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Integration

Calculate the work done by moving along the straight line from (2,1,1) to (3,2,2) in a force field:

F=r/∣r2∣=[xi+yj+zk]/[x2+y2+z2]F=\mathbf{r}/|\mathbf{r^2}| = [ x\mathbf{i}+y\mathbf{j}+ z\mathbf{k} ]/ [x^2+y^2+z^2]
(edited 13 years ago)
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0le
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anyone? :frown:
Original post by djpailo
anyone? :frown:


Were you ever unsure about what you were doing but it worked out so nicely you thought it must be right?

Well that's what just happened to me with this question.

I did it as ∫F⃗dr⃗dtdt\int \vec{F} \frac{d\vec{r}}{dt}dt
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0le
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Original post by Get me off the £\?%!^@ computer
Were you ever unsure about what you were doing but it worked out so nicely you thought it must be right?

Well that's what just happened to me with this question.

I did it as ∫F⃗dr⃗dtdt\int \vec{F} \frac{d\vec{r}}{dt}dt


cool, what answer did you get, i got 119/3
Original post by djpailo
cool, what answer did you get, i got 119/3


I got 12ln⁡176\frac{1}{2}\ln \frac{17}{6} so it seems that at least one of us is wrong.
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0le
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Original post by Get me off the £\?%!^@ computer
I got 12ln⁡176\frac{1}{2}\ln \frac{17}{6} so it seems that at least one of us is wrong.


how did you get that?
The straight line from (2,1,1) to (3,2,2) is x(t)=2+t, y(t)=1+t, z(t)=1+t, 0<t<1.

dx/dt=dy/dt=dz/dt=1.

Subbing in you get

∫012+t+1+t+1+t(2+t)2+(1+t)2+(1+t)2dt\int_0^1 \frac{2+t+1+t+1+t}{(2+t)^2+(1+t)^2+(1+t)^2}dt

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