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Evaluate the line integral \[\int_c F\bullet dr\] where C is given by the vector function r(t). \[F(x,y,z)=(x+y)\hat{i}+(y-z)\hat{j}+z^2\hat{k}\] \[r(t)=t^2\hat{i}+t^3\hat{j}+t^2\hat{k}\] \[0 \le t \le 1\] \[r'(t)=2t\hat{i}+3t^2\hat{j}+2t\hat{k}\] \[\int_c F \bullet dr=\int_a^b F(r(t)) \bullet r'(t)dt\] \[\int_0^1 (t^2\hat{i}+(t^3-t^2)\hat{j} + t^4\hat{k})\bullet (2t\hat{i} +3t^2\hat{j}+2t\hat{k})dt\] how am I doing?

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Oops wrong question, it's \[F(x,y,z)=z\hat{i}+y\hat{j}-z\hat{k}\] \[r(t)=t\hat{i}+sint\hat{j}+cost\hat{k}\] \[0 \le t \le pi\]
for your i, x+y
alright guess we'll start over hahha continues

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Other answers:

would that make \[r'(t)=\hat{i}+cost\hat{j}-sint\hat{k}\] and
\[\int_0^{\pi} (t\hat{i}+sint\hat{j}+cost\hat{k})\bullet (\hat{i}+cost\hat{j}-sint\hat{k})dt\]
So far that looks right to me.
thanks
You're welcome. Once you do the dot product, it looks like it will be very easy to evaluate.
So can you show me what you think the dot product will be?
I can't just multiply the i j and k coefficients can I? \[t+sintcost-sintcost\]
That's exactly what it is. You should end up with \[t+\sin(t)\cos(t)-\sin(t)\cos(t)=t\]
Yaaaaayyyyy!!!!! I just need to build confidence i guess.... sigh
You'll get used to it.
here is the mistake I see: x=t y=sint z=cost that would give us \[\int_0^{\pi} cost dt+\int_0^{\pi}sintcostdt+\int_0^{\pi}tsint dt\]
\[\vec F(x,y,z)=z\hat{i}+y\hat{j}-z\hat{k}\]\[\vec r(t)=t\hat i+\sin t\hat j+\cos t\hat k\]\[\vec r'(t)=\hat i+\cos t\hat j-\sin t\hat k\]\[\int_0^\pi\vec F(\vec r(t))\cdot\vec r'(t)dt=\int_0^\pi\langle\cos t,\sin t,\cos t\rangle\cdot\langle1,\cos t,-\sin t\rangle dt\]do you see a problem with that?
\[=\int_0^\pi\cos t+\sin t\cos t-\sin t\cos tdt=\int_0^\pi\cos tdt=-2\]
oh I see what I did! It's z y -z and not z y -x
thanks!
welcome!
now I go to bed :)
Have a good night's rest!
likewise!

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