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Find the area bounded between the parabola. $x^2-ay \ and \ x^2=-4a(y-2a)$
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$x^2=4ay ------(1) \text { is parabola with vertex at }(0,0)$

$ X^2 = -4a(y-2a) ------(2) \text{ is a parabola with vertex at }(0,2a)$

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The two parabola intersects at:

Solving equation (1) and (2) we get

$4ay = -4a(y-2a) $

$ 4ay = -4ay + 8a^2 $

$ \therefore 8ay = 8 a^2 $

$ ay = a^2$

$ \therefore y=a When\space y=a$

$ X^2 = 4ay $

$ X^2 =4aa $

$X^2 = 4a^2$

$ x= \pm 2a$

$\therefore $ The two parabola intersects at (2a,a) and (-2a,a)

$\therefore$ Now considering vertical strip in order to cover whole bounded region the vertical strip should slide from x=-2a to x=2a

1) Outer limit x

X=-2a to x=2a

2) Inner limit y

a) Upper limit is equation of parabola $x^2=-4a(y-2a)$

$ \therefore y=\dfrac {x^2}{-4a}+2a $

$\therefore y=\dfrac {8a^2-x^2}{4a}$

b) Lower limit is equation of parabola $x^2 = 4ay$

$\therefore y=\dfrac {x^2}{4a}$

$\therefore $ The area bounded by parabola is

$Area =\int\limits_{-2a}^{2a} \int\limits_{\frac {x^2}{4a}}^{\frac {8a^2-x^2}{4a}} dy\space dx$

$ = \int\limits_{-2a}^{2a}[y]_{\frac {x^2}{4a}}^{\frac {8a^2-x^2}{4a}} dx$

$=\int\limits_{-2a}^{2a}\Bigg[\dfrac {8a^2-x^2}{4a}-\dfrac {x^2}{4a}\Bigg]dx$

$Area=\int\limits_{-2a}^{2a}\Bigg(2a-\dfrac {x^2}{4a}-\dfrac {x^2}{4a}\Bigg) dx$

$ =\int\limits_{-2a}^{2a} \Bigg(2a-\dfrac {x^2}{2a}\Bigg) dx$

$ =\Bigg[2ax-\dfrac {x^3}{3\times 2a}\Bigg]_{-2a}^{2a}$

$=\Bigg[2a(2a)-\dfrac {(2a)^3}{3\times 2a}\Bigg]-\Bigg[2a(-2a)-\dfrac{(-2a)^3}{3\times 2a}\Bigg] $

$=\Bigg[4a^2-\dfrac {4a^2}3\Bigg]-\Bigg[-4a^2+\dfrac {4a^2}3\Bigg] $

$=\dfrac {8a^2}3-\Bigg[-\dfrac {8a^2}3\Bigg] $

$ = \dfrac {16a^2}3$

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