# Average distance from a point on a circle to the y-axis.

This is a simple question, but I must be making some mistakes as I don't seem to get the answer in the book.

Question: Determine the average distance from a point on $$x^2+y^2 = 9$$ to the $$y$$-axis.

My attempt: Due to symmetry let us calculate the average distance in the first quadrant, we also see that the distance to $$y$$-axis is the $$x$$ component. The distance function is going to be $$g(y) = \sqrt{9-y^2}$$, now we can integrate the function and divide by integration region area to get the average value of this distance function. $$\frac{4}{9\pi}\int_{0}^{\frac{\pi}{2}} \int_{0}^{3} r\sqrt{9 - r^2\sin^2(\theta)} \, dr \, d\theta = \frac{4}{9\pi}18 = \frac{8}{\pi}$$ In the book the given answer is $$\frac{6}{\pi}$$

Thanks for any ideas or noticing where I go wrong!

• Is the set under consideration the circle $x^2 + y^2 = 9$, or the disk $x^2 + y^2 \leq 9$? Commented Apr 23 at 19:07
• @MatthewLeingang The set I consider for integration is the disk $x^2 + y^2 \leq 9$ If I understood the question correctly. Commented Apr 23 at 19:12
• I would read it the opposite way. Since it says "on", I'd expect all points of the domain to satisfy the relation $x^2+y^2=9$. So the domain is a circle. If it said "region bounded by $x^2+y^2 = 9$", then I would understand it to be the disk. Commented Apr 23 at 19:14
• @MatthewLeingang You are right, the question was in the language I don't speak so I was not 100% sure how to interpret it, thank you. Commented Apr 23 at 19:16

On the circle $$x^2+y^2=9$$ we have identically $$r\equiv 3$$, so in the integral we should consider the length of the circle (which is $$2\pi r=6\pi$$) and integrate only on $$\theta$$. Thus, the integral is actually $$\dfrac{4}{6\pi}\int_{0}^{\frac{\pi}{2}} 3\sqrt{9 - 9\sin^2(\theta)} \, d\theta = \dfrac{4}{6\pi}\int_{0}^{\frac{\pi}{2}} 9\cos(\theta) \, d\theta=\dfrac{6}{\pi}$$
$$x^2 +y^2=a^2$$
$$\bar y= \frac{\int y^2 dx}{\int y~ dx}= \frac{Area First Moment }{Area}.$$