anonymous
  • anonymous
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?
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.