Do you know what "Stoke's Theorem" is? What does it say?
Use Stokes Theorem to evaluate the integral of F*dx , where F(x,y,z)= e^-7x i+e^-4y j+e^5z and C is the boundary of the part of the plane 8x+y+8z=8 in the first octant.
Anyone know how to go about this problem I am beyond lost.
Please help
Thanks!
Possible answers include-
0
23
49
16
69
Stokes theorem is defined by the below function-
link- http://mathworld.wolfram.com/StokesTheorem.html
Okay, good! Since you are asked to use Stoke's theorem to integrate around a path, , this problem must intend you to integrate over the area inside the path.
You are told that so you need to find the "curl" of that.
Also, you are told that C (the in the formula) is the boundary of the portion of the plane 8x+y+8z=8 in the first octant. lThat plane cuts off a triangle in the first quadrant. When x and y are both 0, that is 8z= 8 so z= 1 and (0, 0, 1) is one corner of that triangle. When y and z are both 0, 8x= 8 so x= 1 and (1, 0, 0) is another corner. When x and z are both 0, y= 8 so (0, 8, 0) is the third corner.
What I recommend you do is this: from the equation, 8x+ y+ 8z= 8, we can solve for y= 8- 8x- 8z= 8(1- x- z). That means that we can write that plane a as a vector equation: . The two derivative vectors, and are in the tangent plane to the surface (here, in the plane itself) and their cross product,
. Its length, times "dxdz", gives the "differential of area". (All that is very general. For a plane it reduces to the square root of the sum of squares of the coefficients in the equation- you probably have that formula in your textbook.)
The plane cuts the xz-plane where y= 0: 8x+ 8z= 8 which is the same as x+ z= 1 or z= 1- x. x ranges from 0 up to 1 and, for each x, z ranges from from 0 up to 1- x. Your integral should look like
.