Evaluate the surface integral $$\int\int_S(z+x^2y)ds$$ where $S$ is the part of the cylinder $y^2 + z^2=1$ that lies between the planes $x=0$ and $x=3$ in the first octant $x,y,x\ge 0$
My current workings:
$\mathbf r(x,\theta) = x\mathbf i + \sqrt1 sin\theta \mathbf j + \sqrt1 cos\theta \mathbf k$
$\mathbf r_x(x,\theta) = \mathbf i$
$\mathbf r_\theta (x, \theta) = \sqrt1 sin\theta \mathbf j + \sqrt1 cos\theta \mathbf k$
Cross product
$$=\begin{vmatrix}i & j & k \\ 1 & 0 & 0 \\ 0 & -\sqrt1sin\theta & \sqrt1cos\theta \end{vmatrix}$$
$$=\sqrt1 cos\theta \mathbf i-\sqrt1 sin\theta \mathbf k$$
$$||\mathbf r_x \mathbf r_\theta||=\sqrt{1}$$
An idea:
The surface can be written as $\;z^2=1-y^2\implies z=\sqrt{1-y^2}\;$ , since we're taking only part of the upper part of the cylinder. Thus, projecting on the $\;xy\,-\,$ plane this part of cylinder we have, and calling $\;R\;$ its projection on this plane (this is in fact a $\;3\times1\;$ rectangle...), we get
$$z=g(x,y)=\sqrt{1-y^2}\implies g'_x=0\;,\;\;g'_y=-\frac y{\sqrt{1-y^2}}\implies$$
$$=\iint_S(z+x^2y)dA=\iint_R\left(\sqrt{1-y^2}+x^2y\right)\sqrt{1+(g'_x)^2+(g'_y)^2}\,dA=$$
$$=\iint_R\left(\sqrt{1-y^2}+x^2y\right)\sqrt{1+\frac{y^2}{1-y^2}}dA=\iint_R\left(1+x^2\frac y{\sqrt{1-y^2}}\right)dA=$$
$$\int_0^3\int_0^1\left(1+x^2\frac y{\sqrt{1-y^2}}\right)dy\,dx=\int_0^31\,dx+\int_0^3\left.\frac{x^2}{-2}2\sqrt{1-y^2}\right|_0^1\,dx=$$
$$=3-\int_0^3 x^2\,dx=3-\frac{27}3=-6$$