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Let $C$ be a smooth simple closed curve in the plane $2x + y + 2z = > 3$, equipped with anticlockwise orientation seen from the positive $z$-axis. Let $S$ be the surface bounded by $C$ and contained in the same plane, and let $A$ be the area of $S$.

Is it possible to express the value of the line integral $$\int_{C}(2y\:dx+3z\:dy-x\:dz)$$ in terms of $A$? Calculate the value of $A$.


What I've tried:

This question appears quite similar to the question Calculating line integrals via Stokes' theorem. However, the method used for this question is very different from the question mentioned, because instead of parametrising, the normal is found by taking the unit normal vector $\hat n = \frac{1}{3} \langle 2, 1, 2 \rangle$ and then instead of integrating over the curve $C$; in the previous question, we integrated over the surface $S$ instead.

How are these two questions different, and how do we know when to use the parametrisation method as opposed to some other method?

Also, why must we integrate over the region $S$ instead of under the curve $C$?

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$S$ is simply $$x=x,y=y,z=(3-2x-y)/2$$ with $(x.y)\in D=$ projection of $S$ on the plane $XY$.

The normal vector is $N=\cdots$

and $\text{curl}(2y,3z,-x)=\cdots$

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  • $\begingroup$ yeah seemed pretty easy $\endgroup$ – godonichia Nov 13 '14 at 15:22

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