Question

How do you find the volume of the solid with base region bounded by the curve `y=1-x^2` and the `x`-axis if cross sections perpendicular to the `x`-axis are isosceles triangles with height equal base?

Answer 1

Since its triangular cross-sectional area `A(x)` can be found by
`A(x)=1/2`(Base)(Height)`=1/2(1-x^2)^2`,
the volume `V` of the solid can be found by
`V=1/2int_{-1}^1(1-x^2)^2dx`
by symmetry about the y-axis,
`=int_0^1(1-2x^2+x^4)dx`
`=[x-2/3x^3+1/5x^5]_0^1=1-2/3+1/5=8/15`