"Work" question, calculus, line integrals.

Question

idku · Answer

work done by    $F(x,y) = y{\bf ~i} + y^2{\bf  j}$
as particle moves along the bottom half
of the semi-circle $x^2+y^2=9$, from (-3,0) to (3,0).

anime_lover_levi · Answer

oh my god my brain hurts form reading that

idku · Answer

So, my parametrization is
$r(t)=\langle 3\cos t,3\sin t\rangle$

and the force F in terms of these parameters is
$F(t)=\langle 3\sin t,9\sin^2t\rangle$

My limits are from $\pi$ to $0$, respectively ....

So, the integral I wrote is:
$\displaystyle{\rm Work}=\int_\pi^0\langle 3\sin t,9\sin^2t\rangle•\langle (3\cos t)',(3\sin t)'\rangle ~\mathrm{d}t$

$\displaystyle{\rm Work}=\int_\pi^0\langle 3\sin t,9\sin^2t\rangle•\langle -3\sin t,3\cos t\rangle ~\mathrm{d}t$

idku · Answer

(I ended up getting 9$\pi$/2 ... )
Did I set this up correctly ?

anonymous · Answer

almost--your bounds are incorrect, since $t$ actually progresses from $t=\pi$ to $t=2\pi$, not $t=0$; the situation you found is actually following the upper semicircle while the question calls for the lower one. the actual answer is $-9\pi/2$