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

1. MathSofiya

Oops wrong question, it's $F(x,y,z)=z\hat{i}+y\hat{j}-z\hat{k}$ $r(t)=t\hat{i}+sint\hat{j}+cost\hat{k}$ $0 \le t \le pi$

2. Outkast3r09

3. Outkast3r09

alright guess we'll start over hahha continues

4. MathSofiya

would that make $r'(t)=\hat{i}+cost\hat{j}-sint\hat{k}$ and

5. MathSofiya

$\int_0^{\pi} (t\hat{i}+sint\hat{j}+cost\hat{k})\bullet (\hat{i}+cost\hat{j}-sint\hat{k})dt$

6. KingGeorge

So far that looks right to me.

7. MathSofiya

thanks

8. KingGeorge

You're welcome. Once you do the dot product, it looks like it will be very easy to evaluate.

9. KingGeorge

So can you show me what you think the dot product will be?

10. MathSofiya

I can't just multiply the i j and k coefficients can I? $t+sintcost-sintcost$

11. KingGeorge

That's exactly what it is. You should end up with $t+\sin(t)\cos(t)-\sin(t)\cos(t)=t$

12. MathSofiya

Yaaaaayyyyy!!!!! I just need to build confidence i guess.... sigh

13. KingGeorge

You'll get used to it.

14. MathSofiya

here is the mistake I see: x=t y=sint z=cost that would give us $\int_0^{\pi} cost dt+\int_0^{\pi}sintcostdt+\int_0^{\pi}tsint dt$

15. 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?

16. TuringTest

$=\int_0^\pi\cos t+\sin t\cos t-\sin t\cos tdt=\int_0^\pi\cos tdt=-2$

17. MathSofiya

oh I see what I did! It's z y -z and not z y -x

18. MathSofiya

thanks!

19. TuringTest

welcome!

20. TuringTest

now I go to bed :)

21. MathSofiya

Have a good night's rest!

22. TuringTest

likewise!