Question

Compute the volume of the solid whose bottom face is the circle x^2+y^2<_1 and every cross section of the solid perpendicular to the x -axis is a square

Answer 1

@perl

Answer 2

it is similar to the last problem we did

Answer 3

We are given x^2+y^2=1
Since we want squares perpindicular to x axis we will solve for y
in terms of x

y= +/- sqrt( 1-x^2)
The 'top half' semicircle is the equation
y = sqrt(1-x^2)
the 'bottom half' semicircle is the equation
y = -sqrt(1-x^2)

we use the form
integral A(x)dx

Here A(x) = ((sqrt(1-x^2)-(-sqrt(1-x^2)))^2
which becomes
A(x) = (sqrt(1-x^2) +sqrt(1-x^2))^2
A(x) = (2*sqrt(1-x^2))^2

So we have

integral A(x)dx on x=-1..1

integral (2*sqrt(1-x^2))^2 dy on x=-1..1