# Fermat's Volume Calculation (1636 Repost)

Fermat's volume calculation (1636) Let $R$ be the region bounded by the curve $y=\sqrt{x+a}$ (with $a>0$), the $y$-axis, and the $x$-axis. Let $S$ be the solid generated by rotating $R$ about the $y$-axis. Let $T$ be the inscribed cone that has the same circular base as $S$ and height $\sqrt a$. Show that volume($S$)/volume($T$) = $\frac 85$.

Hi

I tried to be much clearer on my presentation of the problem so I hope that people can see the images. I'm having difficulty with the area of S (A(x)). Once I understand that, I believe I am fine with taking the integral. I am also posting my work so far which is very legible on my side of things so I hope it is for you.I believe that I have the correct answer for the Volume of T but I am not sure. Any help would be appreciated!

## 2 Answers

For a representative disk whose center is located at $(0,y)$, the radius is given by $$r = |x| = |y^2 - a|,$$ hence its differential volume is given by $$dV = \pi r^2 \, dy = \pi (y^2 - a)^2 \, dy.$$ Integrating with respect to $y$ on the interval $0 \le y \le \sqrt{a}$, we readily obtain \begin{align*} S &= \pi \int_{y=0}^\sqrt{a} (y^2 - a)^2 \, dy \\ &= \pi \left[ \frac{y^5}{5} - \frac{2a y^3}{3} + a^2 y\right]_{y=0}^{\sqrt{a}} \\ &= \pi \sqrt{a} \left( \frac{a^2}{5} - \frac{2a^2}{3} + a^2\right) \\ &= \frac{8\pi }{15} a^{5/2}. \end{align*} The volume of the cone is simply one third the area of base times its height; i.e., $$T = \frac{1}{3} \pi a^{5/2},$$ hence the desired ratio is $$\frac{S}{T} = \frac{8}{15} \cdot 3 = \frac{8}{5},$$ as claimed.

Volume for S: are you doing shells or disks.

via shells

$S = 2\pi\int r y \;dx = 2\pi \int_{-a}^0 |x|\sqrt{x+a}\; dx = 2\pi \int_{-a}^0 -x\sqrt{x+a} \;dx$

via disks:

$x= y^2 - a\\ S = \pi \int_{0}^{\sqrt a} x^2 \; dy = \pi \int_{0}^{\sqrt a} (y^2-a)^2 \; dy$

Hopefully either way you get: $\frac {8}{15}\pi a^{\frac 52}$

As for T: you can use simple geometry to get that one. $\frac 13 \pi r^2 h = \frac 13 \pi a^\frac 52$

I see your problem!

$\pi \int_{0}^{\sqrt a} (y^2-a)^2 \; dy \\ \pi (\frac 25 y^5 - \frac {2}{3} a y^3+ a^2y)|_0^{\sqrt a}\\ \pi (\frac 15 a^{\frac 52} - \frac {2}{3} a^{\frac 52}+ a^\frac52)$