Find average distance from line to disk I need to solve the following problem, given in my book. 

Calculus A complete course 9th - Robert A Adams 
14.4    25.
"Find the average distance from points in the quarter disk
   $ x^2 + y^2 \leq a^2 \ , \, x \geq 0 \ , \ y \geq 0, \,$ to the line $x + y = 0$.

I tried drawing an image as shown below

My friend says that the solution can be found by solving 
$$ \large \int_0^a \int_0^\sqrt{a^2-x^2} \frac{x+y}{\sqrt{2}} \,\mathrm{d}x \, \mathrm{d}y $$
But I can not really see why this double integral works, it looks like we are always integrating the distance from $0$ to $a$. But in my eyes the distance changes.
I guess I need to find a line perpendicular to $y = -x$ and find the distance, but could anyone help me out? I've been sitting a few hours with this problem now =(
 A: Let $D(x,y)$ be the distance from the point $(x,y)$ to our line. Call the quarter-disk $Q$.   We calculate 
$$\iint_Q D(x,y) \,dA$$
and divide the result by the area of $Q$.  Because of the symmetry, it seems best to use polar coordinates. But it would be worthwhile to compute using rectangular coordinates, and compare.  
Suppose that point $P$ in our quarter-disk has polar coordinates $(r,\theta)$.  Join the origin to $P$, and drop a perpendicular from $P$ to our line.  The picture shows that the distance from $P$ to our line is $r\sin(\theta+\pi/4)$. So we want 
$$\iint_Q r^2 \sin(\theta+\pi/4)\,dr\,d\theta.$$  
Remark: Add to your picture the point $P$, join it to the origin, and drop a perpendicular as suggested above.  You will see from basic trigonometry that the distance  from $P$ to the line $x+y=0$ is
$r\sin(\theta+\pi/4)$. Then by the Addition Law for the sine function, 
$$r\sin(\theta+\pi/4)=r\sin(\pi/4)\cos\theta+r\cos(\pi/4)\sin \theta=
\frac{1}{\sqrt{2}} \left(r\cos\theta+r\sin\theta\right)=\frac{1}{\sqrt{2}}(x+y).$$ 
This gives an explanation different from the Linear Algebra formula for the fact that the distance from $(x,y)$ to the line is $\frac{x+y}{\sqrt{2}}$.
A: 
I guess I need to find a line perpendicular to y = -x and find the
  distance

That's right. In general the distance $d(x,y)$ from a point $P(x,y)$ to a straight line $r$ whose equation is $Ax+By+C=0$ can be derived algebraically as follows:


*

*Find the equation of the straight line $s$ passing through $P$ and being orthogonal to $r$. Call $Q$ the intersecting point of $r$ and $s$.

*Find the coordinates of $Q(x_{Q},y_{Q})$.

*Find the distance $d(x,y)$ from $P(x,y)$ to $Q(x_{Q},y_{Q})$. We get
the formula
$$\begin{equation*}
d(x,y)=\frac{|Ax+By+C|}{\sqrt{A^{2}+B^{2}}}.\tag{0}
\end{equation*}$$


In the present case,  since the given equation is $x+y=0$, we have $A=B=1,C=0$. So
$$\begin{equation*}
d(x,y)=\frac{|x+y|}{\sqrt{2}}.\tag{1}\qquad 
\end{equation*}$$
Let $R$ be the given quarter disc, centered at $(0,0)$ and radius $a\ge 0$. Its area $A$ is $\pi a^{2}/4$. In the following picture $d(x_k,y_k)$ is the distance of the point $(x_k,y_k)$ to the line $y=-x$.
 
Notice that


*

*$x^{2}+y^{2}=a^{2}$ is equivalent to $y=\pm \sqrt{a^{2}-x^{2}}$. 

*The condition $y\geq 0$ excludes the negative sign. 

*$|x+y|=x+y$ in $R$.


Therefore
$$\begin{equation*}
d(x,y)=\frac{x+y}{\sqrt{2}}.\tag{2}
\end{equation*}$$
Let's decompose $R$ by a partition of $n$ rectangular cells. If the area of the $k^{th}$
cell is $\Delta A_{k}$ and we choose an arbitrary point $\left(
x_{k},y_{k}\right) $ in it (see picture above), then the average distance $d_{\text{avg}}$ is
obtained by a limiting process. It satisfies 
$$\begin{eqnarray*}
A\cdot d_{\text{avg}} &=&\lim_{n\rightarrow \infty }\sum_{k=1}^{n}d\left(
x_{k},y_{k}\right) \Delta A_{k}=\iint_{R}\frac{x+y}{\sqrt{2}}\;\mathrm{d}A \\
\frac{\pi a^{2}}{4}d_{\text{avg}} &=&\iint_{R}\frac{x+y}{\sqrt{2}}\;\mathrm{d}
A \\
&=&\frac{1}{\sqrt{2}}\int_{x=0}^{a}\int_{y=0}^{\sqrt{a^{2}-x^{2}}}x+y\;\mathrm{d}x\;\mathrm{d}y,
\end{eqnarray*}$$
which implies
$$\begin{equation*}
d_{\text{avg}}=\frac{2\sqrt{2}}{\pi a^{2}}\int_{0}^{a}\left( \int_{0}^{\sqrt{a^{2}-x^{2}}}x+y\;\mathrm{d}y\right) \;\mathrm{d}x.
\end{equation*}\tag{3}$$
This corresponds to the integral indicated by your friend divided by the area of the region.

It looks like we are always integrating the distance from $0$ to $a$

The limits of integration are $0\le x\le a$ (with $a\ge 0$) and $0\le y\le \sqrt{a^{2}-x^{2}}$, and the distance is the function of $x$ and $y$ given by $(2)$.   
If we decompose the region $R$ into cells with a shape of sectors of a
circle, which corresponds to using polar coordinates $x=r\cos \theta ,y=r\sin
\theta $, $R$ is limited by $0\leq r\leq a$ and $0\leq \theta \leq \pi /2$.
The Jacobian of the transformation is $r$.
