Prove the equality by introducing a suitable change of variable (double integral) Problem from Apostol's Calculus 2:

Prove the following equality by introducing a suitable change of variable:
$$ \iint_S f(ax + by + c)dxdy = 2\int_{-1}^{1}\sqrt{1-u^2}f(u\sqrt{a^2+b^2} +c)du $$
With $$S= \{(x,y) \space | \space x^2+y^2 \le 1\}$$
And $ a^2+b^2 \neq 0$.


Assuming $a^2+b^2 \neq 0$. Let,
$$ax+by=u \sqrt{a^2+b^2}$$
$$bx-ay=v \sqrt{a^2+b^2}$$
Note then the boundary for our region becomes $$u^2+v^2=\frac{(ax+by)^2+(bx-ay)^2}{a^2+b^2} $$
$$=\frac{a^2(x^2+y^2)+b^2(x^2+y^2)}{a^2+b^2}=1$$
The absolute value of our Jacobian for our transformation is,
$$1$$
And we then have,
$$\iint_S f(ax + by + c)dxdy$$
$$=\iint_{S_2} f(u\sqrt{a^2+b^2}+c) dv du$$
Where $S_2=\{(u,v) : u^2+v^2 \leq 1 \}$.
$$=\int_{-1}^{1} \int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}} f(u\sqrt{a^2+b^2}+c) dv du$$
$$=2 \int_{-1}^{1} \sqrt{1-u^2}f(u\sqrt{a^2+b^2}+c) du$$
