Evaluating line integral in first octant. I need to calculate the integral $$\int_{(0,1,0)}^{(1,0,2)} \frac{z}{y}\mathrm{d}x + (x^2+y^2+z^2)\mathrm{d}z$$ along the curve in the first octant given by $x^2+y^2=1, z = 2x$. I used the following parametrization:
$$
\left\{\begin{aligned}
&x=\sin t\\
&y=\cos t\\
& z=2\sin t\end{aligned}\right.,\quad 0 \leq t \leq \frac{\pi}{2}
$$
However, this does not give me the right answer, as this gives the answer $\frac{20}{3}$ and the correct answer is $2+\frac{3\pi}{2}$. I think I am making a mistake in my parametrization.
 A: Your answer is correct. 
\begin{align}&\int_{(0,1,0)}^{(1,0,2)} \frac{z}{y}\mathrm{d}x + (x^2+y^2+z^2)\mathrm{d}z\\
&=\int_0^\frac{\pi}2 2\tan t \cos t+(1+4\sin^2t)2 \cos t \, dt\\
&=\int_0^\frac{\pi}2 2\sin t + 2\cos t(1+2(2\sin^2 t)) \, dt\\
&=\int_0^\frac{\pi}2 2\sin t + 2\cos t(1+2(1-\cos 2t))\, dt\\
&=\int_0^\frac{\pi}2 2\sin t + 6 \cos t -4\cos t\cos 2t \, dt \\
&= \int_0^\frac{\pi}2 2\sin t + 6 \cos t -2\left( \cos 3t+\cos t\right) \, dt \\
&= 8 -2\left( -\frac13+1\right)\\
&=8-\frac{4}3\\
&=\frac{20}3\end{align}
Now, let me try to guess where is the error made in the book, suppose the author forget to compute $\frac{dz}{dt}$, then his working looks as follows:
\begin{align}&\int_{(0,1,0)}^{(1,0,2)} \frac{z}{y}\mathrm{d}x + (x^2+y^2+z^2)\mathrm{d}z\\
&\color{red}=\int_0^\frac{\pi}2 2\tan t \cos t+(1+4\sin^2t) \, dt\\
&=\int_0^\frac{\pi}2 2\sin t + (1+2(2\sin^2 t)) \, dt\\
&=\int_0^\frac{\pi}2 2\sin t + (1+2(1-\cos 2t))\, dt\\
&=\int_0^\frac{\pi}2 2\sin t + 3 -2\cos 2t \, dt \\
&=2+\frac{3\pi}2\end{align}
Note the I purposely made the mistake at the highlighted equation for the second block of computations.
