Is it true that $\iint_{x^2+y^2 \le 1} f(ax+by+c) dx dy=2 \int_{-1}^1 \sqrt{1-u^2}f(u\sqrt{a^2+b^2} +c)du $?

Question

Is it true that $\iint_S f(ax+by+c) dx dy=2 \int_{-1}^1 \sqrt{1-u^2}f(u\sqrt{a^2+b^2} +c)du $ ?

where $S:=\{(x,y) \in \mathbb R^2 : x^2+y^2 \le 1\}$

I know I need to do a change of variable , but I don't know what change of variable .

Please help . Thanks in advance

Answer 1

Indeed, it is true. Equating the arguments of $f$ we obtain \begin{align*} ax+by+c&=u\sqrt{a^2+b^2}+c\\ \frac{ax+by}{\sqrt{a^2+b^2}}&=u \end{align*}

This suggests a transformation by rotation according to the comment of @achillehui \begin{align*} u=\frac{ax+by}{\sqrt{a^2+b^2}}\qquad&\qquad x=\frac{au-bv}{\sqrt{a^2+b^2}}\\ \tag{1}\\ v=\frac{-bx+ay}{\sqrt{a^2+b^2}}\qquad&\qquad y=\frac{bu+av}{\sqrt{a^2+b^2}} \end{align*}

In order to transform the integral \begin{align*} \iint_{x^2+y^2\leq 1}f(ax+by+c)\,dx\,dy \end{align*}

we first consider the transformation of the disc $x^2+y^2\leq 1$. Since the transformation is a rotation we expect as result the unit disc in the variables $u$ and $v$.

We obtain using (1) \begin{align*} x^2+y^2&= \frac{(au-bv)^2}{a^2+b^2}+\frac{(bu+av)^2}{a^2+b^2}\\ &=\frac{(a^2u^2-2abuv+b^2v^2)+(b^2u^2+2abuv+a^2v^2)}{a^2+b^2}\\ &=u^2+v^2 \end{align*}

We need the Jacobian determinant $\frac{\partial(x,u)}{\partial(u,v)}$ in order to transform the integral:

\begin{align*} \frac{\partial(x,u)}{\partial(u,v)}&=\begin{vmatrix}x_u&x_v\\y_u&y_v\end{vmatrix} =\begin{vmatrix} \frac{a}{\sqrt{a^2+b^2}}&\frac{-b}{\sqrt{a^2+b^2}}\\ \frac{b}{\sqrt{a^2+b^2}}&\frac{a}{\sqrt{a^2+b^2}}\\ \end{vmatrix}\\ &=\frac{1}{a^2+b^2}\begin{vmatrix}a&-b\\b&a\end{vmatrix}\\ &=1 \end{align*}

We obtain \begin{align*} \iint_{x^2+y^2\leq 1}&f(ax+by+c)\,dx\,dy=\\ &=\iint_{u^2+v^2\leq 1}f(u\sqrt{a^2+b^2}+c)\frac{\partial(x,u)}{\partial(u,v)}\,du\,dv\\ &=\int_{-1}^{1}\int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}}f(u\sqrt{a^2+b^2}+c)\,dv\,du\\ &=\int_{-1}^{1}f(u\sqrt{a^2+b^2}+c)\left(\int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}}1\,dv\right)\,du\\ &=2\int_{-1}^{1}\sqrt{1-u^2}f(u\sqrt{a^2+b^2}+c)\,du\\ \end{align*}

Answer 2

Draw a properly scaled $u$-axis in direction of the vector $(a,b)$ trough the origin, and consider the line $\ell_u$ which intersects this axis orthogonally at $u$. If $-1\leq u\leq 1$ this line intersects the disk $x^2+y^2\leq1$ in a segment of length $2\sqrt{1-u^2}$. As $\ell_u$ is orthogonal to $(a,b)$ the linear function $(x,y)\mapsto ax+by+c$ is constant along $\ell_u$, and this constant value is easily seen to be $\sqrt{a^2+b^2}\,u +c$. It is then geometrically obvious that $$\int_{x^2+y^2\leq1} f(ax+by+c)\>{\rm d}(x,y)=2\int_{-1}^1 f\bigl(\sqrt{a^2+b^2}u +c\bigr)\>\sqrt{1-u^2}\>du\ .$$