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I'm studying for a grad-school preliminary exam, and came across this problem which I am unable to solve.

Let $C$ be a closed curve in the plane $ax + by + cz = 0$ (where $a,b,c \in \mathbb{R}$ are not all zero $0$), enclosing a region with area $S$. Evaluate $$ I:= \oint\limits_C \left| \begin{array}{ccc} dx & dy & dz \\ a & b & c \\ x & y & z \end{array} \right| $$

where the integral along $C$ is counterclockwise relative to the normal direction $(a,b,c)$ to the plane.

Expanding the determinant yields $$ I = \oint\limits_C (bz - cy)dx + (cx - az)dy + (ay - bx)dz. $$

If we call $\Omega$ the region bounded by $C$, an application of Green's Theorem (or Stokes Theorem, if you prefer) yields $$ I = 2\iint\limits_\Omega a dydz + bdzdx + c dxdy. $$

From here, I'm not sure what to do; I suspect that it's a simple solution and that I've just forgotten my vector calculus. Anyway, any help is appreciated.

EDIT: Substitute in $z = \frac{1}{c}\left(-ax - by \right)$ and then applying Green's Theorem doesn't bring me any closer. Making such a substitution yields $$ I = \frac{a^2 + b^2 + c^2}{c} \iint_\Omega dxdy $$

If $c \neq 0$ as required to make such a substitution, it is not the case that $\iint dxdy = S$.

SECOND EDIT: I've solved the problem, making use of the following version of Stokes' theorem: $$ \iint_\Omega (\nabla \times \mathbf{F} ) \cdot \mathbf{n} dS = \oint_{\partial \Omega} \mathbf{F} \cdot d\mathbf{r}. $$

The above can be written as \begin{align*} I &= \oint\limits_C (bz - cy,cx - az,ay - bx)\cdot d\mathbf{r} \\ &= \iint_\Omega (\nabla \times (bz - cy,cx - az,ay - bx)) \cdot \left(\frac{1}{\sqrt{a^2 + b^2 + c^2}}(a,b,c)\right) dS \\ &=\iint_\Omega (2a,2b,2c) \cdot \left(\frac{1}{\sqrt{a^2 + b^2 + c^2}}(a,b,c)\right) dS \\ &= 2 \sqrt{a^2 + b^2 + c^2} \iint_\Omega dS \\ &= 2\sqrt{a^2 + b^2 + c^2} (\text{Area of } \Omega) \end{align*}

Marcus M
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1 Answers1

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Use the plane equation as

$$ ax + by + cz = 0 \implies z = -\frac{a}{c}x -\frac{b}{c}y \implies dz= -\frac{a}{c}dx - \frac{b}{c}dy $$

and substitute for $z$ and $dz$ in the integral

$$ \oint\limits_C (bz - cy)dx + (cx - az)dy + (ay - bx)dz. $$

Then Green's theorem is straightforward.

science
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  • Making such a substitution yields $I = \frac{1}{c}\oint_C (-(a^2 + b^2 + c^2)ydx + (a^2 + b^2 + c^2)x dy = \frac{2}{c} \iint (a^2 + b^2 + c^2) dxdy = \frac{2}{c}(a^2 + b^2 + c^2)S$. My intuition tells me that this isn't correct, due to the lack of symmetry; why do we have a $1/c$ in the front rather than a $1/a$ or a $1/b$? – Marcus M Feb 14 '15 at 01:29
  • I think the problem is that $dxdy$ isn't the area element of the plane, and I'm not sure how to fix that. – Marcus M Feb 14 '15 at 17:22