Motivation behind Surface Integral Substitution

I'm sure I'm missing something basic here, but I'm not quite sure what. Here's my problem:

$$\iint_{S}z^2 dS$$

where $S$ is the octant of the sphere of radius 1 centered at the origin. Here's an image:

Of course, the first thing I instinctively did was replace $z^2$ with:

$$\iint_{S} 1-x^2-y^2 dS$$

But ... why am I allowed to do this? Was that $z^2$ not the field I'm integrating over the region $x^2+y^2+z^2=1$? In other words, aren't those two completely different $z$s? Why am I allowed to substitute the region into the field? Hopefully my question makes sense. I appreciate any responses.

This question is from Div, Grad, Curl, and All That Page 26.

Solution 1:

Reason for change of variable: You changed the variable in the integrand to something which exists in the domain of integration. Here you are integrating in the x-y plane as domain.

But why did that work? Since our domain is given as the quarter of hemisphere, our integral simplifies under the domain. It sort of similar to how we can write:

$$ \int_{-1}^0 |x| dx = - \int_{-1}^0 x dx$$

The modulus came out and turned into a minus sign due to how our domain was.

Here we run the integral over the surface of the quarter hemisphere, at each point on that surface, we square it's height about the plane and multiply with dS, repeat that procedure for all the points and add em up. We know how to express the height squared in terms of square of coordinates below it. Hence, we turn the epxression into that of $(x, y)$ but that also allows us to do the integral since integral's domain is in the plane.