Does $\iint_D \frac{x^2}{x^2+y^2} dx dy$ converge on $D= \left\{ (x, y) : x^2+y^2\leq ax \right\}$ ? If yes, what value does it converge to?

Powers equal to $$2$$ entice the polar substitution.

$$D= \left\{ (x, y) : x^2+y^2\leq ax \right\}$$, so $$0 \leq r \leq a \cos \phi.$$ For the domain to make sense, we need either $$\phi \in [0, \frac{\pi}{2}] \cup [\frac{3 \pi}{2}, 2\pi)$$ and $$a \geq 0$$, or $$\phi \in [\frac{\pi}{2}, \frac{3 \pi}{2}]$$ and $$a \leq 0$$.

Everything below was achieved with help from the comments.

$$\iint_D \frac{x^2}{x^2+y^2} dx dy = \iint_{D'} r \cos^2 \phi \ d\phi dr.$$

Case $$1$$:

Since $$(0, 0) \in D$$ (equivalently, $$r=0$$ for any $$\phi$$) makes the integrand indefinite,

$$\lim_{\epsilon \rightarrow 0} \int_{0}^{\frac{\pi}{2}} \cos^2 \phi \ d \phi \int_{\epsilon}^{a \cos \phi} r \ dr + \lim_{\epsilon \rightarrow 0} \int_{\frac{3 \pi}{2}}^{2\pi} \cos^2 \phi \ d \phi \int_{\epsilon}^{a \cos \phi} r \ dr$$

$$2\lim_{\epsilon \rightarrow 0} \int_{0}^{\frac{\pi}{2}} \cos^2 \phi \ d \phi \int_{\epsilon}^{a \cos \phi} r \ dr$$

$$2\lim_{\epsilon \rightarrow 0} \int_{0}^{\frac{\pi}{2}} \left( \frac{a^2 \cos^2 \phi}{2}-\frac{\epsilon^2}{2} \right) \cos^2 \phi \ d \phi$$

$$\lim_{\epsilon \rightarrow 0} \int_{0}^{\frac{\pi}{2}} \left( {a^2 \cos^2 \phi}-{\epsilon^2} \right) \cos^2 \phi \ d \phi$$

Put $$0$$ instead of $$\epsilon$$, because it won't let me integrate: $$a^2 \int_{0}^{\frac{\pi}{2}} { \cos^4 \phi} \ d \phi = \frac{a^2}{2} Β \left(\frac{1}{2}, \frac{5}{2} \right) = \frac{a^2}{2} \cdot \frac{\Gamma(\frac{1}{2}) \Gamma(\frac{5}{2})}{\Gamma(3)} = \frac{a^2}{2} \cdot \frac{\sqrt{\pi} \cdot 0.75 \sqrt{\pi}}{2} = \frac{3 a^2 \pi}{16}$$

For case $$2$$, the result should be same.

Formula $$Β(m, n)=2\int_0^{\pi/2}\sin^{2m-1} \theta \cos^{2n-1} \theta \ d \theta$$ was used, but I'm not sure I can extract it myself, and would be grateful for reference.

The answer would be, that for fixed $$a$$, it's convergent. (I am not sure about the definition of convergence here, and can't look it up, since the classes were missed, and google is blocked).

For Oliver Jones:

let $$r_i=a\frac{i}{i+1}$$, $$\phi_i=\frac{\delta}{i}.$$

$$\iint_{D_\delta} r \cos^2 \phi \ d\phi dr= \lim_{i \rightarrow \infty} \sum_i \left(a\frac{i}{i+1}\right) \left(\cos^2 \frac{\delta}{i} \right) \left( a\frac{i+1}{i+2} - a\frac{i}{i+2} \right) \left( \frac{\delta}{i+1}-\frac{\delta}{i} \right) = \lim_{i \rightarrow \infty} \sum_i \left(a\frac{i}{i+1}\right) \left(1 \right) \left( \frac{a}{(i+1)(i+2)} \right) \left( \frac{-\delta}{i(i+1)} \right) = \sum_{i=0}^{\infty} \left(a^2\frac{1}{i+1}\right) \left(1 \right) \left( \frac{1}{(i+1)(i+2)} \right) \left( \frac{-\delta}{(i+1)} \right).$$

It converges because of the $$4$$th power in the denominator.

Or maybe like this:

$$\lim_{\epsilon \rightarrow 0} \int_{\epsilon}^a x \ dx \int_{-\sqrt{x(a-x)}}^{\sqrt{x(a-x)}} \frac{1}{1+ (y/x)^2} d (y/x)$$

$$\lim_{\epsilon \rightarrow 0} \int_{\epsilon}^a 2 x \arctan \sqrt{x(a-x)} \ dx$$

How to proceed here, I am not sure.

$$\lim_{\delta \rightarrow 0}\iint_{\delta^2 \leq r^2+ar \cos \theta + a^2/4 \leq a^2/4} r \cos ^2 \theta \ dr \ d\theta$$

Let's choose $$\theta \in [-\pi/2, \pi/2], a\geq 0$$.

Then $$\delta^2 \leq r^2 - ar + a^2/4 \leq r^2+ar \cos \theta + a^2/4 \leq r^2 + a^2/4 \leq a^2/4.$$ $$0 \leq r^2 - ar + a^2/4 -\delta^2 \Longrightarrow r\in (0, \frac{a-2\delta}{2}).$$

$$\lim_{\delta \to 0} \int_{-\pi/2}^{\pi/2} \cos ^2 \theta \ d \theta \int_0^{(a-2\delta)/2} r dr = \frac{a^2}{8} В(1/2, 3/2) = \frac{a^2 \pi}{16}$$

• shouldn't the integrand be $r\cos^2 \theta$ after converting to polar? – Doug M Jan 24 at 20:00
• For what it's worth, the integral has a value of $\frac{3a^2\pi}{16}$ (according to computation in Mathematica) – user170231 Jan 24 at 20:29
• @fragileradius This is an improper integral which is defined in a way similar to the one variable case. Let $D_\delta$ be $D$ with a wedge with angle $\delta$ cut out that misses the origin.Then decide on the convergence of $\displaystyle{\lim_{\delta \rightarrow 0} \iint _{D_\delta} \frac{x^2}{x^2+y^2}\, dA }$. – Oliver Jones Jan 24 at 23:49
• The integrand $f$ is bounded (besides, the set of discontinuities of $f$ and the boundary of $D$ have content zero), so the integral exists as the limit of Riemann sums, you do not have to define it as an improper integral. Further, you can apply a change of variables and then evaluate the double integral as an iterated integral (provided that the inner integral exists). Thus you just need to consider the case $a > 0$ and evaluate $$2 \int_0^{\pi/2} \int_0^{a \cos t} r \cos^2 t \, dr dt$$ by using the identity $$\cos^4 t = \frac 3 8 + \frac 1 2 \cos 2 t + \frac 1 8 \cos 4 t.$$ – Maxim Jan 28 at 7:39
• @OliverJones You do not have to introduce the notion of an improper integral at all. Also, the conditions that I listed are sufficient: if $f$ is Riemann integrable on $D$ and $F(v) = \int_a^b f(u, v) du$ exists for all $v$, then $\int_c^d F(v) dv$ also exists and is equal to the double integral. $f$ does not have to be nonnegative. – Maxim Jan 29 at 6:17

I get the same result as your first one. But I don't see the need for special functions, or introducing the $$\epsilon$$ you use.
Consider the region $$x^2+y^2\leq ax$$. This is the same as $$\left(x-\frac{a}{2}\right)^2+y^2\leq\frac{a^2}{4}$$
So the region is a disc centered at $$\left(\frac{a}{2},0\right)$$ with its boundary touching the origin. I'll assume $$a>0$$. A disc like this is described in polar coordinates by $$r\leq a\cos(\theta)$$ for $$\theta\in[-\pi/2,\pi/2]$$.
So the integral is \begin{align} &\int_{\theta=-\pi/2}^{\theta=\pi/2}\int_{r=0}^{r=a\cos(\theta)}r\cos^2(\theta)\,dr\,d\theta\\ &=\int_{\theta=-\pi/2}^{\theta=\pi/2}\cos^2(\theta)\int_{r=0}^{r=a\cos(\theta)}r\,dr\,d\theta\\ &=\int_{\theta=-\pi/2}^{\theta=\pi/2}\cos^2(\theta)\frac{1}{2}a^2\cos^2(\theta)\,d\theta\\ &=\frac{a^2}{2}\int_{\theta=-\pi/2}^{\theta=\pi/2}\cos^4(\theta)\,d\theta\\ &=\frac{a^2}{2}\int_{\theta=-\pi/2}^{\theta=\pi/2}\left(\frac{1+\cos(2\theta)}{2}\right)^2\,d\theta\\ &=\frac{a^2}{4}\int_{\varphi=-\pi}^{\varphi=\pi}\left(\frac{1+\cos(\varphi)}{2}\right)^2\,d\varphi\\ &=\frac{a^2}{4}\int_{\varphi=-\pi}^{\varphi=\pi}\left(\frac14+\cos(\varphi)+\frac{1}{4}\cos^2(\varphi)\right)\,d\varphi\\ &=\frac{a^2}{4}\int_{\varphi=-\pi}^{\varphi=\pi}\left(\frac14+\frac{1}{4}\cos^2(\varphi)\right)\,d\varphi\\ &=\frac{a^2}{4}\int_{\varphi=-\pi}^{\varphi=\pi}\left(\frac14+\frac{1}{8}\left(1+\cos(2\varphi)\right)\right)\,d\varphi\\ &=\frac{a^2}{4}\int_{\varphi=-\pi}^{\varphi=\pi}\left(\frac38+\frac{1}{8}\cos(2\varphi)\right)\,d\varphi\\ &=\frac{a^2}{4}\int_{\varphi=-\pi}^{\varphi=\pi}\frac38\,d\varphi\\ &=\frac{3a^2}{32}(2\pi)=\frac{3a^2\pi}{16} \end{align}