6

Let $\mathbf{F}\colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be given by
$$\mathbf{F}(x,y,z)=(x,y,z).$$
Evaluate $$\iint\limits_S \mathbf{F}\cdot dS$$ where $S$ is the surface of the torus given by
$$\begin{align*} x&=(R+\cos(\phi))\cdot \cos(\theta)\\ y&=(R+\cos(\phi))\cdot \sin(\theta)\\ z&=\sin(\phi) \end{align*}$$ and$$0\leq{\theta}\leq{2\pi},\qquad 0\leq{\theta}\leq{2\pi}.$$
Assume $S$ is oriented outward using the outward unit normal.

My Take

Ok, so I know I have to start by taking the cross product of the partial derivatives of theta and phi, but when it says using the outward unit normal, which vector has to be positive in order for it to be the outward normal? Is it the $i$, $j$ or $k$ vector? How do I know which vector it is? What will my next step be after I find the cross product?

Zev Chonoles
  • 129,973
amanda
  • 425
  • Take a typical point on $S$, say $(R+1,0,0)$ and compute the cross product. By drawing a rough sketch of the surface, you should know what direction the outward-pointing normal actually points at that point. If your answers agree, great. Otherwise, you need to switch the order of $\phi$ and $\theta$. ... Next, you need to go back to your textbook and see the procedure for computing a surface integral. – Ted Shifrin May 18 '13 at 23:19
  • @TedShifrin Thankyou for your help Ted. I am aware of the procedure but I have seen examples for this question where simply take the magnitude of the cross product and then integrate and then I have seen examples where they find the dot product between vector F and the cross product so I am confused as to which is my next step? – amanda May 19 '13 at 04:18
  • When you want to calculate a FLUX integral (the notation $\iint_S \mathbf F\cdot d\mathbf S$), you must do the dot product of $\mathbf F$ with the cross product. Only when it's a scalar integral do you follow your first option. – Ted Shifrin May 19 '13 at 17:49
  • @TedShifrin +1 totally correct; the dot products purpose is to only consider the flux outward (it functions as a projection). – obataku May 21 '13 at 02:03

2 Answers2

4

I don't know if you wanna compute the surface integral the hard way, which is, computing the surface outward normal pointwisely and then integrating. For me this problem looks rather like a standard "applying the divergence theorem" exercise.

Denote the interior of the torus as $T$, and $\partial T = S$, then by Divergence theorem: $$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \iiint_T \nabla \cdot \mathbf{F} \,dV = 3 |T|, $$ where $|T|$ is the volume of $T$. Now we compute the volume using the parametrization you gave: based on the parametrization, if we assume $R>1$, then the torus has $R$ as the distance from the center of the tube to its center, the radius of the tube is 1. Thus the interior's parametrization is: $$ \begin{aligned} &x = (R + r \cos \phi) \cos{\theta} \\ &y= (R + r \cos \phi) \sin{\theta} \\ &z= r \sin \phi \end{aligned} $$ for $(r,\phi,\theta)\in [0,1]\times[0,2\pi]\times[0,2\pi]$. Compute the Jacobian: $$ \left|\frac{\partial (x,y,z) }{\partial (r,\phi,\theta)} \right|= r(R+r\cos\phi). $$ Hence the volume $|T|$ is: $$ |T| = \int^1_0 \int^{2\pi}_0 \int^{2\pi}_0 r(R+r\cos\phi) d\phi d\theta dr = 2\pi^2 R, $$ and $$ \iint_S \mathbf{F}\cdot d\mathbf{S} = 6\pi^2 R. $$ Wikipedia's entry of Torus uses Pappus's centroid theorem to compute the volume, which is much simpler, but I guess we are supposed to use the integral to compute for we are given a parametrization here.

Shuhao Cao
  • 18,935
  • Are you sure you want to integrate over $[0,2\pi)$ for both $\theta,\phi$? – obataku May 21 '13 at 02:09
  • @oldrinb Sure, $\phi$ is rotating in small disk a full $2\pi$ (a plane intersecting the tube to get a disk). $\theta$ is rotating around the tube another full $2\pi$. – Shuhao Cao May 21 '13 at 02:14
2

A related problem. You can use the identity

$$ \iint\limits_S {F}\cdot dS = \iint\limits_D {F}\cdot ({r}_\theta \times {r}_{\phi})\,{dA} = \iint\limits_D {F}\cdot ({r}_\theta \times {r}_{\phi}) {d \phi d\theta}, $$

where

$$ r(\theta,\phi) = (R+\cos(\phi)) \cos(\theta) {i} + ( R +\cos(\phi)) \cos(\theta) {j} + \sin(\phi) {k}, $$

and you can write ${F}$ as

$$ {F} (x,y,z)={F}( {r}(\theta,\phi) )= (R+\cos(\phi)) \cos(\theta) {i} + ( R +\cos(\phi)) \cos(\theta) {j} + \sin(\phi) {k}.$$

I leave it here for you to finish the task.

Community
  • 1
Mhenni Benghorbal
  • 47,431