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How can I calculate this double integral?

Calculate $$\xi=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2b\cos(\theta)}\left(\frac{-2ar\sin(\theta)}{\sqrt{2ar\cos(\theta)-r^{2}}}\right) rdrd\theta$$where $a\in \mathbb{R}^{+}$.

My attempt: We can see $$\begin{align*}\xi=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}-2a\sin(\theta)\int_{0}^{2b\cos(\theta)}\frac{r^{2}}{\sqrt{2ar\cos(\theta)-r^{2}}}drd\theta \end{align*}$$

but I think this double-integral is very difficult, but maybe can I solve it using change the order of the following double integral, but I don't sure how can I to do it.

This double-integral has a relation with that problem (using stokes-theorem): $C$ be curve of intersection of hemisphere $x^2+y^2+z^2=2ax$ and cylinder $x^2+y^2=2bx$ ; to evaluate $\int_C(y^2+z^2)dx+(x^2+z^2)dy+(x^2+y^2)dz$

Context of problem: Using of the information in the link, we can see that $$\boxed{\oint_{C}\vec{F}\cdot d \vec{r}=\int \int_{S}(\nabla \times \vec{F}) \cdot d S=\int \int_{S}(\nabla \times \vec{F})\cdot \vec{n}dS=\int \int_{D}(\nabla \times \vec{F}) \cdot \left(\frac{\partial r}{\partial x} \times \frac{\partial r}{\partial y} \right)dA.}$$

Note that \begin{eqnarray*} \Gamma(x,y)&:=&[(2y-2z)\vec{i}+(2z-2x)\vec{j}+(2x-2y)\vec{k}]\cdot \left( \frac{x-a}{\sqrt{2ax-x^{2}-y^{2}}}\vec{i}+\frac{y}{\sqrt{2ax-x^{2}-y^{2}}}\vec{j}+k \right)\\ &=&\frac{(2y-2z)(x-a)}{\sqrt{2ax-x^{2}-y^{2}}}+\frac{y(2z-2x)}{\sqrt{2ax-x^{2}-y^{2}}}+(2x-2y)\\ &=& \frac{(2y-2\sqrt{a^{2}-y^{2}-(x-a)^{2}})(x-a)}{\sqrt{2ax-x^{2}-y^{2}}}+\frac{y(2\sqrt{a^{2}-y^{2}-(x-a)^{2}}-2x)}{\sqrt{2ax-x^{2}-y^{2}}}+(2x-2y)\\ &:=&\Gamma_{1}(x,y)+\Gamma_{2}(x,y)+\Gamma_{3}(x,y) \end{eqnarray*} So \begin{eqnarray*} \int \int_{D} (\nabla \times \vec{F})\cdot \left(\frac{\partial r}{\partial x} \times \frac{\partial r}{\partial y} \right)dA&=&\int \int_{D} \Gamma(x,y)dA\\ &=&\underbrace{\int \int_{D} \Gamma_{1}(x,y)dA}_{I_{1}}+\underbrace{\int \int_{D} \Gamma_{2}(x,y)dA}_{I_{2}}+\underbrace{\int \int_{D} \Gamma_{3}(x,y)dA}_{I_{3}} \end{eqnarray*}

Note that calculate $I_{3}$ is easy, because \begin{eqnarray*}I_{3}&=&\int \int_{D} \Gamma_{3}(x,y)dA\\ &=& \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2b\cos(\theta)}\Gamma_{3}(r\cos(\theta),r\sin(\theta))rdrd\theta\\ &=&\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2b\cos(\theta)}(2r\cos(\theta)-2r\sin(\theta)) rdrd\theta\\ &=&\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2b\cos(\theta)}2r^{2}(\cos(\theta)-\sin(\theta)drd\theta\\ &=&\frac{2}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left[r^{3} \right]_{0}^{2b\cos(\theta)}(\cos(\theta)-\sin(\theta))d\theta\\ &=&\frac{2}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(2b\cos(\theta))^{3}(\cos(\theta)-\sin(\theta))d\theta\\ &=&\frac{2\cdot 2^{3}b^{3}}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^{3}(\theta)(\cos(\theta)-\sin(\theta))d\theta\\ &=&\frac{2^{4}b^{3}}{3}\cdot \frac{3 \pi}{8}\\ &=&2\pi b^{3} \implies \boxed{I_{3}=2\pi b^{3}} \end{eqnarray*}

but my problem is how can I calculate $I_{1}+I_{2}$?

Add information, we can see $$\boxed{r(x,y)=x\vec{i}+y\vec{j}+\left(\sqrt{a^{2}-y^{2}-(x-a)^{2}}\right)\vec{k}}$$

and $$\boxed{\frac{\partial r}{\partial x}=\vec{i}+\frac{\partial}{\partial x}\left(\sqrt{a^{2}-y^{2}-(x-a)^{2}} \right)\vec{k}=\vec{i}-\frac{x-a}{\sqrt{2ax-x^{2}-y^{2}}\vec{k}}}$$ and $$\boxed{\frac{\partial r}{\partial y}=\vec{j}+\frac{\partial}{\partial y}\left(\sqrt{a^{2}-y^{2}-(x-a)^{2}} \right)\vec{k}=\vec{j}-\frac{y}{\sqrt{2ax-x^{2}-y^{2}}}\vec{k}}$$ so, $$\boxed{\frac{\partial r}{\partial x} \times \frac{\partial r}{\partial y}=\frac{x-a}{\sqrt{2ax-x^{2}-y^{2}}}\vec{i}+\frac{y}{\sqrt{2ax-x^{2}-y^{2}}}\vec{j}+k}$$Now, using information of link, we can see $$\boxed{D=\left\{(r,\theta): \frac{-\pi}{2}\leq \theta \leq \frac{\pi}{2}, 0 \leq r \leq 2b\cos(\theta) \right\}}$$and finally $$\boxed{\vec{F}(x,y,z):=(y^{2}+z^{2})\vec{i}+(x^{2}+z^{2})\vec{j}+(x^{2}+y^{2})\vec{k}}.$$ Note that $$\boxed{\nabla \times \vec{F}=(2y-2z)\vec{i}+(2z-2x)\vec{j}+(2x-2y)\vec{k}}$$

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  • $\begingroup$ @RachidAtmai I will write a little more the context of my problem and how I got to that integral $\endgroup$ – mathproof Sep 7 '20 at 22:44
  • $\begingroup$ argh! instead of editing I deleted my comment...sorry about that. yes please post the context. If $r=0$ then you have an issue with your integral (my comment was about the integral being improper...). $\endgroup$ – Rachid Atmai Sep 7 '20 at 22:47
  • $\begingroup$ I already added the full context of my problem. When trying to use polar coordinates to calculate $I_ {1} + I_ {2}$ the problem becomes very ugly and from there comes that integral that I placed at the beginning that I don't know how to integrate. $\endgroup$ – mathproof Sep 7 '20 at 23:03
  • $\begingroup$ What is the relationship between $a$ and $b$ $\endgroup$ – Ninad Munshi Sep 7 '20 at 23:24
  • $\begingroup$ $0<b<a$ that is the relationship. $\endgroup$ – mathproof Sep 7 '20 at 23:28
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Under the interchange $\theta \leftrightarrow -\theta$

$$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{2b\cos(-\theta)}\frac{-2ar^2\sin(-\theta)}{\sqrt{2ar\cos(-\theta)-r^2}}drd\theta$$

$$= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{2b\cos \theta}\frac{2ar^2\sin \theta}{\sqrt{2ar\cos \theta-r^2}}drd\theta = -I$$

Thus $I = 0$

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  • $\begingroup$ Fantastics! Thank so much! $\endgroup$ – mathproof Sep 7 '20 at 23:29

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