I think the following 3D function is pretty cool. It can be written in Cartesian coordinates $$f(x,y)=\sqrt{x^2+y^2}$$ or in Polar coordinates
$$f(r\cos(\theta),r\sin(\theta))=\sqrt{r^2}.$$
If you plot it in Mathematica, it looks like an ice cream cone.
ParametricPlot3D[{r*Cos[t], r*Sin[t], (r^2)^.5}, {t, 0, 2*Pi}, {r, 0,
4}, BoxRatios -> {1, 1, 1}, Ticks -> False]
Check it 1,2,
recall that azimuthal projection satisfy the equation $T(\phi,\theta)=(r(\phi)cos(\theta),r(\phi)sin(\theta))$. Then $T_{\phi}=((r'(\phi)cos(\theta),r'(\phi)sin(\theta))$ and $T_{\theta}=((r(\phi)(-sin(\theta)),r(\phi)cos(\theta))$. Then $T_{\phi} \cdot T_{\theta} = ((r'(\phi)cos(\theta),r'(\phi)sin(\theta)) \cdot ((r(\phi)(-sin(\theta)),r(\phi)cos(\theta)) = (r'(\phi)cos(\theta))(r(\phi)(-sin(\theta))) + (r'(\phi)sin(\theta))(r(\phi)cos(\theta)) = -(r'(\phi)cos(\theta))(r(\phi)(sin(\theta))) + (r'(\phi)sin(\theta))(r(\phi)cos(\theta)) = 0$
Using the code:
heatSystem = {
D[u[x, t], t] == D[u[x, t], x, x],
u[x, 0] == Sin[5 x],
u[0, t] == 0, u[1, t] == 1};
u[x_, t_] = NDSolveValue[heatSystem,
u[x, t], {x, 0, 1}, {t, 0, 1}]
Manipulate[Plot[u[x, t], {x, 0, 1},
ColorFunction -> "TemperatureMap",
ColorFunctionScaling -> False,
PlotRange -> {-1, 1}],
{t, 0, 0.15}]
pics = Table[Plot[u[x, t], {x, 0, 1},
ColorFunction -> "TemperatureMap",
ColorFunctionScaling -> False,
PlotRange -> {-1, 1}],
{t, 0, 0.5, 0.01}];
Export["anim.gif", pics]
u[0.5, 1]
The graph of the solution is:
The solution to the system $u_{t}=u_{xx}, u(x,0)=sin(5x), u(0,t)=0, u(1,t)=1$ at the midpoint when t=1 is:
$u(0.5,1)=0.0894617$
The equation is as follows.
$$u=2x-1$$
Notice that $u(0)=-1$, $u(2)=3$, and $\nabla^2u=0$.
The graph is quite simple
If we think of the object as a bar, the $u$ value on the graph show the temperature of the bar at every point on the bar. This is a steady state heat flow equation because the temperature will not change with time.
So basically we are given $$\int_C F\cdot dn=\int \int divF=\int \int 2x+3y$$ It is a lot easier to do this in polar coordinates so we have $x=0.1cos(\theta)$ and $y=0.1sin(\theta)$ thus $divF=0.2cos(\theta)+0.3sin(\theta)$. This just becomes a matter of integration so:
$\int_C F\cdot dn= \int_{0}^{2\pi} \int_{0}^{1} 0.2cos(\theta)+0.3sin(\theta) \space drd\theta=$
$\int_{0}^{2\pi} 0.2cos(\theta)+0.3sin(\theta) \space d\theta=0.2 sin(\theta)-0.3cos(\theta)=0$