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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.

$\mathbf{F}=\frac{\langle x, y, z\rangle}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}}$ on the line segment from (1,1,1) to (10,10,10)

$$\frac{3 \sqrt{3}}{10}$$

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All right. Number 45. Given the force field X Y Z over X squared plus y squared. Plus C squared the three for two. Integrate this line of the girl from 111 to 10. 10 turn. Okay. Are tee. It's gonna equal. So we start from one for each of them. The coefficient for the tea is going to 10 minus one. Just give you nine t. Okay. Our priority. Pretty easy. Nine. Coming. Nine. Coming. Nine. No, we plug. This is as good. This is X. This is why this is he. But those into here for our integral. So we have integral c f that our prime if t t t is a good your the in a girl of one puts 90 Call close 90 come one plus nine t over three squared of three. This is just simplifying the ex corpus Oscar plus Z squared to the cube root and ah q power than square root 92 or three DT. And that's from 0 to 1. Oh, God. And dot Our property Very good. TT integrate from 0 to 1. Born over. Suggest doing algebra. Simplifying it. Three Test nine does one was nine t d. T call the Integral from 0 to 1 nine of her. Three times. 1/1 plus 90 square t t equals one of her. Three point that out 90 t over a warm plus 90 squared. So we're gonna do some use substitution. Here stick. And you is one plus 90. The EU is good A 90 t and then converting the limits that we integrate with tickle zero. Then you close one equals one than you equals 10 snow. We're integrating from 1 to 10. Do you? Were you squared? Which is equal to one of our route three. The good 1/4? No, you go one over you from 1 to 10 on. But then a simplify or get route three times 3/10.

Sacramento City College