- #1

kreil

Gold Member

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## Main Question or Discussion Point

The base of the solid is the disc x

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:

[tex]A(y)=\frac{1}{2}bh=(\frac{1}{2})(2\sqrt{1-y^2})(2\sqrt{1-y^2})=2(1-y^2)[/tex]

[tex]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}[/tex]

Thanks

^{2}+y^{2}=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:

[tex]A(y)=\frac{1}{2}bh=(\frac{1}{2})(2\sqrt{1-y^2})(2\sqrt{1-y^2})=2(1-y^2)[/tex]

[tex]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}[/tex]

Thanks