Given that $\textbf{F} = \langle z,x,y \rangle$,

The plane $ z=2x+2y-1$ and the paraboloid $ z= x^2 +y^2$ intersect in a closed curve.

I'm trying to use stokes theorem to find the line integral.


We know that Stokes Theorem is given by: $\iint_S (\nabla \times \textbf{F}) \cdot \textbf{N}\: d\textbf{S} $

$\nabla \times \textbf{F} = \vec{\imath} + \vec{\jmath} + \vec{k} = \langle 1 , 1, 1\rangle$

Now, the problem I have is parameterizing. This is what I did.

$$\vec{r} = \langle x,y,2x+2y-1\rangle$$ $R_x \times R_y = 2\vec{\imath} - 2\vec{\jmath} + 0\vec{k} $

Now, $\iint_S \langle 1 , 1, 1\rangle \cdot \langle 2 , -2, 0\rangle \: d\textbf{S} = 0 $. But this is not right!

Perhaps what I'm struggling most is getting the parameterization right for the surface. Also would I have to convert to polar coordinates to complete the integral?

Any help would be appreciated! Thank you


parameterizing the surface.

$R_x = (1,0,2)\\ R_y = (0,1,2)\\ R_x \times R_y = (-2,-2,1)$

Don't know were you may have lost it.

As a general rule:

$S = (x,y,f(x,y))\\ dS = (-\frac {dz}{dx}, -\frac {dz}{dy}, 1)$

Also consider $dS$ will always be normal to the surface, and the normal line of a plane is just the coefficients of $x,y,z$ in the equation of a plane (when in standard form).

I guess you need to know to scale the $z$ coordinate in $dS$ so that it equals $1.$

  • $\begingroup$ Ah yes. I found the wrong $\textbf{n}$. Now that I can do $\iint_S (\nabla \times \textbf{F}) \cdot \textbf{N}\: d\textbf{S} = \iint_S \langle 1,1,1 \rangle \cdot \langle-2,-2,1 \rangle \: d\textbf{S} = \iint_S -3 d\textbf{S}$. The answer given is $-3\pi$... but I don't understand how they computed that... $\endgroup$ – misheekoh Mar 30 '17 at 0:15
  • 1
    $\begingroup$ @misheekoh: I think your confusion comes from the first equality. In general $\iint_S \mathbf{G} \cdot d\mathbf{S} = \iint_D \mathbf{G}(\mathbf{r}(u,v)) \cdot (\mathbf{r}_x \times \mathbf{r}_y)\,du\,dv$. So you should say $\iint_D (1,1,1) \cdot (-2,-2,1)\,dx\,dy = \iint_D -3\,dx\,dy = -3A(D)$ where $A(D) =$ the area enclosed by the curve projected onto the $xy$-plane (i.e. the area enclosed by the preimage of the curve). This turns out to be the unit circle centered at $(1,1)$ which has area $\pi (1)^2 = \pi$. Hence your integral is $-3\pi$. $\endgroup$ – WB-man Mar 30 '17 at 2:13
  • $\begingroup$ @WB-man Gotcha!! Thank you!!! $\endgroup$ – misheekoh Mar 30 '17 at 2:46

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