anonymous 4 years ago Evaluate the line integral (x2+y2)dx+2xydy where is the path of the semicircular arc of the circle x2+y2=64 starting at (8,0) and ending at (−8,0) going counterclockwise.

1. TuringTest

did you want to check an answer, or are you stuck?

2. anonymous

$512\int\limits_{0}^{\pi} \cos^2t sint - \sin^3t$ im stuck at how to integrate this.

3. TuringTest

use$u=\cos t$and$\sin^2t=1-\cos^2t$

4. anonymous

so du =-sinx

5. TuringTest

how did you get cos^2t*sint for x^2+y^2dx ?

6. anonymous

i keep getting confused but heres where i started $\int\limits_{0}^{\pi} ((8cost)^2 + (8sint)^2)(-8sint) + 2(8cost)(8sint)(8cost)dt$

7. TuringTest

yes, and that can be written$\int_0^\pi512(\sin^2t+\cos^2t)(-\sin t)dt+\int_0^\pi512(2\cos^2t\sin t) dt$

8. anonymous

$-512\int\limits_{0}^{?\pi} sint dt +1024\int\limits_{0}^{\pi} \cos^2tsint$ which becomes this

9. TuringTest

yes, which you can recombine to form this$512\int_0^\pi2\cos^2t\sin t-\sin tdt$

10. anonymous

so would you use u=cost and du=-sint ?

11. TuringTest

yep

12. anonymous

or could you do $sint(\cos^2t-1)$ which becomes $\sin^3t$ ?

13. TuringTest

and how to you propose to integrate that?

14. anonymous

oh yeah.haha.

15. anonymous

$-512\int\limits_{1}^{-1}- 2u^2du - du$ ?

16. anonymous

i mean + du

17. TuringTest

yeah, looks good

18. TuringTest

oh wait, how did 512 get negative?

19. anonymous

im confused because du=-sint and the sint is positive so (-1/-1) would need to be multiplied right?

20. TuringTest

yeah, but you already covered that by changing thethe signs og the terms

21. TuringTest

however the double du notation is highly unorthodox and looks screwy to me, so I'd avoid it

22. TuringTest

$512\int_0^\pi2\cos^2t\sin t-\sin tdt$$u=\cos t\implies du=-\sin tdt$$512\int_{1}^{-1}- 2u^2du +\int_1^{-1} -\sin tdt$

23. TuringTest

sorry I messed up the bounds on the last integral, they should be 0 to pi

24. anonymous

and the answer is 1706.66666666667?

25. anonymous

nevermind thats wrong.

26. TuringTest

$512\int_0^\pi2\cos^2t\sin t-\sin tdt$$u=\cos t\implies du=-\sin tdt$$512\int_{1}^{-1}- 2u^2du +\int_0^{\pi} -\sin tdt$let me see what I get...

27. TuringTest

$512\left(\left.-\frac23u^3\right|_1^{-1}+\left.\cos t\right|_0^\pi\right)$$512(\frac43-2)=512(-\frac23)$but I always could have made a mistake...

28. anonymous

its right! yay!

29. TuringTest

sweet, I just hope you find your (likely algebra-based) mistake

30. anonymous

thats what i got to too! thanks so much!

31. TuringTest

welcome!