## anonymous 5 years ago find ∫G(x,y)ds ??? , on the indicated curve G(X,Y)=2XY, C: X=5*COST, Y=5*SINT, 0≤t≤π÷4

1. Akshay_Budhkar

ok i will give u a hint, u should solve it yourself :D

2. anonymous

ok no problem

3. Akshay_Budhkar

$2 \sin(x) \cos(x)=\sin(2x)$

4. Akshay_Budhkar

5. Akshay_Budhkar

$2 \times 5 \times \cos(x) \times 5 \times \sin(x)$

6. Akshay_Budhkar

= $25 \sin (2x)$

7. Akshay_Budhkar

you can intergerate that :D

8. anonymous

i know how to find $\int\limits_{C}^{} G(x,y) dx$ and $\int\limits_{C}^{} G(x,y) dy$ what i want is $\int\limits_{C}^{} G(x,y) ds$

9. Akshay_Budhkar

one min i need to get a better help 1 second

10. anonymous

i am thinking that i should add $\int\limits_{C}^{} G(x,y) dx + \int\limits_{C}^{} G(x,y) dy$ but i am not sure

11. Akshay_Budhkar

@turing ds???

12. anonymous

what do you think ?

13. TuringTest

14. anonymous

this is how the question written exactly ... what does he mean by ds?

15. Akshay_Budhkar

i don't know that is why i called turing

16. Akshay_Budhkar

ds? are u sure that is what is asking for?

17. anonymous

yes i am sure

18. anonymous

the answer will be 125/2 .. but i do not know how to get it

19. Akshay_Budhkar

why do you think you need to add?

20. Akshay_Budhkar

ds generally in my questions is the area

21. anonymous

really i am not sure i tried everything but i could not reach the answer

22. Akshay_Budhkar

i am waiting for turing to respond lol.. he will surely help.. i m thinking of possibilities till then

23. anonymous

ok I appreciate your help bro .. thanks.

24. Akshay_Budhkar

no no its not done! i wont sleep till u get ur answer :P

25. anonymous

Lol thanks so much.. me too cuz i have to submit this assignment tomorrow :D

26. TuringTest

ok ds=sqrt[(dx/dt)^2+(dy/dt)^2]=5 2xy=50cost*sint got it from there?

27. anonymous

Integrating with respect to ds means that you are integrating with respect to the arc length (which is often denoted s) of the curve C. Note that $ds=\sqrt{dx^2+dy^2}=\sqrt{(-5\cos (t) dt)^2+(5\sin (t) dt)^2}=5dt$ Thus, you have that $\int_C G(x,y)ds=\int_0^{\pi/4}25\sin(2t)\cdot 5dt = \int_0^{\pi/4}125\sin(2t)dt = \frac{125}{2}$

28. Akshay_Budhkar

ds?

29. Akshay_Budhkar

arc length is what it signifies?

30. TuringTest

yes, you need ds in line integrals, arc length differential

31. anonymous

thanks guys I get it .. I appreciate your help .