Question

Find the area between the curves y^2=ax and ay^2=x^3(a>0), at the points of intersection other than the origin.

Answer 1

Area between two curves f and g is given by:

A = $\int^b_a (f(y) -g(y)) dy$ where a<y<b

Here,

g = y²=ax

$x= \frac{y^2}{a}$

f = ay²=x³

$x = a^{1/3} y^{2/3}$

A = $\int^a_{-a} (-\frac{y^2}{a} +a^{1/3} y^{2/3}) dy$

$=\int^a_{-a} -\frac{y^2}{a} dy + \int^a_{-a} a^{1/3} y^{2/3} dy$

$=\left. \frac{-y^3}{3a} \right|^{a}_{-a} + \left. \frac{3y^{5/3}a^{1/3}}{5} \right|^{a}_{-a}$

=$-\frac{a^2}{3} + \frac{3a^2}{5} -\frac{a^2}{3} + \frac{3a^2}{5}$

= $-\frac{2a^2}{3} + \frac{6a^2}{5}$

= $\frac{-10a^2+18a^2}{15}$

=$\frac{8a^2}{15}$