Work of a vector field along a curve

Question

I'm having doubts about this exercise:

Evaluate the work of the following field: $$F(x,y) = (\cos(y+2) + 1, -x\sin(y+2)-2) $$ along the curve: $$ \begin{split} \delta &:[0, 3\pi] \to \mathbb{R}^2 \\ t &\mapsto (t\cos(t), t\sin(t)) \end{split} $$ I've tried to solve it using this: $$ L_{\delta}(F) = \int_0^{3\pi}{F(\delta(t))\ \cdot \ \delta'(t) \ dt}$$ but the resulting integral looks a bit too hard to evaluate by hand so I'm wondering if i used the wrong formula but i can't seem to find another method. Thanks.

Answer 1

After some thinking I worked out the solution. I leave it here for anyone to use.

The tricky part was to understand that i needed to use the work of conservative fields along a curve. The formula states: $$ L_{\delta}(F) = \int_{\delta}F\cdot d\delta = U(\delta(b))-U(\delta(a))$$ Using this i could easily evaluate the field potential $$ U(x,y) = x\cos(y+2)+x-2y$$ And use the formula to obtain as following $$L_{\delta}(F) = \int_{\delta}F\cdot d\delta = U(\delta(b))-U(\delta(a)) = U(-3\pi,0)-U(0,0) = -3\pi(1+\cos(2))$$

Answer 2

Well, the formula for work when given a vector field and path is just a line integral, so your formula is not wrong by any means.

Anyhow let's see,

so we have $$\vec{F} = \langle \cos(y+2)+1, -\sin(y+2)-2 \rangle,\qquad \vec{\delta} = \langle t\cos(t), t\sin(t)\rangle$$ $$\implies \vec{F}(\vec{\delta}) = \langle \cos(t\sin(t)+2)+1, -\sin(t\sin(t)+2)-2 \rangle$$

We also have $$\vec{\delta}\,' = \langle \cos(t)-t\sin(t), \sin(t)+t\cos(t)\rangle$$

So $$\vec{F}(\vec{\delta}) \cdot \vec{\delta}\,'= -t\sin(t)-2\sin(t)-t\sin(t+t\sin(t)+2) - 2t\cos(t)+\cos(t)+\cos(t+t\sin(t)+2)$$

The antiderivative of this function is not expressible in terms of frequently used functions (from the existence of the nested cosine and sine terms), so either

• You were supposed to solve this integral numerically
• The problem was given wrong/you misread the problem

Numerically the work done is around -2.5.