I know the answer but cannot for the love of everything figure out how the book got the answer and the solution is only the answer nothing step-by step at all. $$ F(D) = F*dr $$calculate curve integral using the parametrisation where F is $$ F(x,y,z) = yz, xz, xy $$ over domain D, where D is $$ (x,y,z) = cos(t), sin(t), t $$ where t is between (0, pi/4)
So im supposed to use the parametrization that is already there
I get this:
$$ F(x,y,z) =tsin(t), tcos(t), sin(t)cos(t) $$ $$ r'(t) =-sin(t), cos(t), 1 $$
I know im supposed to multiply F with r'(t) but I can't simplify in any way that get the answer pi/8 that should be the final answer for the integral $$ \int_0^{\pi/4}-tsin^2(t) + tcos^2(t) + cos(t)sin(t)$$
Any idea how to get from there to pi/8?
You are given the vector field ${\bf F}(x,y,z):=(yz,xz,xy)$ and the path $$\gamma:\quad {\bf r}(t):=(\cos t,\sin t, t)\qquad\left(0\leq t\leq{\pi\over4}\right)\ .$$ The line integral of thie field ${\bf F}$ along $\gamma$ is given by $$\int_\gamma{\bf F}\cdot d{\bf r}=\int_0^{\pi\over4}{\bf F}\bigl({\bf r}(t)\bigr)\cdot{\bf r}'(t)\>dt=\int_0^{\pi\over4}\bigl(-t\sin^2 t+t\cos^2 t+\sin t\cos t)\>dt={\pi\over8}\ .$$ For doing the integral you can use $t(\cos^2 t-\sin^2 t)=t\cos(2t)$ and proceed by partial integration; furthermore $\sin t\cos t={1\over2}\sin(2t)$.