Is there a reason not to use \(u=\sin(t)\)? Otherwise there is a negative sine involved.What happens if you use u = cos(t)
If \(0\leq t\leq\frac{\pi}{2}\) then the given interval for the circle does not work out.What if t goes from 0 to pi/2?
Since 3xy is negative at every point along the given curve, I would say "yes, it does!"I've been working on a line integral, and I got a negative answer. Does this make sense?
That's a perfectly good answer.Here is the question and my work.
View attachment 18768
I got the same answer using \(u=\cos(t)\).What happens if you use u = cos(t)?
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 \(\displaystyle dy= 2 cos(t)dt\).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
So instead of putting the arc length in, I should put in the derivative of y?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
\(\displaystyle -24\int_0^{-1} u^2 du= 24\int_{-1}^0 u^2du= \left[8 u^3\right]_{-1}^0= 8\).