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Idealist

  • 2 years ago

Evaluate the line integral 3x ds, where C is the quarter-circle x^2+y^2=4 from (2, 0) to (0, 2).

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  1. SithsAndGiggles
    • 2 years ago
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    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?

  2. Idealist
    • 2 years ago
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    You're right.

  3. SithsAndGiggles
    • 2 years ago
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    Are you absolutely sure? Because it could also be \(\frac{dx}{dt}, \text{ or }\frac{dy}{dx}.\)

  4. SithsAndGiggles
    • 2 years ago
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    ^^ instead of \(\frac{dy}{dt},\) I mean.

  5. Idealist
    • 2 years ago
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    But where did you get the x(t)=2/sin(t) and y(t)=2/cos(t)?

  6. SithsAndGiggles
    • 2 years ago
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    Those are the parametric equations of a circle with radius 2 (centered about the origin). They describe the path C.

  7. Idealist
    • 2 years ago
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    What's the next step I should do?

  8. SithsAndGiggles
    • 2 years ago
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    Set up your integral. Depending on what dS is supposed to be, I'm pretty sure it'd be \[\large\int_C3x\;dS=\int_0^{\frac{\pi}{2}}3\cos t\sqrt{\cdots}\;dt\]

  9. Idealist
    • 2 years ago
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    But how to find the ds?

  10. SithsAndGiggles
    • 2 years ago
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    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 arc-length-mindset. So, the integral is \[\large\int_0^{\frac{\pi}{2}}3\cos t\sqrt{\left(-2\sin t\right)^2+\left(2\cos t\right)^2}\;dt\]

  11. Idealist
    • 2 years ago
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    Thanks.

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