Let $ \ \vec F(x,y)=xy^2 \hat i+x^2y \hat j \ $ . Evaluate $ \ \int_C \vec F \cdot d \vec s \ $ , where $ \ C \ $ is the upper half of the circle $ \ x^2+y^2=1 \ $
Answer:
Let,
$ x=\cos(t) , \\ y=\sin(t) \ $
$ d \vec s=dx \hat i+ dy \hat j= -\sin (t) dt \hat i+\cos (t) dt \hat j \ $
Thus,
$ \vec F \cdot d \vec ds =-\frac{1}{4} \sin (4t ) \ $
Therefore,
$ \int_C \vec F \cdot d \vec s=-\frac{1}{4} \int_{0}^{\pi} \sin (4t) dt \ =0 \ $
Am I right ?
Help me out
We have: $$ \vec F \cdot \vec ds=\cos t \sin^2 t (-\sin t)dt + \cos^2 t\sin t(\cos t) dt=$$ $$= \sin t \cos t(\cos^2 t-\sin^2 t)dt= $$ $$ =\frac{1}{2}\sin (2t) \cos(2t)=\frac{1}{4}\sin(4t) $$
So there is a sign error in your result, but it does not matter for the final result.