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Evaluate the line integral 3x ds, where C is the quartercircle x^2+y^2=4 from (2, 0) to (0, 2).
 one year ago
 one year ago
Evaluate the line integral 3x ds, where C is the quartercircle x^2+y^2=4 from (2, 0) to (0, 2).
 one year ago
 one year ago

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SithsAndGigglesBest ResponseYou've already chosen the best response.1
Parameterize \(C\) by \(x(t)=2\cos t,\;y(t)=2\sin t, \text{ with }0\le t\le\frac{\pi}{2}.\) I'm not sure what the "ds" means, though. Is that supposed to mean \(\sqrt{1+\left(\frac{dy}{dt}\right)^2}\), or something like that?
 one year ago

SithsAndGigglesBest ResponseYou've already chosen the best response.1
Are you absolutely sure? Because it could also be \(\frac{dx}{dt}, \text{ or }\frac{dy}{dx}.\)
 one year ago

SithsAndGigglesBest ResponseYou've already chosen the best response.1
^^ instead of \(\frac{dy}{dt},\) I mean.
 one year ago

IdealistBest ResponseYou've already chosen the best response.0
But where did you get the x(t)=2/sin(t) and y(t)=2/cos(t)?
 one year ago

SithsAndGigglesBest ResponseYou've already chosen the best response.1
Those are the parametric equations of a circle with radius 2 (centered about the origin). They describe the path C.
 one year ago

IdealistBest ResponseYou've already chosen the best response.0
What's the next step I should do?
 one year ago

SithsAndGigglesBest ResponseYou've already chosen the best response.1
Set up your integral. Depending on what dS is supposed to be, I'm pretty sure it'd be \[\large\int_C3x\;dS=\int_0^{\frac{\pi}{2}}3\cos t\sqrt{\cdots}\;dt\]
 one year ago

IdealistBest ResponseYou've already chosen the best response.0
But how to find the ds?
 one year ago

SithsAndGigglesBest ResponseYou've already chosen the best response.1
Ah, according to this link (http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtI.aspx), \[dS=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\;dt\] In my earlier posts, I was stuck in an arclengthmindset. So, the integral is \[\large\int_0^{\frac{\pi}{2}}3\cos t\sqrt{\left(2\sin t\right)^2+\left(2\cos t\right)^2}\;dt\]
 one year ago
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