I'm trying to evaluate the following integral:

$ \int_C(y+\sin x) dx +(z^2+\cos y)dy+(x^3)dz$

Where $C$ is the curve: $c(t) = (\sin t, \cos t, \sin 2t) $. Note that $C$ lies on the surface $z= 2xy$.

Question: Could I just use the parameterization of the surface S $\phi(u,v)=(u, v,2uv) $ with $-1\leq u\leq 1 $ and $-1\leq v \leq 1$ and use Stokes' Theorem to evaluate $ \int_S \nabla \times F\cdot dS$ which should be equal to the required integral?

My only trepidation about the above method is that $c(t)$ may not necessarily be on the boundary of $S$, which would be required to use Stokes' Theorem.


Your instinct to use Stokes's Theorem is a good one. But you need to use the region in that surface whose boundary is $C$, as you understood. Think polar coordinates, since you've got a good start with the parametrization of the curve $C$. Try using $\phi(r,t) = (r\sin t, r\cos t, r^2\sin 2t)$. (Note that this is a parametrization of a region in your surface!) I'll let you figure out the $r$ and $t$ bounds.

  • $\begingroup$ The curve is given in the first place. $\endgroup$ – Shuhao Cao May 18 '13 at 20:33
  • $\begingroup$ Sure, so of course one could do the integral directly. But applying Stokes's Theorem and taking full advantage of the symmetry in the problem makes for an easy computation. $\endgroup$ – Ted Shifrin May 18 '13 at 20:38

The answer is Yes.

Your parametrization is correct.

Now you need to compute the surface normal $$ \nu = \frac{\partial \phi}{\partial u}\times \frac{\partial \phi}{\partial v}. $$ Applying the Stokes theorem then the surface integral becomes: $$ \oint_C F\cdot dr = \iint_S \nabla \times F\cdot dS = \iint_D \nabla \times F\big(x(u,v),y(u,v),z(u,v)\big) \cdot \nu/\|\nu\| \,du dv, $$ where $D = \{u^2 + v^2 \leq 1\}$.

$C$ which is giving by: $$ \begin{aligned} &x = \sin t \\ &y = \cos t \\ &z = \sin (2t) \end{aligned} $$ assuming $t\in [0,2\pi)$, $C$ lies on $z = 2xy$ simply because the parametrization satisfies the equation of the surface: $$ z = \sin(2 t) = 2\sin t\cos t = 2xy, $$ therefore, all the points on $C$, must be on $S$.

The boundary of $S$ is artificial chosen by cut the unbounded surface $z = 2xy$ using the cylinder $x^2+y^2 = 1$ (notice the parametrization of the curve also satisfies this), the intersection curve of $z = 2xy$ and $x^2+y^2 = 1$ is $C$.

  • $\begingroup$ This is nonsense, I'm sorry. The projection of $C$ on the $xy$-plane is the unit circle $\{x^2+y^2=1\}$, not the boundary of the square $[-1,1]\times [-1,1]$. The OP's trepidation was totally well-founded. In addition, one ordinarily refers to the surface normal as the unit normal, which this is not. But, when you use the parametrization, this is the correct formula for evaluating the surface integral. $\endgroup$ – Ted Shifrin May 18 '13 at 21:12
  • 1
    $\begingroup$ @TedShifrin Thanks, I was being careless. $\endgroup$ – Shuhao Cao May 18 '13 at 21:54

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