Curve for a line integral - direction confusion

[laser]

Homework Statement

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Relevant Equations

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When I take $x = 2\cos(t)$ and $y = 2\sin(t)$, the integral becomes $\int_{t=\frac{\pi}{2}}^0 4(2\cos(t))^2 \cdot 2 dt = -8\pi$. The final answer is $8\pi$. Why is my method wrong?

I played around with desmos and the parameterisation seems correct: https://www.desmos.com/calculator/fgid1zbbir (starting at $\frac{\pi}{2}$ and ending at $0$.

Question/Answer source: https://tutorial.math.lamar.edu/Solutions/CalcIII/LineIntegralsPtI/Prob9.aspx

 

[Orodruin]

You are integrating counter-clockwise. Not clockwise as specified

 

[laser]

You are integrating counter-clockwise. Not clockwise as specified

But if $t$ starts at $\frac{\pi}{2}$ and ends at $0$ it is clockwise, no?

 

[docnet]

But if $t$ starts at $\frac{\pi}{2}$ and ends at $0$ it is clockwise, no?

It's counter-clockwise.

$(2\sin(0),2\cos(0))=(0,2)$ and $(2\sin(\frac{\pi}{2}),2\cos(\frac{\pi}{2}))=(2,0).$

 

[Orodruin]

But if $t$ starts at $\frac{\pi}{2}$ and ends at $0$ it is clockwise, no?

Yes you are right. I read a bit fast. However, the integral is wrt ds, not $d\theta$

 

[laser]

It's counter-clockwise.

$(2\sin(0),2\cos(0))=(0,2)$ and $(2\sin(\frac{\pi}{2}),2\cos(\frac{\pi}{2}))=(2,0).$

Those are not my $x$ and $y$! I chose $x=2\cos(x)$, NOT $x=2\sin(x)$!!

 

[laser]

Yes you are right. I read a bit fast. However, the integral is wrt ds, not $d\theta$

$ds$ is always positive though - I don't know why my answer is incorrect. i.e. $ds=2dt$ and my limits for $t$ are as shown.

 

[Orodruin]

$ds$ is always positive though - I don't know why my answer is incorrect. i.e. $ds=2dt$ and my limits for $t$ are as shown.

Because if you integrate from pi/2 to 0, dt is negative

 

[docnet]

Those are not my $x$ and $y$! I chose $x=2\cos(x)$, NOT $x=2\sin(x)$!!

oh sorry, I'm very tired. I'm going to make myself log off right now.

 

[pasmith]

$ds$ is always positive though - I don't know why my answer is incorrect. i.e. $ds=2dt$ and my limits for $t$ are as shown.

But if you take (x,y) = (2\cos t, 2 \sin t) then the curve starts with s = 0 at t = \frac \pi 2 and ends with s = 2(\pi/2) = \pi when t = 0. The parametrization in terms of arclength is then not s = 2t but s = \pi - 2t so that ds/dt = -2. Hence \begin{split}<br /> \int_0^{\pi} f(x(s),y(s))\,ds &amp;= \int_{\pi/2}^0 f(x(s(t)),y(s(t))) \frac{ds}{dt}\,dt \\<br /> &amp;= - \int_0^{\pi/2} f(x(s(t)),y(s(t))) (-2)\,dt \\<br /> &amp;= 2\int_0^{\pi/2} f(x(s(t)),y(s(t)))\,dt.\end{split}