If curve $C$ is the curve on both cylinder $x^2+y^2=1$ and plane $z=ax+by-1$ with $a,b\neq0$, calculate $\int_Cf\,ds$ for $f(x,y,z)=x^2+y^2+z$.
I used this parametrization:
- $x=\cos t$
- $y=\sin t$
- $z=a \cos t+ b\sin t-1$
So it becomes:
$\vec{r}=(\cos t,\,\sin t,\,a\cos t+b\sin t-1)$
$d\vec{r}=(-\sin t,\,\cos t,\,-a\sin t+b\cos t)\,dt$
$ds=\sqrt{\sin^2t+\cos^2t+(b\cos t-a\sin t)^2}\,dt=\sqrt{1+(b\cos t-a\sin t)^2}\,dt$
So the integral becomes: $$\int_0^{2\pi}(a\cos t+b\sin t)\sqrt{1+(b\cos t-a\sin t)^2}\,dt$$ With variable change $u=b\cos t-a\sin t$, it becomes: $$-\int\sqrt{1+u^2}\,du$$ $$=-\dfrac{(1+u^2)u+\ln(u+(1+u^2))}{2}$$
But actually, for $t=0$, $u=b$ and for $t=2\pi$, $u=b$. Hence the integral must be zero. Am I right? If it's true that answer is zero, is there a better way (e.g. by symmetry) that we can understand that the answer is zero from the beginning?