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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area of the region bounded by the ellipse \( \frac{\Large x^2}{\Large 16} + \frac{\Large y^2}{\Large 9} = 1\)

$\begin{array}{1 1} 12 \pi\;sq.units \\ 16 \pi \;sq.units \\ 8 \pi \;sq.units \\ 4 \pi \;sq.units \end{array} $

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Toolbox:
  • Area of the region bounded by the ellipse is 4 times the are bounded by it in the first quadrant.
  • If it is bounded by the major axis and the lines ,x=0 and x=a,then the area of the region is\[A=\int_0^a\sqrt {1-\frac{x^2}{a^2}}\]
As shown in the fig,the ellipse is symmetrical about both x-axis and y-axis.
Hence the required area is 4 times the area bounded by it in the first quadrant and x-axis and x=0 and x=4.
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow y=\sqrt {1-\frac{x^2}{a^2}}$
Hence $A=4\times \int_0^4 y\;dx$
$\quad\;\;\;\;\quad\;=4\times\frac{3}{4}\int_0^4\sqrt{16-x^2}dx$
$\quad\;\;\;\;\quad\;=3\int_0^4\sqrt {16-x^2} \;dx$
on integrating we get,
$\quad\;\;\;\;\quad\;=3\begin{bmatrix}\frac{x}{2}\sqrt {16-x^2}+\frac{16}{2}\sin^{-1}(\frac{x}{4})\end{bmatrix}_0^4$
on applying limits we get,
$\quad\;\;\;\;\quad\;=3\begin{bmatrix}2\sqrt {16-16}+8\sin^{-1}(1)\end{bmatrix}$
$\quad\;\;\;\;\quad\;=3\begin{bmatrix}\frac{8\pi}{2}\end{bmatrix}=12\pi sq.units.$
Hence the required area is $12\pi$ sq.units.
answered Dec 20, 2013 by yamini.v
 

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