# Question #526a3

Apr 4, 2017

${V}_{\left(a\right)} = \frac{16}{3} {a}^{3}$

${V}_{\left(b\right)} = \frac{4 \sqrt{3}}{3} {a}^{3}$

#### Explanation:

Establish a system of reference with origin in the center of the square such that the diameter $A B$ lies on the $x$ axis.

(a) For every $x \in \left(- a , a\right)$ the plane perpendicular to the $x$ axis intercepts the solid in a square whose side has the length of the chord orthogonal to $A B$ in the point $\left(0 , x\right)$, which is:

$l = 2 \sqrt{{a}^{2} - {x}^{2}}$

So the area of the square is:

$S = {l}^{2} = 4 \left({a}^{2} - {x}^{2}\right)$

The element of the volume of the solid generated by such squares between $x$ and $x + \mathrm{dx}$ is then:

$\mathrm{dV} = {l}^{2} \mathrm{dx} = 4 \left({a}^{2} - {x}^{2}\right) \mathrm{dx}$

Integrating over the interval:

$V = {\int}_{- a}^{a} 4 \left({a}^{2} - {x}^{2}\right) \mathrm{dx}$

using the linearity of the integral:

$V = 4 {a}^{2} {\int}_{- a}^{a} \mathrm{dx} - 4 {\int}_{- a}^{a} {x}^{2} \mathrm{dx}$

$V = {\left[4 {a}^{2} x - 4 {x}^{3} / 3\right]}_{- a}^{a}$

$V = 4 {a}^{3} - \frac{4}{3} {a}^{3} + 4 {a}^{3} - \frac{4}{3} {a}^{3} = \frac{16}{3} {a}^{3}$

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

$S = \frac{\sqrt{3}}{4} {l}^{2} = \sqrt{3} \left({a}^{2} - {x}^{2}\right)$

and the volume is:

$V = {\int}_{- a}^{a} \sqrt{3} \left({a}^{2} - {x}^{2}\right) \mathrm{dx} = \frac{4 \sqrt{3}}{3} {a}^{3}$