A solid has a circular base of radius r=4. Find in each case the volume of the solid if all intersection between the solid and a plane perpendicular to a set diameter is:

a) A square

b) A triangle rectangle isosceles which one of its sides is on the circular base.

The first I thought was in some kind of cylindroid. I know I must find an equation that I can integrate. So, the sides of the squares will be given by:

$\displaystyle z=2\sqrt{r^2-x^2}$

Is the integration given by: $\displaystyle \displaystyle\int_{-4}^{4}[2\sqrt{r^2-x^2}]^2dx=4\displaystyle\int_{-4}^{4}(r^2-x^2)dx$ for the square?

And how can I find the volume of the other solid b)?