# Path Integral problems

• May 5th 2009, 10:10 PM
monster
Path Integral problems
Have a problem that i have had a go at but am completely unsure if am anywhere near on the right track any help would be appreciated.

Evaluate the path integral;

$
\int_{C} f . ds

$

Where f = xy;

And the path C is the part of the circle centred at the origin starting at (0,2) going anticlockwise ending at (-2,0).

I did that x=2cost, y=2sint;
so that c(t) = (2cost , 2sint),
where 0< t < pi/2

then did;

$

\int f.ds = \int c(t).[c'(t)].dt

$

but am not sure i have set this up properly am a bit lost here.
• May 5th 2009, 10:41 PM
redsoxfan325
Quote:

Originally Posted by monster
Have a problem that i have had a go at but am completely unsure if am anywhere near on the right track any help would be appreciated.

Evaluate the path integral;

$
\int_{C} f . ds

$

Where f = xy;

And the path C is the part of the circle centred at the origin starting at (0,2) going counterclockwise ending at (-2,0).

I did that x=2cost, y=2sint;
so that c(t) = (2cost , 2sint),
where 0< t < pi/2

then did;

$

\int f.ds = \int c(t).[c'(t)].dt

$

but am not sure i have set this up properly am a bit lost here.

Well, one thing that's a bit unclear is that you have $f(x,y)=xy$, yet you are only parameterizing two variables.

I think you want $c(t)=(2\cos(t),2\sin(t),0), \frac{\pi}{2}\leq t\leq \pi$
• May 5th 2009, 11:02 PM
TwistedOne151
Actually, I think the first problem is in the limits for t. As you go from (0,2) to (-2,0), that is, with $c(t)=(2\cos{t},2\sin{t})$, then $\frac{\pi}{2}\le{t}\le\pi$.
Next $x(t)=2\cos{t}$, and $y(t)=2\sin{t}$, so $f(c(t))=xy=4\sin{t}\cos{t}$.
And since this is the line integral of a scalar function, ds is the arc length parameter:
$ds^2=dx^2+dy^2$, so in terms of t, $ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{ dy}{dt}\right)^2}\,dt=\sqrt{(-2\sin{t})^2+(2\cos{t})^2}\,dt$
$ds=\sqrt{4\sin^2{t}+4\cos^2{t}}\,dt=2\,dt$
and you have
$\int_{\pi/2}^{\pi}(4\sin{t}\cos{t})2dt=8\int_{\pi/2}^{\pi}\sin{t}\cos{t}\,dt$.

--Kevin C.
• May 5th 2009, 11:03 PM
redsoxfan325
Quote:

Originally Posted by redsoxfan325
Well, one thing that's a bit unclear is that you have $f(x,y)=xy$, yet you are only parameterizing two variables.

I think you want $c(t)=(2\cos(t),2\sin(t),0), \frac{\pi}{2}\leq t\leq \pi$

$ds = \sqrt{[dx/dt]^2+[dy/dt]^2+[dz/dt]^2}\,dt = 2\,dt$

$8\int_{\pi/2}^{\pi}\sin(t)\cos(t)\,dt = 4\int_{\pi/2}^{\pi}\sin(2t)\,dt = -2\cos(2t)|_{\pi/2}^{\pi} = -2-2 = \boxed{-4}$
• May 5th 2009, 11:08 PM
TwistedOne151
Where are you getting the z term, redsoxfan325? This is a two-dimensional problem (in the xy plane), so only x and y are involved; there's no z.

--Kevin C.
• May 5th 2009, 11:10 PM
redsoxfan325
He/she said $f=xy$, but it doesn't change the answer either way because $z=0$.
• May 6th 2009, 01:27 AM
monster
Quote:

Originally Posted by TwistedOne151
Actually, I think the first problem is in the limits for t. As you go from (0,2) to (-2,0), that is, with $c(t)=(2\cos{t},2\sin{t})$, then $\frac{\pi}{2}\le{t}\le\pi$.
Next $x(t)=2\cos{t}$, and $y(t)=2\sin{t}$, so $f(c(t))=xy=4\sin{t}\cos{t}$.
And since this is the line integral of a scalar function, ds is the arc length parameter:
$ds^2=dx^2+dy^2$, so in terms of t, $ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{ dy}{dt}\right)^2}\,dt=\sqrt{(-2\sin{t})^2+(2\cos{t})^2}\,dt$
$ds=\sqrt{4\sin^2{t}+4\cos^2{t}}\,dt=2\,dt$
and you have
$\int_{\pi/2}^{\pi}(4\sin{t}\cos{t})2dt=8\int_{\pi/2}^{\pi}\sin{t}\cos{t}\,dt$.

--Kevin C.

I went;

c(t) = (2cos(t) , 2sin(t))
so c'(t) = ( - 2sin(t), 2cos(t))

so [c'(t)] = 2
so;
$

\int_C f.ds = \int_{\pi/2}^{\pi} (8cos(t)sin(t)).dt
$

then made sub u = cost to give
$

\int_{0}^{-1} -8u .du
$

= -4

Is the way i have gone about it correct?
and thank you for all help guys.
• May 6th 2009, 10:56 AM
redsoxfan325
Yeah, you did it correctly.