The area bounded by the parabola y2 = 4ax and x2 = 4ay is

This problem is the generalized form of question 2.

Let’s proceed to solve it similarly –

We need the bounds, i.e., where the area starts and where it ends. At the points of intersection, both the equations are satisfied.

This means that at points of intersection

y2 = 4ax

(from the other equation)

Let us solve this.

x4 = 64a3x

x(x3 – 64a3) = 0

x = 0 or x3 = 64a3, i.e, x = 4a

So the points of intersection are (0,0) and (4a,4a)

Now, let’s compute the area.

If we integrate w.r.t x, we’ll have to integrate the space between the two curves from x = 0 to x =4a.