written 7.8 years ago by | • modified 6.0 years ago |
$x^2=4ay ------(1) \text { is parabola with vertex at }(0,0)$
$ X^2 = -4a(y-2a) ------(2) \text{ is a parabola with vertex at }(0,2a)$
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$