Evaluating the integral of $f(x, y)=yx$ Evaluate the integral $I = \int_C f(x,y) ds$ where $f(x,y)=yx$. and the curve $C$ is given by $x=\sin(t)$ and $y=\cos(t)$ for $0\leq t\leq \frac{pi}{2}$. I got the answer for this as $\frac{\sqrt{2}}{2}$ is that right?

The line integral of the curve $f(x,y)$ is given by $\int_a^b{f(x,y)ds}$ where $ds=\sqrt{{(\frac{dx}{dt})}^2+{(\frac{dy}{dt})}^2}dt$. In this case $f(x,y)=xy=sin(t)cos(t)$ & $ds=1dt$ so $$\int_0^{\pi/2}{sin(t)cos(t)dt}=\frac12$$.
