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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area of the region bounded by \(x^2 = 4y, y = 2, y = 4\) and the \(y\) - axis in the first quadrant.

$\begin{array}{1 1} \frac{32-8\sqrt 2}{3}\;sq.units. \\ \frac{16-8\sqrt 2}{3}sq.units. \\ \frac{32-4\sqrt 2}{3}sq.units. \\ \frac{16-4\sqrt 2}{3}\;sq.units\end{array} $

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  • The area A of the region bounded by the curve x=g(y),y-axis and the lines y=c,y=d is given by\[A=\int_c^d xdy=\int_c^d g(y)dy.\]
Given curve is $x^2=4y\Rightarrow x=2\sqrt y.$
The required area is bounded by the curve $x^2=4y,y=2,y=4 and the y-axis in the first quadrant.
Hence $A=2\int_2^4 \sqrt y\;dy.$
On integrating we get $A=2\begin{bmatrix}\frac{y^\frac{3}{2}}{\frac{3}{2}}\end{bmatrix}_2^4$
on applying limits we get,
$\;\;\;=\frac{3}{2}[8-2\sqrt 2]$
$\;\;\;=\frac{32-8\sqrt 2}{3}sq.units.$
Hence the required area is $\frac{32-8\sqrt 2}{3}$sq.units.
answered Jan 24, 2013 by sreemathi.v
edited Dec 20, 2013 by balaji.thirumalai

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