I'm trying to solve the following problem. (10.13.10 from Apostol Vol. 2.)
Let
$$ F(x,y) = (3y^2 + 2)i + 16xj $$
be a force field. I want to find the work done by $F(x,y)$ from
$(-1,0)$ to $(1,0)$ along the upper half of the ellipse $b^2 x^2 + y^2
> = b^2$.
Here's my solution attempt. The ellipse can be parametrized by $a(t) = \cos(t)i + b\sin(t)$ for $t \in [-\pi, \pi]$. So
$$ F(a(t)) = (3(b \sin(t) )^2 + 2)i + 16\cos(t) j $$
$$ a'(t) = -\sin(t) i + b\cos(t) j .$$
Thus the work done is
$$ \int_C F ds = \int_{-\pi}^\pi F(a(t)) \cdot a'(t) dt $$
which is equal to
$$ \int_{-\pi}^\pi -3b^2 \sin(t)^3 – 2\sin(t) + 16b \cos^2(t) dt. $$
This integral evaluates to $16b\pi$. I checked it both by hand and using Mathematica. However, the solution is supposed to be $4b^2 – 8\pi b + 4$. Where is my solution attempt incorrect?
Best Answer
All your work ;-) is ok but the limits of integration.
$$\int_{\pi}^0 -3b^2 \sin(t)^3 - 2\sin(t) + 16b \cos^2(t) dt=4b^2 - 8\pi b + 4$$
The particle is moving from $(-1,0)$ where $\theta=\pi$ to $(1,0)$ where $\theta=0$ (if it had to move by the lower half, the limits had to be from $-\pi$ to $0$