## Idealist Group Title Evaluate the line integral 3x ds, where C is the quarter-circle x^2+y^2=4 from (2, 0) to (0, 2). one year ago one year ago

1. SithsAndGiggles Group Title

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 Group Title

You're right.

3. SithsAndGiggles Group Title

Are you absolutely sure? Because it could also be $$\frac{dx}{dt}, \text{ or }\frac{dy}{dx}.$$

4. SithsAndGiggles Group Title

^^ instead of $$\frac{dy}{dt},$$ I mean.

5. Idealist Group Title

But where did you get the x(t)=2/sin(t) and y(t)=2/cos(t)?

6. SithsAndGiggles Group Title

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

7. Idealist Group Title

What's the next step I should do?

8. SithsAndGiggles Group Title

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 Group Title

But how to find the ds?

10. SithsAndGiggles Group Title

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 Group Title

Thanks.