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

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

Mathematics
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ok i will give u a hint, u should solve it yourself :D
ok no problem
\[2 \sin(x) \cos(x)=\sin(2x)\]

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Other answers:

so your G(x,y) =
\[2 \times 5 \times \cos(x) \times 5 \times \sin(x)\]
= \[25 \sin (2x)\]
you can intergerate that :D
i know how to find \[\int\limits_{C}^{} G(x,y) dx \] and \[\int\limits_{C}^{} G(x,y) dy \] what i want is \[\int\limits_{C}^{} G(x,y) ds \]
one min i need to get a better help 1 second
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
@turing ds???
what do you think ?
I'm sorry I'm multitasking...
this is how the question written exactly ... what does he mean by ds?
i don't know that is why i called turing
ds? are u sure that is what is asking for?
yes i am sure
the answer will be 125/2 .. but i do not know how to get it
why do you think you need to add?
ds generally in my questions is the area
really i am not sure i tried everything but i could not reach the answer
i am waiting for turing to respond lol.. he will surely help.. i m thinking of possibilities till then
ok I appreciate your help bro .. thanks.
no no its not done! i wont sleep till u get ur answer :P
Lol thanks so much.. me too cuz i have to submit this assignment tomorrow :D
ok ds=sqrt[(dx/dt)^2+(dy/dt)^2]=5 2xy=50cost*sint got it from there?
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 \[\int_C G(x,y)ds=\int_0^{\pi/4}25\sin(2t)\cdot 5dt = \int_0^{\pi/4}125\sin(2t)dt = \frac{125}{2}\]
ds?
arc length is what it signifies?
yes, you need ds in line integrals, arc length differential
thanks guys I get it .. I appreciate your help .

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