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I would like a second opinion on a strategy.

Problem:

For $F(u,v)= (u+v, 2u-3v, u+5v+1)$ with the restriction that $\frac{(u-c_1)^2}{a^2}+\frac{(v-c_2)^2}{b^2}=1$.

We want to show that the work $W= \int F \cdot \frac{dr}{dt}dt$ is independent of the values of $c_1$ and $c_2$. My thinking was that what I need to do is to solve for say $v$ using the restriction and plug that back in and that would my parameterization and then I can calculate. My question is there a better way to do this or not really because working with $v= \sqrt{b^2-b^2\left[\frac{u-c_1}{a}\right]^2}+c_2$ does not look appealing.

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    You can just take $u=a\cos t+c_1$ and $v=b\sin t+c_2$ with $t \in [0,2\pi]$ if you don't want to use stokes, that would be easier than what you have written. – 2017-01-07
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    Is F not a force then what is the force? @Kori – 2017-01-07

1 Answers 1

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Consider the force given by,

$$F(x,y,z)=(x+y,2x-3y,x+5y+1)$$

With $z=0$

Use Stokes theorem to show that,

$$w=\iint_{D} 1 dA=A(D)$$

Now all you have to show is that,

$$\frac{(x-c_1)^2}{a^2}+\frac{(y-c_2)^2}{b^2}=1$$ Encloses the same area if we vary $c_1$ and/or $c_2$.