theEric 2 years ago Calculus 3 - Find work done along curve in force field. How do I set up the integral? The curve is the upper half of a circle with radius 3, centered at the origin, from (-3,0) to (3,0).

1. theEric

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2. theEric

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3. theEric

So I think it works nicely if I chose the parametric equations for the curve as$x(t)=cos(t)$$y(t)=sin(t)$and so$\frac{dx(t)}{dt}=-sin(t)$and$\frac{dy(t)}{dt}=cos(t)$

4. theEric

$\text {So then the parameter, }t\text{, goes from }\pi \text{ to }0 \text{.}$ $\text{Is it okay to write:} \int_\pi ^0 F(t)dt\text{, or must it be }-\int_0 ^ \pi F(t)dt \text{?}$ Don't they equal the same thing?

5. Zarkon

those two integrals are the same

6. Zarkon

you might want to look at your parametric equations again

7. theEric

$\text{Thanks.. I've considered rewriting them to go from }0\text{ to }\pi \text{. In which case:}$ $x(t)=-cos(t)$$y(t)=sin(t)$ Is this what you mean? Or were my original equations incorrect?

8. Zarkon

neither are correct

9. Zarkon

10. theEric

Oh... Yes.. Thank you! I tend to not carry all of the math into the next step. I tend to carry over my immediate thoughts and stop there...

11. theEric

$x(t)=3cos(t)$$y(t)=3sin(t)$$\text{where }t \text{ goes from }0\text{ to }\pi \text{.}$ or$x(t)=-3cos(t)$$y(t)=3sin(t)$$\text{where }t\text{ goes from }0 \text{ to } \pi \text{.}$

12. theEric

The first one is from $\pi$ to 0, I mean.

13. Zarkon

ok

14. theEric

Those work, then?

15. theEric

And so they're the same?

16. Zarkon

What is your force field $$F$$?

17. Zarkon

$W=\int\limits_{a}^{b}F(c(t))\cdot c'(t) dt$

18. Zarkon

$$c(t)$$ is the particular path

19. theEric

To continue the problem I'm given,$F(x,y)=(2y+x^2)i + (x^2 - 2x)j$

20. theEric

yielding... $W=-\int _0 ^\pi ((6sin(t)+9cos(t))(-3sint)+(9cos^2(t)-6cos(t))(3cos(t)))dt$

21. theEric

Which is ugly.. I've simplified it a lot, and I think I've solved it, but I'm not sure. I'll just check with Wolfram Alpha. Thank you for clearing up the integration!

22. Zarkon

np

23. theEric

Take care!