Use the result of Exercise 39 to calculate the area of the triangle with vertices $(1,0,0),(0,1,0)$, and $(0,0,1)$ as a line integral. Verify your result using geometry.

Data From Exercise 39

Prove the following generalization of Eq. (1). Let \(C\) be a simple closed curve in the plane \(\mathcal{S}\) with equation \(a x+b y+c z+d=0\) (Figure 8). Then the area of the region \(R\) enclosed by \(C\) is equal to
\[
\frac{1}{2\|\mathbf{N}\|} \oint_{C}(b z-c y) d x+(c x-a z) d y+(a y-b x) d z
\]
where \(\mathbf{N}=\langle a, b, cangle\) is the normal to \(\mathcal{S}\), and \(\mathcal{C}\) is oriented as the boundary of \(\mathcal{R}\) (relative to the normal vector \(\mathbf{N})\). Apply Stokes' Theorem to \(\mathbf{F}=\langle b z-c y, c x-a z, a y-b xangle\).

Transcribed Image Text:

X 1 R Plane S N = a, b, c