Parametric surface with Bezier cross section B(t), bounded by a polar curve in the x-y plane

Question

I am trying to create a parametric 'petal' surface with a vertical cross section defined by an arbitary bezier curve, and bounded by the area enclosed by a polar curve in the x-y plane.

I can achieve this by defining a cross section implicity in terms of $r$. For instance, if I want a cross section $r^2$ and a polar curve $sin^2 2\theta$, I can define the following:

$r = {t\sin^2 2\theta}$

$x = {t\sin^2 2\theta \cos\theta}$

$y = {t\sin^2 2\theta \sin\theta}$

$z = {r^2}=({t\sin^2 2\theta})^2$

Where $r = \sqrt{x^2+y^2}$

Visualisation here : Each 'petal' is lifted along the curve $z = {r^2}$.

Suppose I would like to do the same using a 2-D bezier curve ${B(t)}$ as the cross section; I know how to create a surface 'sweeping' a full revolution i.e:

$x=B_r(t)\cos\theta$
$y=B_r(t)\sin\theta$
$z=B_z(t)$

Where $r = \sqrt{x^2+y^2}$

However, I am not sure how to make it bounded by a polar curve such as $r = {\sin^2 2\theta}$.

Thanks!

Answer 1

If you want your petal to be "lifted" on the sphere with center $(0,0,R)$ and radius $R$, therefore with equation

$$\underbrace{x^2+y^2}_{r^2=\sin^2(2 \theta)}+(z-R)^2=R^2,\tag{1}$$

it suffice to extract $z$ from (1) under the form:

$$(z-R)^2=R^2-\sin^2(2 \theta) \ \implies \ z=R\color{red}{-}\sqrt{R^2-\sin^2(2 \theta)} \tag{2}$$

(please note that we have taken the $\color{red}{minus}$ sign in order to have the initial point at the origin (0,0,0))

(with the meaning that any point (x,y,0) defined by polar equation $r=\sin(2 \theta)$ on (x,y) horizontal plane, is "lifted at height given by (2).

Answer 2

Suppose your height curve is defined in the $xz$-plane, say, $z = h(x)$. This can be any function you like, including a real-valued Bézier curve.

And the let $r = k(\theta)$ be your 2D cross-section curve, in polar coordinates.

Then the parametric equation of the lifted surface is $$ x = r \cos\theta = k(\theta)\cos\theta. \\ y = r \sin\theta = k(\theta)\sin\theta. \\ z = h(r) = h\big(\sqrt{x^2 +y^2}\big) $$