Find the volume of the solid

Find the volume of the solid whose base is the region between the curve $\displaystyle y=x^3$ and y-axis from $\displaystyle y=0$ to $\displaystyle y=1$, and whose cross sections taken perpendicular to the y-axis are squares.

I don't understand "whose cross sections taken perpendicular to y-axis are squares".

How can I solve this problem? If possible, would you please explain step by step?

Thank you very much.

Picture a bunch of squares stacked up perp.to the y-axis in your given region.

This is NOT a volume of revolution. Since we are perp to the y-axis we will integrate wrt y.

$\displaystyle x=y^{\frac{1}{3}}$

The area of a square is $\displaystyle x^{2}$

So, we have $\displaystyle \int_{0}^{1}y^{\frac{2}{3}}dy$