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TN XII Math
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Integral Calculus and its applications
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Find the area of the region bounded $x^{2}=36y , y$-axis,$y=2$ and $y=4$
tnstate
class12
bookproblem
ch7
sec-1
exercise7-4
p117
q5
aims-12-math
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asked
Apr 27, 2013
by
poojasapani_1
edited
Apr 19, 2014
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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
How to draw a graph for this....do the needful...
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