If you project S onto the (x,y)-plane, it casts a "shadow" corresponding to the trapezoidal region
T = {(x,y) : 0 ≤ x ≤ 10 - y and -4 ≤ y ≤ 4}
Let z = f(x, y) = √(16 - y²) and z = g(x, y) = -√(16 - y²), each referring to one half of the cylinder to either side of the plane z = 0.
The surface element for the "positive" half is
dS = √(1 + (∂f/∂x)² + (∂f/dy)²) dx dy
dS = √(1 + 0 + 4y²/(16 - y²)) dx dy
dS = √((16 + 3y²)/(16 - y²)) dx dy
The the surface integral along this half is
![\displaystyle \iint_T xz \,dS = \int_{-4}^4 \int_0^{10-y} x \sqrt{16-y^2} \sqrt{\frac{16+3y^2}{16-y^2}} \, dx \, dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciint_T%20xz%20%5C%2CdS%20%3D%20%5Cint_%7B-4%7D%5E4%20%5Cint_0%5E%7B10-y%7D%20x%20%5Csqrt%7B16-y%5E2%7D%20%5Csqrt%7B%5Cfrac%7B16%2B3y%5E2%7D%7B16-y%5E2%7D%7D%20%5C%2C%20dx%20%5C%2C%20dy)
![\displaystyle \iint_T xz \,dS = \int_{-4}^4 \int_0^{10-y} x \sqrt{16+3y^2}\, dx \, dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciint_T%20xz%20%5C%2CdS%20%3D%20%5Cint_%7B-4%7D%5E4%20%5Cint_0%5E%7B10-y%7D%20x%20%5Csqrt%7B16%2B3y%5E2%7D%5C%2C%20dx%20%5C%2C%20dy)
![\displaystyle \iint_T xz \,dS = \frac12 \int_{-4}^4 (10-y)^2 \sqrt{16+3y^2} \, dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciint_T%20xz%20%5C%2CdS%20%3D%20%5Cfrac12%20%5Cint_%7B-4%7D%5E4%20%2810-y%29%5E2%20%5Csqrt%7B16%2B3y%5E2%7D%20%5C%2C%20dy)
![\displaystyle \iint_T xz \,dS = 416\pi](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciint_T%20xz%20%5C%2CdS%20%3D%20416%5Cpi)
You'll find that the integral over the "negative" half has the same value, but multiplied by -1. Then the overall surface integral is 0.