Parametric Sweeping of a Rose 
My thought process: 
Create the 2D representation of the rose, which I have defined as :
For Domain (u,v) [0,1]
x(u,v) = 1 + sin(8π u) cos(2π u)
y(u,v) = 1 + sin(8π u) sin(2π u)
z(u,v) = 0
May I know how do I manipulate 3π/2 and a vertical displacement of 2 to complete the sweep ?
 A: Let $\sqrt{x^2+z^2}=:\rho$. We first draw your rose in the $(\rho,y)$-halfplane:
$$\rho=1+{1\over2}\sin(8\pi u)\cos (2\pi u),\qquad y=-1+{1\over2}\sin(8\pi u)\sin (2\pi u)\ .\tag{1}$$
Here I have put $\alpha:=2\pi u$ in order to obey the condition $0\leq u\leq 1$. The $1$ and the $-1$ I have read off from the right part of your figure. If we rotate this rose around the $y$-axis the $\rho$ becomes distributed among $x$ and $z$ via a new variable $\phi={3\pi v\over2}$ $(0\leq v\leq 1)$. In this way the first formula of $(1)$ will be replaced by the pair
$$\eqalign{x&=\rho\cos\phi=\bigl(1+{1\over2}\sin(8\pi u)\cos (2\pi u)\bigr)\cos{3\pi v\over2}\cr
z&=-\rho\sin\phi=-\bigl(1+{1\over2}\sin(8\pi u)\cos (2\pi u)\bigr)\sin{3\pi v\over2}\ .\cr}$$
But the rose is not only rotated. It is also lifted in the $y$-direction by the amount $2$ while $v$ runs from $0$ to $1$. It follows that the final representation of your surface is
$$\left.\eqalign{x&=\bigl(1+{1\over2}\sin(8\pi u)\cos (2\pi u)\bigr)\cos{3\pi v\over2}\cr
y&=-1+{1\over2}\sin(8\pi u)\sin (2\pi u)+2v\cr
z&=-\bigl(1+{1\over2}\sin(8\pi u)\cos (2\pi u)\bigr)\sin{3\pi v\over2}\cr}\right\}
\qquad(0\leq u\leq 1, \ 0\leq v\leq 1)\ .$$
