Line integrals

[burt]

I've been working on a line integral, and I got a negative answer. Does this make sense?
Here is the question and my work.

 

[Steven G]

What happens if you use u = cos(t)?

What if t goes from 0 to pi/2? Try it with u=sin(x) AND u = cos(x).

What results are you getting from these 4 methods?

 

[burt]

What happens if you use u = cos(t)

Is there a reason not to use \(u=\sin(t)\)? Otherwise there is a negative sine involved.

 

[burt]

What if t goes from 0 to pi/2?

If \(0\leq t\leq\frac{\pi}{2}\) then the given interval for the circle does not work out.

 

[HallsofIvy]

I've been working on a line integral, and I got a negative answer. Does this make sense?

Since 3xy is negative at every point along the given curve, I would say "yes, it does!"

Here is the question and my work.
View attachment 18768

That's a perfectly good answer.

 

[burt]

What happens if you use u = cos(t)?

I got the same answer using \(u=\cos(t)\).

 

[HallsofIvy]

I've been working on a line integral, and I got a negative answer. Does this make sense?
Here is the question and my work.
View attachment 18768

Looking at this more closely, although this specifically says "line integral", so that we would normally expect "ds", the differential of arc length, here it says "dy"! If that is not a typo then instead of \(\displaystyle ds= \sqrt{4(sin^2(t)+ cos^2(t))}dt\) you should have \(\displaystyle dy= 2 cos(t)dt\).

If that is correct then the integral should be
\(\displaystyle \int_{\frac{\pi}{2}}^\pi 3(2cos(t))(2 sin(t))(2 cos(t)dt)= \int_{\frac{\pi}{2}}^\pi 24 cos^2(t)sin(t)dt\)
To integrate that let \(\displaystyle u= cos(t)\) so that \(\displaystyle du= -sin(t)dt\) and the integral becomes
\(\displaystyle -24\int_0^{-1} u^2 du= 24\int_{-1}^0 u^2du= \left[8 u^3\right]_{-1}^0= 8\).

 

[burt]

Looking at this more closely, although this specifically says "line integral", so that we would normally expect "ds", the differential of arc length, here it says "dy"! If that is not a typo then instead of \(\displaystyle ds= \sqrt{4(sin^2(t)+ cos^2(t))}dt\) you should have \(\displaystyle dy= 2 cos(t)dt\).

If that is correct then the integral should be
\(\displaystyle \int_{\frac{\pi}{2}}^\pi 3(2cos(t))(2 sin(t))(2 cos(t)dt)= \int_{\frac{\pi}{2}}^\pi 24 cos^2(t)sin(t)dt\)
To integrate that let \(\displaystyle u= cos(t)\) so that \(\displaystyle du= -sin(t)dt\) and the integral becomes
\(\displaystyle -24\int_0^{-1} u^2 du= 24\int_{-1}^0 u^2du= \left[8 u^3\right]_{-1}^0= 8\).

So instead of putting the arc length in, I should put in the derivative of y?
Does that mean this is not really a line integral?