# Curve for a line integral - direction confusion

• laser
laser
Homework Statement
description
Relevant Equations
description

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##.

docnet
You are integrating counter-clockwise. Not clockwise as specified

Orodruin said:
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?

laser said:
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).##

laser said:
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##

docnet
docnet said:
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)##!!

docnet
Orodruin said:
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.

laser said:
##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
laser said:
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.

laser said:
##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 $$\begin{split} \int_0^{\pi} f(x(s),y(s))\,ds &= \int_{\pi/2}^0 f(x(s(t)),y(s(t))) \frac{ds}{dt}\,dt \\ &= - \int_0^{\pi/2} f(x(s(t)),y(s(t))) (-2)\,dt \\ &= 2\int_0^{\pi/2} f(x(s(t)),y(s(t)))\,dt.\end{split}$$

laser

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