An object moves in the force field $F=yz\hat{i}+zx\hat{j}+xy\hat{k}$ starting at the origin and ending at some point $A(\xi,\eta,\zeta)$ that lies on the surface $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$. What is the maximum possible value of the work done $W = \displaystyle\oint F \cdot dr$?
There are several ways that I've tried, and all are a little problematic, so far:
Since the curve that the object is moving along starts at $(0,0,0)$ and ends at $A$, this is not a closed curve, so we can't apply Stokes' theorem.
I computed the curl of $F$ and got $(0,0,0)$, so $F$ is a gradient field. And it's easy to see the corresponding potential function: $f(x,y,z)= xyz$.
The Fundamental Theorem of Calculus for Line Integrals, which is specifically for gradient fields, tells us that the integration will be path-independent.
However, I don't know how to parameterize this curve.
Also, should I use the integrand $f(\gamma(t))(\gamma'(t))dt$, assuming I have a parametrization for the curve, or should I use the "exact differential" as my integrand, $f_xdx + f_ydy + f_zdz$?
Thanks,
As you mentioned, the force field $F(x,y,z) = yz\hat{i}+zx\hat{j}+xy\hat{k}$ is a conservative force field with a potential function of $f(x,y,z) = xyz$. Hence, the work done in moving from the origin $O(0,0,0)$ to a point $A(\xi,\eta,\zeta)$ is simply $f(\xi,\eta,\zeta) - f(0,0,0) = \xi\eta\zeta$.
Since the point $A(\xi,\eta,\zeta)$ is on the surface of the ellipsoid $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2} = 1$, we must have $\dfrac{\xi^2}{a^2}+\dfrac{\eta^2}{b^2}+\dfrac{\zeta^2}{c^2} = 1$.
So, you simply need to maximize $\xi\eta\zeta$ subject to the constraint $\dfrac{\xi^2}{a^2}+\dfrac{\eta^2}{b^2}+\dfrac{\zeta^2}{c^2} = 1$.
This is a simple problem which can be solved by Lagrange Multipliers. Note: If all of the Greek letters are confusing, you can simply maximize $xyz$ subject to $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2} = 1$.