Evaluate the surface integral $\displaystyle\int_{\mathcal S}xz\,\mathrm dS$ where $\mathcal S$ is the boundary of the region enclosed by the cylinder $y^2+z^2=25$ and the planes $x=0$ and $x+y=10$.
I arrive at a solution of $0$, but only after going through the trouble of splitting up the integration region into subsets (the part of $\mathcal S$ in the plane $x=0$, in the plane $x+y=10$, and the cylindrical portion), then integrating along each sub-region and seeing that each integral vanishes. Is there a faster way to arrive at this just by "eyeballing" the integrand and the domain of integration?
Or is this answer incorrect? My work is shown below.
Denote the sub-regions of $\mathcal S$ as listed above by $\mathcal S_i$, $i\in\{1,2,3\}$, respectively. Then parameterize each region by $$\mathbf s_i(u,v)= \begin{cases} \langle0,u\cos v,u\sin v\rangle&\text{for }i=1\\ \langle10-u\cos v,u\cos v,u\sin v\rangle&\text{for }i=2\\ \langle u,5\cos v,5\sin v\rangle&\text{for }i=3 \end{cases}$$ with $0\le u\le5,0\le v\le2\pi$ when $i\in\{1,2\}$, and $0\le u\le10-5\cos v,0\le v\le2\pi$ when $i=3$.
Then $$\mathrm dS=\begin{cases}u\,\mathrm du\,\mathrm dv&\text{for }i=1\\\sqrt 2u\,\mathrm du\,\mathrm dv&\text{for }i=2\\5\,\mathrm du\,\mathrm dv&\text{for }i=3\end{cases}$$ The integrals are then $$\begin{align*} \iint_{\mathcal S_1}xz\,\mathrm dS&=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=5}0u^2\sin v\,\mathrm du\,\mathrm dv\\[1ex] \iint_{\mathcal S_2}xz\,\mathrm dS&=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=5}(10-u\cos v)u\sin v(\sqrt2u)\,\mathrm du\,\mathrm dv\\[1ex] \iint_{\mathcal S_3}xz\,\mathrm dS&=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=10-5\cos v}5\,\mathrm du\,\mathrm dv \end{align*}$$ The first integral clearly vanishes since $x=0$. The other two take a bit more effort but also apparently vanish, courtesy of the periodic functions.