Double integral $\iint_{D} \left(x^2+y^2\right)\mathrm{d}x \mathrm{d}y$ over the domain $D$: $x\le x^2+y^2 \le 2x,\ \ y\ge 0$

Question

$$\iint\limits_{D} \left(x^2+y^2\right)\mathrm{d}x \mathrm{d}y$$

Where the domain is $D$: $x\le x^2+y^2 \le 2x,\ \ y\ge 0$

So far I have tried getting the bounds by working on the domain and taking the right part i could determine that $y^2+(x-1)^2 = 1$ which would be a circle, but other than that i have no idea how to work the first part of the inequality

Answer 1

The two conditions $$ x\le x^2+y^2\le 2x$$ collect all points $(x,y)$

• with $(x-1)^2+y^2\le 1^2$ (second inequality), which are the points inside the closed disk $D$ centered in $(1,0)$ with radius one,
• and eliminates those with $\left(x-\frac 12\right)^2 +y^2 <\left(\frac 12\right)^2$, so we cut the disk $D'$ centered in $\left(\frac 12,0\right)$ and radius $\frac 12$.

The picture for these two conditions only is:

sage: x, y = var('x,y');
sage: region_plot([ x <= x^2 + y^2 , x^2 + y^2 <= 2*x ], (x, -3, 3), (y, -3, 3) )
Launched png viewer for Graphics object consisting of 1 graphics primitive

We further have to restrict to the part $y\ge 0$. But note the symmetry of both the blue domain $D$ above, and of the function under the integral w.r.t. the $Ox$ axis. So we may integrate on $D-D'$, and take half of the result.

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The integral to be computed, call it $$ J \ ,$$ is a difference of two integrals. Each of the two integrals can be then computed

• by either splitting the variables via Cavalieri / Fubini, and to have a simpler bound among the two, we may want to use the symmetry w.r.t. the $Ox$-axis, and start in $y=0$,

• or by passing to polar coordinates

  • on $D$ substitute $\left\{\begin{aligned}x &= 1+r\sin t\\y&=r\cos t\end{aligned}\right.$ where $r\in(0,1)$ and $t\in(0,2\pi)$, $dx\wedge dy=(\sin t\; dr+r\cos t\; dt) \wedge(\cos t\; dr-r\sin t\; dt)=r\;dt\wedge dr$
  • on $D'$ substitute $\left\{\begin{aligned}x &= 1/2+r\sin t\\y&=r\cos t\end{aligned}\right.$ where $r\in(0,1/2)$ and $t\in0,2\pi)$, $dx\wedge dy=(\sin t\; dr+r\cos t\; dt) \wedge(\cos t\; dr-r\sin t\; dt)=r\;dt\wedge dr$:

  We will integrate on the whole blue part, so that in case the sine integral appearing is done on a full period, and vanishes $\int_0^{2\pi}\sin t\; dt=0$. Alternatively, we may switch $\sin$ and $\cos$ in the polar coordinate parametrizations. $$ \begin{aligned} J&=\frac 12\int_{D-D'}(x^2 +y^2)\; dx\; dy \\ &= \frac 12\int_{D}(x^2 +y^2)\; dx\; dy - \frac 12\int_{D'}(x^2 +y^2)\; dx\; dy \\ &= \int_{D\ :\ x\ge 0}(x^2 +y^2)\; dx\; dy - \int_{D'\ :\ x\ge 0}(x^2 +y^2)\; dx\; dy \\ &= \int_{x=0}^{x=2}dx\int_{y=0}^{y=\sqrt{2x-x^2}}(x^2+ y^2)\; dy - \int_{x=0}^{x=1}dx\int_{y=0}^{y=\sqrt{x-x^2}}(x^2+ y^2)\; dy \\ &= \int_{x=0}^{x=2}dx\left[x^2y +\frac 13y^3\right]_{y=0}^{y=\sqrt{2x-x^2}} - \int_{x=0}^{x=1}dx\left[x^2y +\frac 13y^3\right]_{y=0}^{y=\sqrt{x-x^2}} \\ &= \int_0^2 dx\left[x^2 +\frac 13(2x-x^2)\right]\sqrt{2x-x^2} - \int_0^1 dx\left[x^2 +\frac 13(x-x^2)\right]\sqrt{x-x^2} \\ &= \cdot\frac{3\pi}{4} - \cdot\frac{3\pi}{64} = \frac{3(16-1)\pi}{64} = \bbox[yellow]{\ \frac{45\pi}{64}\ }\ . \\[4mm] &\qquad\qquad\text{ Alternatively:} \\[4mm] J&=\frac 12\int_{D-D'}(x^2 +y^2)\; dx\; dy \\ &= \frac 12\int_{D}(x^2 +y^2)\; dx\; dy - \frac 12\int_{D'}(x^2 +y^2)\; dx\; dy \\ &= \frac 12\int_0^1\int_0^{2\pi}(1+2r\sin t+r^2)\; r\; dt\; dr - \frac 12\int_0^{1/2}\int_0^{2\pi}\left(\frac 14+r\sin t+r^2\right)\; r\; dt\; dr \\ &= \pi\int_0^1(1+r^2)\; r\; dr - \pi\int_0^{1/2}\left(\frac 14+r^2\right)\; r\; dr \\ &= \pi\left[\frac 12r^2+\frac 14r^4\right]_0^1 - \pi\left[\frac 18r^2+\frac 14r^4\right]_0^{1/2} \\ &= \pi\left[\frac 12+\frac 14\right] - \pi\left[\frac 1{32}+\frac 1{64}\right] = \bbox[yellow]{\ \frac{45\pi}{64}\ }\ . \end{aligned} $$

Answer 2

The idea is writing the region D as the region between two functions of $x$, to integrate with Fubini's theorem. From the first innequality we have: $$x \leq x^2 + y^2$$ $$ y \geq\sqrt{x - x^2}$$ for $0\leq x \leq 1$, else the radicand would be negative. Now the second inneuquality yields: $$2x \geq x^2 + y^2$$ $$ y \leq\sqrt{2x-x^2}$$ for $0\leq x \leq 2$ Now we can write $$D = \{(x,y) \in \mathbb{R}^2| 0 \leq x \leq 1, \sqrt{x - x^2} \leq y \leq \sqrt{2x-x^2}\} \bigcup \{(x,y) \in \mathbb{R}^2| 1 \leq x \leq 2, \leq y \leq \sqrt{2x-x^2}\}$$

Now you can use Fubini's theorem to evaluate the integral as:

$$\int_{0}^{1}\int_{\sqrt{x - x^2}}^{\sqrt{2x-x^2}} (y^2 + x^2)dydx + \int_{1}^{2}\int_{0}^{\sqrt{2x-x^2}} (y^2 + x^2)dydx$$ Can you take it from here?

Answer 3

After plotting the region I find it can be expressed in polar coordinates as $\cos\theta<r<2\cos\theta$ and $0<\theta<\pi/2$. Integrating in polar over this region we get $$\int_0^{\pi/2} \int_{\cos\theta}^{2\cos\theta} r^2 \cdot rdr.$$ This is not too hard except for when you have to integrate $\cos^4\theta$. At this point you can repeatedly apply the power-reduction identity $\cos^2 x = (1+\cos(2x))/2$ (first to $\cos\theta$ and then to $\cos(2\theta))$.
My answer ends up being $45\pi/64$.