Egg

Relevant wiki: Volume of Revolution - Disc Method - Easy

$\begin{aligned} V & = \lim_{n\to \infty} \sum_{k=1}^n A_k \delta y & \small \color{#3D99F6} \text{See figure.} \\ & = \lim_{n\to \infty} \sum_{k=1}^n \pi x_k^2 \delta y \\ & = \int_0^1 \pi x^2 dy & \small \color{#3D99F6} \text{Since }4x^2 = y(1-y)(2-y)^2 \\ & = \int_0^1 \frac \pi 4 y(1-y)(2-y)^2 dy \\ & = \frac \pi 4 \int_0^1 \left(4y-8y^2+5y^3-y^4\right) dy \\ & = \frac \pi 4 \bigg[2y^2 - \frac 83 y^3 + \frac 54 y^4 - \frac 15 y^5\bigg]_0^1 \\ & = \frac \pi 4 \times \frac {23}{60} \\ & \approx \boxed{0.301} \end{aligned}$

We can assume the 'egg' to be made of many infinitesimally small cylindrical discs of varying radii and thickness $dy$ stacked on top on each other. Since the egg is symmetrical about the $y-$ axis, the radii of each of the discs is the magnitude of abscissa at at a given point on the curve as shown in the 2-D representation above.

The infinitesimal volume $dV$ can be expressed for each cylindrical disc as

$dV = \pi x^2 dy$

The required volume is the sum of all infinitesimally small volumes $dV$ , i.e., the integral from $y=0$ to $y=1$ .

$\begin{aligned} V &= \sum dV \\ &= \sum \pi x^2 dy \\ &= \int_0^1 \pi \left[ \dfrac{y(1-y)(2-y)^2}{4} \right] dy \\ &= \dfrac{23 \pi}{240} \\ &= \boxed{0.301069} \end{aligned}$

First of all, let's rewrite the problem in more standard notation for cylindrical polar coordinates: we have $4r^2=z(1-z)(2-z)^2$ so $r$ is real-valued between $z=0$ and $z=1$ .

Now, we can evaluate the volume as a triple integral. The volume element for cylindrical polar coordinates is $r\mathrm{d}r \mathrm{d}z \mathrm{d} \phi$ . So our integral is

$V=\int_{\phi=0}^{2\pi}\int_{z=0}^1 \int_{r'=0}^{r(z)}r' \mathrm{d}r' \mathrm{d}z\mathrm{d}\phi$

The integrals in $r$ and $\phi$ are straightforward, giving us $V=\pi \int_0^1 r(z)^2 \mathrm{d}z$ We can then use the given expression for $r^2$ to get an integral

$\begin{aligned}V&=\frac{\pi}{4} \int_{z=0}^1 4z-8z^2+5z^3-z^4\mathrm{d}z \\ &= \frac{\pi}{4}\left[2z^2-\frac{8}{3}z^3+\frac{5}{4}z^4 -\frac{1}{5}z^5\right]^1_0 \\ &=\frac{\pi}{4}\cdot\frac{23}{60}\approx 0.301069\end{aligned}$