Evaluate \[ \iint_{\mathcal R}(x^3-3xy^2)\,dx\,dy, \] where \[ \mathcal R=\{(x,y)\in\mathbb R^2 : (x+1)^2+y^2\le 9,\ (x-1)^2+y^2\ge 1\}. \]
Solution
Let \[ D_-=\{(x,y):(x+1)^2+y^2\leq9\}, \qquad D_+=\{(x,y):(x-1)^2+y^2\leq1\}. \]
1. One has\[ D_+\subseteq D_-, \]and therefore\[ \mathcal R=D_-\setminus D_+. \]
The centers of \(D_-\) and \(D_+\) are \((-1,0)\) and \((1,0)\), whose distance is \(2\). Since \[ 2+1=3, \] the disk of radius \(1\) centered at \((1,0)\) lies inside the disk of radius \(3\) centered at \((-1,0)\); the two boundary circles are internally tangent at \((2,0)\). The description of \(\mathcal R\) is then immediate from its two defining inequalities.
2. If\[ D(c,R)=\{(x,y):(x-c)^2+y^2\leq R^2\}, \]then\[ \iint_{D(c,R)}(x^3-3xy^2)\,dx\,dy = \pi R^2c^3. \]
Make the translation \[ x=c+u. \] Then the disk becomes \(u^2+y^2\leq R^2\), and \[ \begin{aligned} (c+u)^3-3(c+u)y^2 &= c^3+3c^2u+3cu^2+u^3\\ &\quad-3cy^2-3uy^2. \end{aligned} \] On the disk centered at the origin, the terms containing an odd power of \(u\) integrate to zero by the symmetry \(u\mapsto-u\). Also rotational symmetry gives \[ \iint_{u^2+y^2\leq R^2}u^2\,du\,dy = \iint_{u^2+y^2\leq R^2}y^2\,du\,dy. \] Hence the contributions \(3cu^2\) and \(-3cy^2\) cancel after integration. Only the constant term remains, so \[ \iint_{D(c,R)}(x^3-3xy^2)\,dx\,dy = c^3\operatorname{area}(D(c,R)) = \pi R^2c^3. \]
3. The integral over \(D_-\) is\[ -9\pi, \]and the integral over \(D_+\) is\[ \pi. \]
Apply step 2 with \((c,R)=(-1,3)\) and \((c,R)=(1,1)\): \[ \iint_{D_-}(x^3-3xy^2)\,dx\,dy = \pi\cdot 3^2\cdot(-1)^3 = -9\pi, \] while \[ \iint_{D_+}(x^3-3xy^2)\,dx\,dy = \pi\cdot1^2\cdot1^3 = \pi. \]
4. Therefore\[ \boxed{ \iint_{\mathcal R}(x^3-3xy^2)\,dx\,dy=-10\pi }. \]
By step 1 and additivity of the integral, \[ \begin{aligned} \iint_{\mathcal R}(x^3-3xy^2)\,dx\,dy &= \iint_{D_-}(x^3-3xy^2)\,dx\,dy -\iint_{D_+}(x^3-3xy^2)\,dx\,dy\\ &= -9\pi-\pi\\ &= -10\pi. \end{aligned} \]
5. Q.E.D.
Step 4 is the required evaluation.