find the area given y^2-2y+x^2=0 ; y^2-6y+x^2=0 ; y=x/(3)^1/3 and x=0
First you need to draw the curves and shade the required area.
$\displaystyle y^2 - 2y + x^2 = 0 \Rightarrow (y - 1)^2 + x^2 = 1$ which is a circle of radius 1 and centre at (0, 1).
$\displaystyle y^2 - 6y + x^2 = 0 \Rightarrow (y - 3)^2 + x^2 = 9$ which is a circle of radius 3 and centre at (0, 3).
There are a number of approaches that can now be taken in setting up the integral(s) that gives the area bounded by the given curves.
Although you can do this using Cartesian coordinates (you need two seperate integrals - why?), it will be much less work I think to work in polar coordinates if you've been taught how to do this.