A community for students.
Here's the question you clicked on:
 0 viewing
MathSofiya
 3 years ago
Evaluate the line integral \[\int_c F\bullet dr\]
where C is given by the vector function r(t).
\[F(x,y,z)=(x+y)\hat{i}+(yz)\hat{j}+z^2\hat{k}\]
\[r(t)=t^2\hat{i}+t^3\hat{j}+t^2\hat{k}\]
\[0 \le t \le 1\]
\[r'(t)=2t\hat{i}+3t^2\hat{j}+2t\hat{k}\]
\[\int_c F \bullet dr=\int_a^b F(r(t)) \bullet r'(t)dt\]
\[\int_0^1 (t^2\hat{i}+(t^3t^2)\hat{j} + t^4\hat{k})\bullet (2t\hat{i} +3t^2\hat{j}+2t\hat{k})dt\]
how am I doing?
MathSofiya
 3 years ago
Evaluate the line integral \[\int_c F\bullet dr\] where C is given by the vector function r(t). \[F(x,y,z)=(x+y)\hat{i}+(yz)\hat{j}+z^2\hat{k}\] \[r(t)=t^2\hat{i}+t^3\hat{j}+t^2\hat{k}\] \[0 \le t \le 1\] \[r'(t)=2t\hat{i}+3t^2\hat{j}+2t\hat{k}\] \[\int_c F \bullet dr=\int_a^b F(r(t)) \bullet r'(t)dt\] \[\int_0^1 (t^2\hat{i}+(t^3t^2)\hat{j} + t^4\hat{k})\bullet (2t\hat{i} +3t^2\hat{j}+2t\hat{k})dt\] how am I doing?

This Question is Closed

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.3Oops wrong question, it's \[F(x,y,z)=z\hat{i}+y\hat{j}z\hat{k}\] \[r(t)=t\hat{i}+sint\hat{j}+cost\hat{k}\] \[0 \le t \le pi\]

Outkast3r09
 3 years ago
Best ResponseYou've already chosen the best response.0alright guess we'll start over hahha continues

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.3would that make \[r'(t)=\hat{i}+cost\hat{j}sint\hat{k}\] and

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.3\[\int_0^{\pi} (t\hat{i}+sint\hat{j}+cost\hat{k})\bullet (\hat{i}+cost\hat{j}sint\hat{k})dt\]

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1So far that looks right to me.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1You're welcome. Once you do the dot product, it looks like it will be very easy to evaluate.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1So can you show me what you think the dot product will be?

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.3I can't just multiply the i j and k coefficients can I? \[t+sintcostsintcost\]

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1That's exactly what it is. You should end up with \[t+\sin(t)\cos(t)\sin(t)\cos(t)=t\]

MathSofiya
 3 years ago
Best ResponseYou've already chosen the best response.3Yaaaaayyyyy!!!!! I just need to build confidence i guess.... sigh

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1You'll get used to it.

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.3here 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
 2 years ago
Best ResponseYou've already chosen the best response.0\[\vec F(x,y,z)=z\hat{i}+y\hat{j}z\hat{k}\]\[\vec r(t)=t\hat i+\sin t\hat j+\cos t\hat k\]\[\vec r'(t)=\hat i+\cos t\hat j\sin t\hat k\]\[\int_0^\pi\vec F(\vec r(t))\cdot\vec r'(t)dt=\int_0^\pi\langle\cos t,\sin t,\cos t\rangle\cdot\langle1,\cos t,\sin t\rangle dt\]do you see a problem with that?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.0\[=\int_0^\pi\cos t+\sin t\cos t\sin t\cos tdt=\int_0^\pi\cos tdt=2\]

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.3oh I see what I did! It's z y z and not z y x

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.3Have a good night's rest!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.