Confusion over a Polar Coordinates Problem

The question given is the following:

Let $\Omega$ be the domain bounded by the paraboloid z = 4 - ($x^2$ + $y^2$) and the plane $z = 0$.

Let $f$ be the scalar field $f(x,y,z) = ax + by + cz$, where a, b and c are constants. Find $∫_{\Omega}$ $f$ $dV$.

I know that the answer to this is 32$\pi$$c$/3, I calculated it myself.

The lecturer made an aside comment for an exercise; that being

"Without doing the integral, how does the answer depend on a, b and c"

I'm very confused about this, and have no idea, any help?

The volume is symmetric about the z axis so the $ax$ and $by$ terms in the field, which will be positive on one side and negative on the other cancel themselves out. So they can be nulled. So the problem is the same as $f(x,y,z)=cz$.

Because the field only depends on z, the field is constant for a particular value of z. Also, cross sections of the paraboloid at a particular z value are circles. We could integrate the area of the circles from z=0 to z=4 multiplied by the field to get the answer. But we are avoiding integration so let's just scale the area by the field and see what happens.

First let's get an expression for the cross section area from the original paraboloid equation.

$z = 4 - (x^{2} + y^{2})$

$4-z = (x^{2} + y^{2})$

$\pi(4-z) = \pi(x^{2} + y^{2}) = \pi r^{2}$ where $r^{2} = x^{2} + y^{2}$

Okay so now that we have the expression for the area of the cross sections, lets multiply the area by the field.

$\pi(4-z) * f(x,y,z) = \pi(x^{2} + y^{2})$

Substitute in the new expression for the field.

$\pi(4-z) * cz = \pi(x^{2} + y^{2})$

So now it looks like we can have an $x^{2}$, $y^{2}$, and $z^{2}$ terms so this equation is looking like that of a sphere or an ellipsoid. Lets do some manipulation.

$(4-z) * cz = (x^{2} + y^{2})$

$c(4z-z^{2}) = (x^{2} + y^{2})$

$-c(z^{2}-4z+4-4) = (x^{2} + y^{2})$

$-c(z^{2}-4z+4)+4c = (x^{2} + y^{2})$

$-c(z-2)^{2}+4c = (x^{2} + y^{2})$

$4c = x^{2} + y^{2}+c(z-2)^{2}$

$1 = \frac{x^{2}}{(2\sqrt{c})^{2}} + \frac{y^{2}}{(2\sqrt{c})^{2}}+\frac{(z-2)^{2}}{(2)^{2}}$

This is now in the form of an ellipsoid. The standard form for an ellipsoid is:

$1 = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}$

The formula for the volume of an ellipsoid is

$V = \tfrac{4}{3}\pi abc$

So the volume of our ellipsoid is

$V = \tfrac{4}{3}\pi (2\sqrt{c})*(2\sqrt{c})*(2) = 32\pi c/3$

So this is the final result

$∫_{\Omega}$ $f$ $dV = 32\pi c/3$

I admit I would not have found this route of solving without integration if your question did not say it was possible. I think this problem is interesting and it makes me wonder if a field can be thought of as a fourth dimension.