Integral over a ball

Question

Let $a=(1,2)\in\mathbb{R}^{2}$ and $B(a,3)$ denote a ball in $\mathbb{R}^{2}$ centered at $a$ and of radius equal to $3$.
Evaluate the following integral: $$\int_{B(a,3)}y^{3}-3x^{2}y \ dx dy$$

Should I use polar coordinates? Or is there any tricky solution to this?

robjohn · Accepted Answer · 2021-02-06T20:13:00.223

12

Note that $\Delta(y^3-3x^2y)=0$, so that $y^3-3x^2y$ is harmonic. By the mean value property, we get that the mean value over the ball is the value at the center. Since the area of the ball is $9\pi$ and the value at the center is $2$, we get $$ \int_{B(a,3)}\left(y^3-3x^2y\right)\mathrm{d}x\,\mathrm{d}y=18\pi $$

edited Feb 06 '21 at 20:13

answered May 26 '13 at 20:06

robjohn

345,667

2

On the ball, as always. +1 – Potato May 26 '13 at 20:16
Super slick, +1 – icurays1 May 26 '13 at 20:19
Thanks a lot! Very tricky :) – Paweł Orliński May 26 '13 at 21:10
2

Not so tricky when you recognize $y^3-3x^2y=-\mathrm{Im}\left((x+iy)^3\right)$. – robjohn May 26 '13 at 21:14

score 2 · Answer 2 · answered May 26 '13 at 20:42

Here is the brute force approach. Let $f(x,y) = y^3-3x^2y$. Shift so the center of the ball is at the origin and go to polar coordinates. The integral is then
$$\int_0^{2\pi}\int_0^3 f(r\cos\theta+1,r\sin\theta+2) r dr d\theta.$$ The integral is straightforward but tedious, with value $18\pi$ as found by @robjohn in his much slicker answer.

Integral over a ball

2 Answers2