Question

Question #526a3

Answer 1

`V_((a)) = 16/3a^3`

`V_((b)) = (4sqrt3)/3a^3`

Explanation:

Establish a system of reference with origin in the center of the square such that the diameter `AB` lies on the `x` axis.

(a) For every `x in (-a,a)` the plane perpendicular to the `x` axis intercepts the solid in a square whose side has the length of the chord orthogonal to `AB` in the point `(0,x)`, which is:

`l = 2sqrt(a^2-x^2)`

So the area of the square is:

`S = l^2 = 4(a^2-x^2)`

The element of the volume of the solid generated by such squares between `x` and `x+dx` is then:

`dV = l^2dx = 4(a^2-x^2)dx`

Integrating over the interval:

`V = int_(-a)^a 4(a^2-x^2)dx`

using the linearity of the integral:

`V = 4 a^2 int_(-a)^a dx -4 int_(-a)^a x^2dx`

`V = [4 a^2x - 4x^3/3 ]_(-a)^a`

`V = 4a^3 -4/3a^3 +4a^3 -4/3 a^3 = 16/3a^3`

(b) In this case the length of the base is the same, so the area of the triangle is:

`S = sqrt3/4 l^2 = sqrt(3)(a^2-x^2)`

and the volume is:

`V = int_(-a)^a sqrt3 (a^2-x^2)dx = (4sqrt3)/3 a^3`