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Find the common area enclosed by the parabolas $4y^{2}=9x$ and $3x^{2}=16y$

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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 $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
Solving $3x^2 =16 y, 4y^2 =9x$
$4 \bigg(\large\frac{3x^2}{16}\bigg)^2 $$=9x$
$\large\frac{36x^4}{256}$$=9x$
$x^4= 64x$
$x(x^3-4^3)=0=>x=0,4$
Required area= Area $4y^2=9x$ between $x=0\;x=4$ -area $3x^2=16 y$ between $ x=0,x=4$
$\qquad= \large\int \limits_0^4 \bigg(\large\frac{3}{2}$$x^{1/2}-\large\frac{3x^2}{16}\bigg)$$dx=\large\frac{3}{2}.\frac{2}{3} $$x^{3/2}-\large\frac{x^3}{16}\bigg]_0^4$
$\qquad= x^{\large\frac{3}{2}} -\large\frac{x^3}{16}\bigg]_0^4$
$\qquad= 4 \sqrt 4 -4$
$\qquad=8-4=4 \;sq\;units$
answered Aug 16, 2013 by meena.p
 

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