Let $S$ be the intersection of $x^2+y^2 \le 1$ and $x+y+z=1$.

Calculate the area of the intersection.

Method 1:

$S$ is defined as by $g(x,y,z)=x+y+z-1=0$ over the domain $A=\Big\{(x,y) | x^2+y^2\le 1\Big\}$

The area is derived from the formula

$$S=\iint_A \sqrt{\frac{g_x'+g_y'+g_z'}{g_z'}}dxdy = \sqrt{3}\iint_Adxdy=\sqrt3\cdot\pi$$

Method 2:

Define the vector field $F=(3z,3x,3y)$ and let $\gamma=\partial S$.

Then by stokes theorem:

$$\oint_\gamma \vec F\cdot d\vec r=\iint_S \nabla\times \vec F dS$$

The $\hat n$ is $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$

The curl of $F$ is $$\nabla \times \vec F = \begin{vmatrix}\hat i & \hat j & \hat k\\ \partial_x & \partial_y & \partial_z\\ 3z & 3x& 3y\end{vmatrix}=(3,3,3)$$

Therefore it follows that $$\iint_S \nabla \times \vec F \cdot \hat n = 3\iint_Sdxdy = 3Area(S)$$

Therefore, by Stokes; the surface area is equal to

$$\frac{1}{3}\oint_\gamma \vec F \cdot d\vec r$$

The parametrization of $\gamma$ is ($0\le t \le 2\pi$) $$r(t)=(\cos t,\sin t, 1-\cos t - \sin t)$$ $$r'(t) = (-\sin t, \cos t, \sin t - \cos t)$$

Therefore we are looking to find

$$\frac{1}{3} \int_0^{2\pi}(\cos t,\sin t, 1-\cos t - \sin t) \cdot (-\sin t, \cos t, \sin t - \cos t) = 0$$

I got two different answers; where did I go wrong?


1 Answer 1


It's just a tiny misstep. Your vector field $F$ is not defined proportional to $(x,y,z)$ but $(z,x,y)$ so the correct line integral should be $$\require{cancel}\oint_\gamma \vec F \cdot d\vec r \propto \int_0^{2\pi}\underbrace{{\cancel{(\cos t,\sin t, 1-\cos t - \sin t)} }}_{\displaystyle (1-\cos t - \sin t,\, \cos t,\,\sin t)} \cdot (-\sin t, \cos t, \sin t - \cos t) \,\mathrm{d}t $$

I believe you can figure out the rest.


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