Calculating Volume of Solid with Isosceles Right Triangular Cross-Sections

[kreil]

The base of the solid is the disc x²+y²=1. The cross-section by planes perpendicular to the y-axis between y=-1 and y=1 are isosceles right triangles with 1 leg in the disc. Find the volume of the solid.

I did the problem, but I am not sure if I did it correctly, and if I did I really just need the problem explained. I understand most of it, I think:

$$A(y)=\frac{1}{2}bh=(\frac{1}{2})(2\sqrt{1-y^2})(2\sqrt{1-y^2})=2(1-y^2)$$

$$V_s=\int_{-1}^1A(y)dy=2\int_{-1}^1(1-y^2)dy=2(y-\frac{1}{3}y^3]_{-1}^1)=2(\frac{4}{3})=\frac{8}{3}$$


Thanks

 

[fresh_42]

Hard to tell without image. I got a triangle with base $2x$ and height $x$. That would mean $A(y)=x^2=(1-y^2)$ and a volume $\displaystyle{\int_{-1}^{1}}(1-y^2)\,dy= \dfrac{4}{3}$ but I'm not sure I got it right.