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Work integrals in $\mathbb{R}^{3}$ Given the force field $\mathbf{F}$, find the work required to move an object on the given oriented curve.

$\mathbf{F}=\frac{\langle x, y, z\rangle}{x^{2}+y^{2}+z^{2}}$ on the line segment from (1,1,1) to (8,4,2)

$$w=1.666$$

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remember 46 in the field. X Y C over X squared plus y squared plus Z squared. Integrated this line over the segments from 0.1 on 1 to 8 for two. Uh, this are of tea of a line. They'll start with one. It's going to eat minus one at 70. It's been four months. One is three t two minutes. Bonus won t That's our tea. Our prime of tea. Seven come of three. Come on, One. Okay, You know, this is X. This is why this is easy Plug all the and to hear when we write out our line and grow after our promise T d t in a girl from 0 to 1 one plus 70 one plus three t one pushed t over one plus 70 squared plus one plus three t square plus one plus t squared. Thought that with seven. Come in 321 t All right. During a little bit algebra. When we got these two together and add them, we get 11 plus 15 9 t over one for 70 square plus one plus three t squared plus one plus t square T t now again More algebra just simplify the bottom. We get 50 90 squared plus 22 t plus three d t. Okay, now we can pull out a factor too. 0 to 1 of 22 +11 80 over 50. 90 square plus 22 T plus three d t. Sara, we can recognize this form. Um, it's a form of the integral f prime of ref of X equals D X. You can see the derivative of this term here Appears right here the dirt of this term. Here, Piers. Right here. So? So this form equals Ln of f of X, so basically equals l in the bottom. So one half of Ellen of 59 por una 15 90 squared plus 22 t plus three. Battle waited From what? It zero. That gives us half of Ellen of 84 minus Eleanor three equals half of Ellen of 28. That's a log function. So you divide these two. When you're subtracting clogs, you get 84/3, which is 28

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