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

- anonymous

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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- jamiebookeater

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- Akshay_Budhkar

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

- anonymous

ok no problem

- Akshay_Budhkar

\[2 \sin(x) \cos(x)=\sin(2x)\]

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## More answers

- Akshay_Budhkar

so your G(x,y) =

- Akshay_Budhkar

\[2 \times 5 \times \cos(x) \times 5 \times \sin(x)\]

- Akshay_Budhkar

= \[25 \sin (2x)\]

- Akshay_Budhkar

you can intergerate that :D

- 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 \[\int\limits_{C}^{} G(x,y) ds \]

- Akshay_Budhkar

one min i need to get a better help 1 second

- 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

- Akshay_Budhkar

@turing ds???

- anonymous

what do you think ?

- TuringTest

I'm sorry I'm multitasking...

- anonymous

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

- Akshay_Budhkar

i don't know that is why i called turing

- Akshay_Budhkar

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

- anonymous

yes i am sure

- anonymous

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

- Akshay_Budhkar

why do you think you need to add?

- Akshay_Budhkar

ds generally in my questions is the area

- anonymous

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

- Akshay_Budhkar

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

- anonymous

ok I appreciate your help bro .. thanks.

- Akshay_Budhkar

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

- anonymous

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

- TuringTest

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

- 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 \[\int_C G(x,y)ds=\int_0^{\pi/4}25\sin(2t)\cdot 5dt = \int_0^{\pi/4}125\sin(2t)dt = \frac{125}{2}\]

- Akshay_Budhkar

ds?

- Akshay_Budhkar

arc length is what it signifies?

- TuringTest

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

- anonymous

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

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