Double integral of maximum function.

Question

Let $D:= \lbrace (x,y) \in [0,\infty)^2: 1 \le x^2+y^2\le9 \rbrace$. Determine the integral : $$\int\int_D \max(3x^2,y^2)\;dx\,dy.$$ I have a little problem, because I'm not sure where the maximum is $3x^2$ or $y^2$.

Answer 1

Let $x=r\cos\theta$ and $y=r\sin\theta$, then $$ \iint_D \max\{3x^2,y^2)\ \mathsf dx\ \mathsf dy = \iint_D \det J\cdot\max\{3r^2\cos^2\theta, r^2\sin^2\theta \}\ \mathsf d\theta\ \mathsf d r, $$ where $$ J = \begin{bmatrix}\frac{\partial x}{\partial r}&\frac{\partial x}{\partial \theta}\\ \frac{\partial y}{\partial r}&\frac{\partial y}{\partial \theta} \end{bmatrix} = \begin{bmatrix} \cos\theta&-r\sin\theta\\ \sin\theta& r\cos\theta \end{bmatrix}. $$ The determinant of the Jacobian matrix is $$ \det J = \cos\theta\cdot r\cos\theta - (-r\sin\theta\cdot\sin\theta) = r(\sin^2\theta + \cos^2\theta) = r, $$ and so the integral in question is $$ \int_1^3\int_0^{2\pi} r\cdot\max\{3r^2\cos^2\theta,r^2\sin^2\theta\}\ \mathsf d\theta\ \mathsf dr = \int_1^3 r^3\int_0^{2\pi} \max\{3\cos^2\theta,\sin^2\theta\}\ \mathsf d\theta\ \mathsf dr. $$ To compute $\max\{3\cos^2\theta,\sin^2\theta\}$, first note that \begin{align} 3\cos^2\theta = \sin^2\theta &\iff \sqrt 3|\cos\theta| = |\sin\theta|\\ &\iff \frac{|\sin\theta|}{|\cos\theta|} = \sqrt 3\\ &\iff \theta \in \left\{\frac\pi3, \frac{2\pi}3,\frac{4\pi}3,\frac{5\pi}3 \right\}. \end{align} Hence \begin{align} &\quad\int_1^3r^3\int_0^{2\pi} \max\{3\cos^2\theta,\sin^2\theta\}\ \mathsf d\theta\ \mathsf dr\\ &= \int_1^3 r^3 \bigg(\int_0^{\frac\pi3}3\cos^2\theta\ \mathsf d\theta + \int_{\frac\pi3}^{\frac{2\pi}3} \sin^2\theta\ \mathsf d\theta + \int_{\frac{2\pi}3}^{\frac{4\pi}3}3\cos^2\theta\ \mathsf d\theta \\ &\quad\quad\quad+\int_{\frac{4\pi}3}^{\frac{5\pi}3}\sin^2\theta\ \mathsf d\theta + \int_{\frac{5\pi}3}^{2\pi} 3\cos^2\theta \bigg) \mathsf d\theta\\ &=40\sqrt3 + \frac{140\pi}3. \end{align}