# Area with double integral.

$f:\mathbb{R}^2\to\mathbb{R}, f(x,y)=(x^2+y^2)^2-8(x^2-y^2).$ Find the critical points of $f$ and the area of $X=\{(x,y):f(x,y)\leq 0\}$.

To find the critical points I just have to find $(x,y):\nabla f(x,y)=0.$ The partial derivatives are given by $\displaystyle\frac{\partial f}{\partial x}= 4x(x^2+y^2)-16x, \displaystyle\frac{\partial f}{\partial y}= 4y(x^2+y^2)+16y$. It must be $\displaystyle\frac{\partial f}{\partial x}=\displaystyle\frac{\partial f}{\partial y}=0$ then $4x[(x^2+y^2)-4]=0,4y[(x^2+y^2)+4=0]\implies (x,y)=\{(0,0),(-2,0),(2,0)\}$.

But how can I use it to find the area of $X$?. I believe this could be computed with a double integral using polar coordinates. If I set $x=\rho\cos\theta,y=\rho\sin\theta$ then $(x,y)\in X \iff r^4-8r^2\cos2\theta\leq0 \iff \frac{1}{8}r^2\leq\cos2\theta$. Then I can get the are by $\displaystyle\int_0^{\pi}\displaystyle\int_0^{2\sqrt{2\cos2\theta}}\rho\;d \rho d\theta?.$

Then the area is given by $\displaystyle\int_0^{\pi}\displaystyle\int_0^{2\sqrt{2\cos2\theta}}\rho\;d \rho d\theta=4\displaystyle\int_0^{\pi}\cos2\theta\;d\theta=2\sin2\theta|_{\theta=0}^{\theta=\pi}=0$

Where did I go wrong?.

• As John said, your interval isn't correct. Notice that for $\theta \in [\pi/4, 3\pi/4]$ we have $\cos (2 \theta) \leq 0$, for which the inequality isn't valid. – Mark Fantini Feb 1 '14 at 22:10

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\sss}[1]{\scriptscriptstyle{#1}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \color{#00f}{\large{\cal A}_{\rm X}}&\equiv\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \Theta\pars{8\bracks{x^{2} - y^{2}} - \bracks{x^{2} + y^{2}}^{2}}\,\dd x\,\dd y =\int_{0}^{2\pi}\dd\theta\int_{0}^{\infty}\dd\rho\,\rho\, \Theta\pars{8\rho^{2}\cos\pars{2\theta} - \rho^{4}} \\[3mm]&= \half\int_{0}^{2\pi}\dd\theta\int_{0}^{\infty}\dd\rho\, \Theta\pars{8\cos\pars{2\theta} - \rho} = \half\int_{0}^{2\pi}\dd\theta\,\Theta\pars{\cos\pars{2\theta}}\int_{0}^{8\cos\pars{2\theta}}\dd\rho \\[3mm]&= 4\int_{0}^{2\pi}\Theta\pars{\cos\pars{2\theta}}\cos\pars{2\theta}\,\dd\theta = 2\int_{0}^{4\pi}\Theta\pars{\cos\pars{\theta}}\cos\pars{\theta}\,\dd\theta \\[3mm]&= 4\int_{0}^{2\pi}\Theta\pars{\cos\pars{\theta}}\cos\pars{\theta}\,\dd\theta = -4\int_{-\pi}^{\pi}\Theta\pars{-\cos\pars{\theta}}\cos\pars{\theta}\,\dd\theta \\[3mm]&=-8\int_{0}^{\pi}\Theta\pars{-\cos\pars{\theta}}\cos\pars{\theta}\,\dd\theta = -8\int_{\pi/2}^{\pi}\cos\pars{\theta}\,\dd\theta = \left.-8\sin\pars{\theta}\right\vert_{\pi/2}^{\pi} \\[3mm]&=-8\sin\pars{\pi} + 8\sin\pars{\pi \over 2} = \color{#00f}{\LARGE 8} \end{align}
The problem is in evaluating the polar integral. The area under the curve $r^2=8\cos2\theta$ is $0$ if you just integrate over the whole domain since there are equal amounts of "positive" and "negative" area. Use symmetry and break it into one small piece you can evaluate and get a positive answer from.