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Find the area enclosed by the curve $x^2+y^2=|x|+|y|$ on the coordinate plane.

I have no idea how this curve looks like, therefore can't find the area. Please help. Thank you.

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4 Answers 4

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To get an idea of what the curve looks like, try different cases. For example, when $x\ge 0$ and $y\le 0$, you get $$x^2+y^2=x-y\iff x^2-x+y^2+y=0\\\iff\left(x-\frac 12\right)^2-\frac 14+\left(y+\frac 12\right)^2-\frac 14=0\\ \iff \left(x-\frac 12\right)^2+\left(y+\frac 12\right)^2=\left(\frac 1{\sqrt 2}\right)^2$$ So here the curve is part of a circle centered at $(1/2,-1/2)$ with radius $1/\sqrt 2$.

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If you have a Mac, it came with wonderful free software called Grapher which can graph things like this easily.

Anyway, I think this one is best tackled in polar. We can work in the first quadrant to get rid of the absolute value signs and then multiply our answer by 4. We have $$ x^2 + y^2 = x + y $$ so $$ r^2 = r(\cos(\theta) + \sin(\theta)) $$ i.e. $$ r = \cos(\theta) + \sin(\theta). $$ The area is therefore $$ 4\int_0^{\frac{\pi}{2}}\frac{1}{2}(\cos(\theta) + \sin(\theta))^2 d\theta. $$

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In the first place $S:=\{(x,y)\ |\ x^2+y^2=|x|+|y|\}$ is just a set. As $S$ is symmetric with respect to the axes we begin by determining the points of $S$ lying in the (closed) first quadrant $Q$. In $Q$ we have $|x|=x$ and $|y|=y$. Therefore $$S\cap Q=\{(x,y)\in Q|x^2+y^2=x+y\}\ .$$ Now the equation $x^2+y^2=x+y$ is equivalent with $(x-{1\over2})^2+(y-{1\over2})^2={1\over2}$ and describes a circle $C$ with center $({1\over2},{1\over2})$ and radius ${1\over\sqrt{2}}$.

It follows that $S\cap Q=C\cap Q$, and $S$ becomes the set you get by reflecting this intersection on the two axes. Now it should not be too difficult to compute the area enclosed by $S$.

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Wolfram|Alpha shows you how the curve looks like. To obtain the area, first note that by symmetry we can concentrate on the first quadrant. There the equation is equivalent to $x^2 + y^2 = x+y\iff y^2 - y + (x^2-x) = 0$, so \[ y_{1,2} = \frac 12 \pm \sqrt{\frac 14 - (x^2-x)} \] The randicand is nonnegative for $x^2 - x \le \frac 14$, that is $0 \le x \le \frac 12(1 + \sqrt 2)$ (note that we are in the first quadrant, hence $x \ge 0$). $y_2$ is positive if moreover $x^2-x \ge 0$, that is $x \ge 1$. So the area is \begin{align*} A &= 4 \left(\int_0^1 \left(\frac 12 + \sqrt{\frac 14 - (x^2-x)}\right)\,dx + \int_1^{\frac 12(1 +\sqrt 2)} 2\sqrt{\frac 14 - (x^2 - x)}\, dx\right) \end{align*}

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You can also use polar coordinates for $r(t)=\cos t + \sin t$, which is a circle. See a plot. –  lhf Nov 7 '12 at 11:41
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