Prove that the volume of the solid bounded

Prove that the volume of the solid bounded by the surfaces: $$ z=0,\qquad x^2+y^2=c^2,\qquad z\big[\varphi(x)+\varphi(y)\big]=a\varphi(x)+b\varphi(y), $$ where $\varphi$ is an arbitrary positive integrable function and $a,b>0$, is: $$ V=\frac12\pi c^2(a+b), $$ independent of the particular choice of $\varphi$.

Explanation of Method

The key is not to evaluate the integral directly (which would require knowing $\varphi$) but to exploit the symmetry of the disk under $x\leftrightarrow y$. Splitting the height function into two pieces, one weighted by $\varphi(x)$ and the other by $\varphi(y)$, the $x\leftrightarrow y$ swap shows the two pieces integrate to the same value, and their sum is simply the area of the disk.

Solve for $z$ and split the volume integral into two pieces.
Show, via the symmetry $(x,y)\to(y,x)$, that the two pieces are equal.
Compute their sum directly (it no longer depends on $\varphi$), then solve for each piece.
Assemble the volume.

Step 1 · Solve for $z$ and split the integral

Since $\varphi>0$, $\varphi(x)+\varphi(y)>0$ everywhere, so $z=\dfrac{a\varphi(x)+b\varphi(y)}{\varphi(x)+\varphi(y)}$ is well-defined on the disk $D=\{x^2+y^2\leqslant c^2\}$. The volume is:

$$ V=\iint_Dz\,dA = a\underbrace{\iint_D\frac{\varphi(x)}{\varphi(x)+\varphi(y)}\,dA}_{I_1}+b\underbrace{\iint_D\frac{\varphi(y)}{\varphi(x)+\varphi(y)}\,dA}_{I_2}. $$

Step 2 · $I_1=I_2$ by symmetry

The disk $D$ is invariant under the swap $(x,y)\mapsto(y,x)$ (a measure-preserving change of variables, Jacobian $1$). Applying this swap to $I_1$:

$$ I_1=\iint_D\frac{\varphi(x)}{\varphi(x)+\varphi(y)}\,dx\,dy \xrightarrow{x\leftrightarrow y} \iint_D\frac{\varphi(y)}{\varphi(y)+\varphi(x)}\,dy\,dx = I_2. $$

Since the region and the measure are unchanged by this substitution, $I_1=I_2$.

Step 3 · Compute $I_1+I_2$

$$ I_1+I_2 = \iint_D\frac{\varphi(x)+\varphi(y)}{\varphi(x)+\varphi(y)}\,dA = \iint_D1\,dA = \pi c^2, $$

the area of the disk — independent of $\varphi$. Combined with $I_1=I_2$:

$$ I_1=I_2=\frac{\pi c^2}2. $$

Step 4 · Assemble the volume

$$ V = aI_1+bI_2 = a\cdot\frac{\pi c^2}2+b\cdot\frac{\pi c^2}2 = \frac12\pi c^2(a+b). $$

Since this derivation never used a specific form of $\varphi$ — only that $\varphi>0$ so the swap argument and the sum $\varphi(x)+\varphi(y)$ cancelling against itself both go through — the result holds for any positive integrable $\varphi$. $\blacksquare$

Additional Notes

This proof matches the source's argument exactly, and the logic is sound: the swap symmetry of the disk under $x\leftrightarrow y$ is the only property of $D$ used, and positivity of $\varphi$ is exactly what's needed for $z$ to be well-defined.
As a concrete numerical check (not required for the proof, but a useful sanity check), evaluating the double integral directly for a specific test function, $\varphi(t)=t^2+1$, with $c=1.7,\,a=2.1,\,b=0.9$, gives $\approx13.6188$, matching $\frac12\pi c^2(a+b)$ at those parameters exactly — consistent with the formula's independence from the choice of $\varphi$.

Conclusion

For any positive integrable $\varphi$, the volume is:

$$ V = \frac12\pi c^2(a+b). $$