Question

Find the area of the region bounded by the curves $y^2=4ax$ and $x^2=4ay$.

Answer 1

The given curves are $y^2=4ax$ [right parabola] .....(1)

And, $x^2=4ay$ [upward parabola] .......(2)

On squaring both sides of equation 1, we get,

$(y^2{)}^2=16a^2{x}^2$

$y^4=16a^2\left(4ay\right)$ [From eq. 2]

$y(y^3-64a^3)=0$

$y=0,y=4a$

When $y=0,x=0$

When $y=4a,x=4a$

Thus, the given parabolas intersect each other at O(0,0) and A(4a,4a). Then,

the shaded part in the figure is the required area.

For the curve $y^2=4ax$

$y=2\sqrt{a}\sqrt{x}=f\left(x\right)$

And, for the curve $x^2=4ay$,

$y=\frac{x^2}{4a}=g\left(x\right)$

Therefore,

Required area = Area of shaded region OACO

$=\underset{0}{\overset{4a}{\int }}\left[f\right(x)-g(x\left)\right]dx$

$=\underset{0}{\overset{4a}{\int }}2\sqrt{a}\sqrt{x}dx-\underset{0}{\overset{4a}{\int }}\frac{x^2}{4a}dx$

$=\frac{4}{3}\sqrt{a}(4a{)}^{3/2}-\frac{1}{12a}(4a{)}^3$

$=\frac{32a^2}{3}-\frac{16a^2}{3}=\frac{16a^2}{3}sq.units$