So Im new to using polar coordinates to work out double integrals and was wondering if anyone could help me.

So I am looking at the question:

$I_{6}=\int\int_{Q}x/\pi(x^{2}+y^{2}) dxdy$ where $Q=\{(x,y)|(x-2)^{2}+y^{2}\leq 4, y\geq0\}$

Looking at the first equation:

$(x-2)^{2}+y^{2} \leq4$ which gives $x^{2} + y^{2} = 4x$

and in polar form:

$r^{2}cos^{2}\theta +r^{2}sin^{2}\theta =4rcos\theta$

which simplifies to :

$r^{2}=4rcos\theta$ which equals $r=4cos\theta$

So my first integral limits will be between $0$ and $4cos\theta$

So the first part of the equation looks like:

$\int_{0}^{4cos\theta} (rcos\theta)/\pi(r^2cos^2\theta+r^2sin^2\theta) rdr$

Which simplifies to:

$\int_{0}^{4cos\theta}(cos\theta/\pi) dr$

which gives me: $\large[(cos\theta/\pi)r\large]_{0}^{4cos\theta}$

and gives:


but im unsure on what my next set of limits will be. I know from a previous question that ive asked that I need to draw a diagram with the centre of the circle at (2,0) but I dont undestand how to utilise this to get my second limit.

Any help would be greately appreciated.


I rather would use polar coordinates adapted to the given disk centered in $(2,0)$, so we have $$ \begin{aligned} x &= 2+r\cos t\ ,\\ y &= 0+r\sin t\ ,\\ dx &= \cos t\; dr - r\sin t\; dt\ ,\\ dy &= \sin t\; dr + r\cos t\; dt\ ,\\ dx\wedge dy &= (\cos t\; dr - r\sin t\; dt)\wedge(\sin t\; dr + r\cos t\; dt)\\ &=r\cos^2 t\; dr\wedge dt - r\sin^2 t\; dt\wedge dr\\ &=r\cos^2 t\; dr\wedge dt + r\sin^2 t\; dr\wedge dt\\ &=r\; dr\wedge dt\ ,\qquad\text{ so the Jacobian is $|r|=r$}\\ dx\; dy &=r\; dr\; dt\\[3mm] % I_6 &= \iint_D\frac x{\pi(x^2+y^2)}\; dx\; dy\\ &= \frac 1\pi \iint_{\substack{r\in[0,2]\\t\in[0,\pi]}} \frac {2+r\cos t}{(2+r\cos t)^2+(r\sin t)^2}\; r\; dr\; dt\\ &= \frac 1\pi \int_0^2 r\; dr\; \int_0^\pi \frac {2+r\cos t}{r^2+ 4r\cos t + 4}\; dt\\ &= \frac 1\pi \int_0^2 r\; dr\; \left[ \frac 14 t + \frac 12\arctan\frac{(2-r)\sin t}{(2+r)(1+\cos t)} \right]_0^\pi \\ &= \frac 1\pi \int_0^2 r\; dr\; \left[ \frac 14 t + \frac 12\arctan\left(\frac{2-r}{2+r}\tan \frac t2\right)\right]_0^\pi \\ &= \frac 1\pi \int_0^2 r\; dr\; \left[ \frac \pi4 + \frac \pi4 \right] \\ &=1\ . \end{aligned} $$

CAS check. Using polar coordinates as above. The computer algebra system (CAS) used is sage:

sage: var('r,t');
sage: INT = integral
sage: X, Y = 2 + r*cos(t), r*sin(t)
sage: I6 = 1/pi * INT( INT( X/(X^2+Y^2) * r, r, 0, 2), t, 0, pi)
sage: I6

CAS check. Using Fubini. sage again:

sage: var('x,y');
sage: integral( integral( x/(x^2+y^2), y, 0, sqrt(x*(4-x)) ), x, 0, 4) / pi

Human check. (Using Fubini again.) $$ \begin{aligned} I_6 &= \frac 1\pi \int_0^4 dx \int_0^{\sqrt{4-(x-2)^2}} \frac {x\; dy}{x^2+y^2} \\ &= \frac 1\pi \int_0^4 dx \left[ \arctan\frac yx \right]_0^{\sqrt{4-(x-2)^2}} \\ &= \frac 1\pi \int_0^4 \arctan\frac {\sqrt{4-(x-2)^2}}x \; dx \\ &\qquad\text{(Substitution: $x=2+2\cos u$, $u\in[0,\pi]$)} \\ &= \frac 1\pi \int_0^\pi \arctan\frac {2\sin u}{2(1+\cos u)}\; 2\sin u\; du \\ &= \frac 1\pi \int_0^\pi \arctan\frac {4\sin \frac u2\cos \frac u2}{2\cos^2 \frac u2)}\; 2\sin u\; du \\ &= \frac 1\pi \int_0^\pi \frac u2\; 2\sin u\; du \\ &= 1\ . \end{aligned} $$


If you draw a picture of the circular region you will see that the given region is $\{r,\theta): r\leq 4\cos \theta, 0\leq \theta \leq \frac {\pi}2\}$. [ $\theta$ ranges from $0$ to $\frac {\pi}2$].


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