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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).

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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\[\frac{dy(t)}{dt}=cos(t)\]

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Other answers:

\[\text {So then the parameter, }t\text{, goes from }\pi \text{ to }0 \text{.}\] \[\text{Is it okay to write:} \int_\pi ^0 F(t)dt\text{, or must it be }-\int_0 ^ \pi F(t)dt \text{?}\] Don't they equal the same thing?
those two integrals are the same
you might want to look at your parametric equations again
\[\text{Thanks.. I've considered rewriting them to go from }0\text{ to }\pi \text{. In which case:}\] \[x(t)=-cos(t)\]\[y(t)=sin(t)\] Is this what you mean? Or were my original equations incorrect?
neither are correct
"circle with radius 3"
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...
\[x(t)=3cos(t)\]\[y(t)=3sin(t)\]\[\text{where }t \text{ goes from }0\text{ to }\pi \text{.}\] or\[x(t)=-3cos(t)\]\[y(t)=3sin(t)\]\[\text{where }t\text{ goes from }0 \text{ to } \pi \text{.}\]
The first one is from \[\pi\] to 0, I mean.
Those work, then?
And so they're the same?
What is your force field \(F\)?
\[W=\int\limits_{a}^{b}F(c(t))\cdot c'(t) dt\]
\(c(t)\) is the particular path
To continue the problem I'm given,\[F(x,y)=(2y+x^2)i + (x^2 - 2x)j\]
yielding... \[W=-\int _0 ^\pi ((6sin(t)+9cos(t))(-3sint)+(9cos^2(t)-6cos(t))(3cos(t)))dt\]
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!
Take care!

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