Determining the volume of an egg 
This is the equation of a perfect egg, is there anyway that one can find the volume using multiple integrals?. I have tried something,I just want t validate whether what  have done makes any sense.
 A: HINT:
It is a surface of revolution, no need for $u$. The coordinates $r,z$  are functions of a single parameter $v$
$$ r = (1 + .2 v)  \sin v ; z = 1.65 \cos v ; $$
Volume $$ = \int \pi r^2 dz = (...) dv $$
A: Call the region surrounded by the eggshell $E$, then the volume $V$ is given by the triple integral,
$$V=\iiint_E\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$
You can use the provided parameterization for the surface to convert to a different set of coordinates somewhat resembling spherical coordinates. Let
$$\begin{cases}x(r,u,v)=r\left(1+\dfrac{1}{5}v\right)\cos u\sin v\\[1ex]
y(r,u,v)=r\left(1+\dfrac{1}{5}v\right)\sin u\sin v\\[1ex]
z(r,u,v)=\dfrac{33r}{20}\cos v\end{cases}$$
so that the Jacobian is given by
$$|J|=\frac{33r^2}{500}\sin v\left(5+v\right)(\sin v\cos v+v+5)$$
then compute the integral:
$$\begin{align*}
V&=\iiint_E\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z\\[1ex]
&=\iiint_E|J|\,\mathrm{d}r\,\mathrm{d}u\,\mathrm{d}v\\[1ex]
&=\int_0^\pi\int_0^{2\pi}\int_0^1|J|\,\mathrm{d}r\,\mathrm{d}u\,\mathrm{d}v
\end{align*}$$
which - barring any mistake with the setup on my part - should give a volume of
$$\frac{11\pi}{250}\left(\pi^2+10\pi+\frac{410}{9}\right)\approx 12.0041$$
(If anyone sees a mistake, please let me know.)
Added: To compute the Jacobian, you have
$$\begin{align*}
|J|&=\begin{vmatrix}
\dfrac{\partial x}{\partial r}&\dfrac{\partial y}{\partial r}&\dfrac{\partial z}{\partial r}\\[1ex]
\dfrac{\partial x}{\partial u}&\dfrac{\partial y}{\partial u}&\dfrac{\partial z}{\partial u}\\[1ex]
\dfrac{\partial x}{\partial v}&\dfrac{\partial y}{\partial v}&\dfrac{\partial z}{\partial v}
\end{vmatrix}\\[2ex]
&=\begin{vmatrix}
\left(1+\dfrac{v}{5}\right)\cos u\sin v&\left(1+\dfrac{v}{5}\right)\sin u \sin v&\dfrac{33\cos v}{20}\\[1ex]
-r\left(1+\dfrac{v}{5}\right)\sin u\sin v&r\left(1+\dfrac{v}{5}\right)\cos u\sin v&0\\[1ex]
r\cos u\left(\dfrac{\sin v}{5}+\left(1+\dfrac{v}{5}\right)\cos v\right)&r\sin u\left(\dfrac{\sin v}{5}+\left(1+\dfrac{1}{5}v\right)\cos v\right)&-\dfrac{33r\sin v}{20}
\end{vmatrix}\\[2ex]
&=\frac{33}{20}\cos v
\begin{vmatrix}-r\left(1+\dfrac{v}{5}\right)\sin u\sin v&r\left(1+\dfrac{v}{5}\right)\cos u\sin v\\r\cos u\left(\dfrac{\sin v}{5}+\left(1+\dfrac{v}{5}\right)\cos v\right)&r\sin u\left(\dfrac{\sin v}{5}+\left(1+\dfrac{v}{5}\right)\cos v\right)\end{vmatrix}\\[2ex]
&\quad\quad\quad\quad-\dfrac{33r}{20}\sin v
\begin{vmatrix}\left(1+\dfrac{v}{5}\right)\cos u\sin v&\left(1+\dfrac{v}{5}\right)\sin u \sin v\\[1ex]
-r\left(1+\dfrac{v}{5}\right)\sin u\sin v&r\left(1+\dfrac{v}{5}\right)\cos u\sin v\end{vmatrix}
\end{align*}$$
and so on. For the last step I computed the determinant of the 3x3 matrix via a cofactor expansion along the third column, since the $0$ entry makes one of the terms in the expansion disappear.
