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Math Help - Line Integral

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    Line Integral

    If C is any curve from (0,0,0) to (1,1,1) and F(x,y,z)=(z+4y){i}+(2z+4x){j}+(2y+x){k}. Compute the line integral \int_{C}F\,dr
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    Let be r=(t,t,t) the curve. Then, dr=(1,1,1). Note that the curve passes through the given points. Now, F\cdot dr=(t+4t)+(2t+4t)+(2t+t)dt now, integrate

    \int_0^1 F\cdot dr and the problem will be done.
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  3. #3
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    Quote Originally Posted by pkrleads View Post
    If C is any curve from (0,0,0) to (1,1,1) and F(x,y,z)=(z+4y){i}+(2z+4x){j}+(2y+x){k}. Compute the line integral \int_{C}F\,dr
    Here's a different way to do it: The question implies that the path does not matter and a little checking of "mixed derivatives" verifies that. That means that this is an "exact differential"- there exist a function, F(x,y,z), such that \nabla F= \frac{\partial F}{\partial x}\vec{i}+ \frac{\partial F}{\partial y}\vec{j}+ \frac{\partial F}{\partial z}\vec{k}= (z+4y)\vec{i}+ (2z+4x)\vec{j}+ (2y+x)\vec{k}. Find that "anti-derivative" and evaluate at (1,1,1) and (0,0,0).

    From \frac{\partial F}{\partial x}= z+ 4y we get F(x,y,z)= xz+ 4xy+ g(y,z). Notice that the "constant of integration" may be a function of y and z since they are treated as constants in differentiation with respect to x.

    Differentiating that with respect to y, \frac{\partial F}{\partial y}= 4x+ \frac{\partial g}{\partial y}= 2z+ 4x. The "4x" terms cancel (as they had to do- otherwise the integral would depend on the path) so now we have \frac{\partial g(y,z)}{\partial y}= 2z. Taking the anti-derivative of that with respect to y, g(y,z)= 2yz+ h(z) where now the "constant of integration" may depend on z. That is, we now have [tex]F(x,y,z)= xz+ 4xy+ 2yz+ h(z).

    Differentiating that with respect to z, \frac{\partial F}{\partial z}= x+ 2y+ \frac{dh}{dz}= 2y+ x, so \frac{dh}{dz}= 0- it really is a constant.

    F(x,y,z)= xz+ 4xy+ 2yz+ C.
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