How can I find the implicit equations of a surface if I have the parametric equations? For example, if the surface $(S)$ is given by:
$$x = u+\sin v$$
$$y=u+\cos v$$
$$z = u+a$$
what are the implicit equations of this surface?
Write the implicit equation of a surface when given the parametric equations
surfaces
Related Solutions
Alright, I'm kind of embarrassed at how difficult I made this problem, but I'm glad that the solution is very elegant.
A "toroidal knot patch" can be made simply by using one of the following equations:
$$ \vec r_1 (t, u) = (R+r\cos (p(t+u)))\cos qt \mathbf i + (R+r\cos(p(t+u)))\sin qt \mathbf j + r\sin (p(t+u)) \mathbf k $$
or
$$ \vec r_2 (t, u) = (R+r\cos (pt))\cos(t+u) \mathbf i + (R+r\cos(pt))\sin(t+u)\mathbf j + r\sin(pt)\mathbf k $$
As $u$ increases in the first equation, the patch is "increased" in the poloidal direction. As $u$ increases in the second equation, the patch is "increased" in the toroidal direction.
To find the extents $u$ should range over to correspond to the orthogonal projection from the tube diameter to the torus surface, we project the curve $u$ (at constant $t$) to the tangent plane to the torus at $t$.
This just turns into a problem of projecting a vector to a plane. Namely, for tube diameter $d$, solve
$$ ((\vec r(t,u)-\vec r(t,0))-((\vec r(t,u)-\vec r(t,0))\cdot \hat N(t))\hat N(t)) \cdot \hat B(t) = d/2 $$
for $u$. In the above, $\vec r$ can be either $\vec r_1$ or $\vec r_2$, and $\vec r(t,u)-\vec r(t,0)$ is the vector pointing from the toroidal knot centered at $t$ to $u$ in the direction of $u$ and $\hat N(t)$ and $\hat B(t)$ are the Darboux frame normal and binormal to the torus knot at $t$.
Below are graphs plotting using WinPlot
First Equation (Poloidal Direction):
$$ p = 10, R = 5, r = 2, q = 1 $$ such that $$ \vec r(t) = (5+2\cos (10(t+u)))\cos t \mathbf i + (5+2\cos (10(t+u)))\sin t \mathbf j + 2\sin (10(t+u)) \mathbf k $$
$ 0 \le u \le 0.1 $
$ 0 \le u \le 0.3 $
$ 0 \le u \le 0.5 $
Second Equation (Toroidal Direction):
$$ p = 10, R = 5, r = 2, q = 1 $$ such that $$ \vec r(t) = (5+2\cos (10t))\cos (t+u) \mathbf i + (5+2\cos (10t))\sin (t+u) \mathbf j + 2\sin (10t) \mathbf k $$
$ 0 \le u \le 0.1 $
$ 0 \le u \le 0.3 $
$ 0 \le u \le 0.5 $
Since in the end you have to integrate over the unit disc in the $(x,y)$-plane your source proposes to use polar coordinates instead of $x$ and $y$ as parameters. Then $x=s\cos t$, $y=s\sin t$. In this way the idea $z=3xy$ $\>(x^2+y^2\leq1)$ translates into $${\bf r}(s,t)=(s\cos t,s\sin t,3s^2\cos t\sin t)\qquad(0\leq s\leq 1, \ 0\leq t\leq2\pi)\ .$$ In order to find the area of this floppy disc $F$ we have to compute $${\bf r}_s=(\cos t,\sin t, 6s\cos t\sin t),\quad {\bf r}_t=\bigl(-s\sin t,s\cos t ,3s^2\cos(2t)\bigr)$$ and then $${\bf r}_s\times{\bf r}_t=(\ldots,\ldots,\ldots)\ .$$ The area is then finally given as $${\rm area}(F)=\int_0^1\int_0^{2\pi}\bigl|{\bf r}_s\times{\bf r}_t\bigr|\>dt\>ds\ .$$ The resulting integral will be simpler than dreaded.
Best Answer
Hint.
Essentially you want to cancel out the parameter $u$ and $v$.
The parameter $u$ disappears immediately if you substitute $u=z-a$ into the first and second equations: $$ x=z-a+\sin v,\quad y=z-a+\cos v\tag{1} $$
This should remind you of a circle in poloar coordinates.
Now apply the identity $\sin^2 v+\cos^2 v=1$ to combine the two equations in (1).