## theEric Group Title 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). one year ago one year ago

1. theEric Group Title

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2. theEric Group Title

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3. theEric Group Title

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 Group Title

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

5. Zarkon Group Title

those two integrals are the same

6. Zarkon Group Title

you might want to look at your parametric equations again

7. theEric Group Title

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

8. Zarkon Group Title

neither are correct

9. Zarkon Group Title

10. theEric Group Title

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 Group Title

$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 Group Title

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

13. Zarkon Group Title

ok

14. theEric Group Title

Those work, then?

15. theEric Group Title

And so they're the same?

16. Zarkon Group Title

What is your force field $$F$$?

17. Zarkon Group Title

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

18. Zarkon Group Title

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

19. theEric Group Title

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

20. theEric Group Title

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

21. theEric Group Title

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 Group Title

np

23. theEric Group Title

Take care!