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Find the area of the region bounded $x^{2}=36y , y$-axis,$y=2$ and $y=4$

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  • Area bounded by the curve $t=f(x),$ the x-axis and the ordinates $x=a,x=b$ is $\int \limits_a^b f(x) dx $ or $ \int \limits _a^b y dx $
  • If the curve lies below the x-axis for $a \leq x \leq b,$ then the area is $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
Area bounded by $x^2=36y$ and $y-axis,y=4,y=2$
$\qquad=\int \limits_2^4 x\;dy$
$\qquad=\int \limits_2^4 6y^{\large\frac{1}{2}}\;dy$
$\qquad= 6\bigg[\large\frac{2}{3}$$ y^{\large\frac{3}{2}}\bigg]_2^4$
$\qquad= 4 [4 \sqrt 4- 2\sqrt 2]$
$\qquad= 4 [8- 2\sqrt 2]$
$\qquad= 8 [4 -\sqrt 2] sq.units$
answered Aug 16, 2013 by meena.p