Example of using Stokes's Theorem

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For more information see: www.chapman.edu/~jipsen/asciimath.html. Example of using Stokes's Theorem to transform an integral
Let C be the oriented triangle lying in the first octant in the plane 4x+2y+z=12 [i.e., the triangle from (3,0,0) to (0,6,0) to (0,0,12) and back to (3,0,0)]. Use Stokes's Theorem to express `int_C bbF•dr` as an iterated integral, giving the limits of integration for each integral. Use the definition of `bbF` as `-8y^2bbi+6zbbj+5xbbk`.

We know that `int_C bbF•dr = int_Sint (curl\quadbbF)•bbNds = int_Rint (curl\quadbbF)•(-g_x(x,y)bbi - g_y(x,y)bbj + bbk) dA`

We need to express the surface in the form `z=g(x,y)`, but we can do this since we have the equation for the plane of the triangle is `4x+2y+z=12` so `g(x,y)= -4x - 2y + 12` and that makes `-g_x(x,y)bbi - g_y(x,y)bbj + bbk = 4bbi + 2bbj +1bbk = <4,2,1>`.

From the definition of `bbF` we have that

`M(x,y,z)=-8y^2, N(x,y,z)=6z, P(x,y,z)=5x` so

`curl\quad bbF =(0-6)bbi -(5-0)bbj +(0-(-16y))bbk = -6bbi-5bbj+16ybbk = <-6,-5,16y>` which gives the associated dot product with `<4,2,1>` as
`-24-10+16y = -34+16y`.

The limits of integration across the x axis will be 0 to 3 and the functional values of the lower bounds will be `y=0` and the upper bounds `y=-2x+6`.

All of which means that

`int_C bbF•dr = int_0^3int_0^(-2x+6)-34+16y\quaddydx`

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