Estimate the surface integral over a hemisphere of a scalar function, given four points of the function.

Let H be the part of the surface $x^2+y^2+z^2=50$ defined for $z ≥ 0$, and suppose $f$ is a continuous function with $f (3, 4, 5) = 7$, $f(3, −4, 5) = 8$, $f(−3, 4, 5) = 9$, and $f(−3, −4, 5) = 12$.

By dividing H into four patches, estimate the value of $\iint_H f(x,y,z)dS$.

I tried to solve this problem given, that for a scalar function $f(x,y,z)$,

$$\iint_H f(x,y,z)dS=\iint_Df(x,y,z) \lvert r_u\times r_v\rvert dudv$$ where $r(u,v)$ is a parametrisation of the surface.

I used spherical coordinates to parameterise the surface by $r(\theta,\phi)=\sqrt{50}\cos(\theta)\sin(\phi)i+\sqrt{50}\sin(\theta)\sin(\phi)j+\sqrt{50}\cos(\phi)k$ for $0≤\theta≤2\pi$ and $0≤\phi≤\pi/2$.

However I am not sure how to use the given points of $f$ to estimate this integral.

Why parameterize?! You don't have $f$ at all, so why even think about computation. The given sample points are uniformly distributed over the sphere. So divide the sphere into 4 equal patches each containing one of the sample points. By the very definition of integral, a reasonable approximation to $\int_Sf$ will be $$f(point_1)*Area_1 + f(point_2)*Area_2+\cdots +f(point_4)*Area_4 \ .$$ Since we have a nice sphere, each area is $1/4$ the surface area of sphere of radius $\sqrt{50}$. Plug things in.