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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}_2^4A=2\times\frac{2}{3}\begin{bmatrix}y^\frac{3}{2}\end{bmatrix}$on applying limits we get,$A=\frac{4}{3}\begin{bmatrix}4^\frac{3}{2}-2^\frac{3}{2}\end{bmatrix}\;\;\;=\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.
edited Dec 20, 2013