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Use Stokes' Theorem to evaluate $\mathbf{F} \cdot d \mathbf{r}$ where $\mathbf{F} =(2x+3y, 1y+5z, 9z+1x)$ and the curve is the triangle with vertices $(4,0,0), (0,6,0)$ and $(0,0,24)$ oriented so that the vertices are traversed in the specified order.

I first found $\operatorname{curl} F$ which was $-5i -j-3k$.

I let $p=(4,0,0)$ and $q=(0,6,0)$ and $r=(0,0,24)$. So $PQ = (-4,6,0)$ and $QR = (0,-6,24)$. Taking the cross product of the two I got $144i + 96j + 24k$, so the plane is $$144(x-4) + 96(y-0) + 24(z-0).$$I solved for $z$ and got my plane to be $z=24-6x-4y$. I let $z=0$ and solved for $y$ and got $6-(3/2)x$.

I took the integral from $0$ to $4$ and from $0$ to $6-(3/2)x$ of $-9 \, dy \, dx$. My final answer was $-108$ but that is incorrect.

Mark Fantini
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Ayoshna
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1 Answers1

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$\int_C \textbf{F}\cdot d\textbf{r} = \int\int_S curl\textbf{F}\cdot d\textbf{A} $

$d\textbf{A} = \textbf{N} \cdot dA = \frac{\nabla G(x,y,z)}{|G_3|} $

$ G(x,y,z) = -24+6x+4y+z$

$ \nabla G(x, y, z) = 6i +4j +k$

$ |G_3| = 1$

$\int\int_S curl\textbf{F}\cdot d\textbf{A} = \int_0^4\int_0^{6-1.5x} (-5, -1, -3) \cdot (6, 4, 1)dydx$

$\int_C \textbf{F}\cdot d\textbf{r} = -444 $

mzee99
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