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Change the order of integration in:

$$\int_0^{2a}\int_0^\sqrt{2ax-x^2}\frac{\phi'(y)(x^2+y^2)x }{\sqrt{4a^2x^2-(x^2+y^2)^2}}dxdy$$ and hence evaluate it.


I changed the order of integration and got the limits $y=0$ to $a$ and $x=a-\sqrt{a^2-y^2}$ to $a+\sqrt{a^2-y^2}$. I then tried changing to polar coordinates, but it is only making the integrand more complex.

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  • $\begingroup$ when are you allowed to change order of integration? $\endgroup$ – Alvin Lepik Aug 7 '17 at 9:08
  • $\begingroup$ At the start. The evaluation is to be done only after changing the order $\endgroup$ – user467745 Aug 7 '17 at 9:10
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    $\begingroup$ What is $\phi$? $\endgroup$ – mickep Aug 7 '17 at 9:11
  • $\begingroup$ Any function of y. The final answer will include $\phi$. $\endgroup$ – user467745 Aug 7 '17 at 9:14
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For each positive real number $a\in\mathbb{R}_{>0}$ and continuously differentiable function $\phi\in C^{1}{\left[0,a\right]}$, define the function $\mathcal{F}$ via the double integral,

$$\mathcal{F}{\left[\phi\right]}{\left(a\right)}:=\int_{0}^{2a}\mathrm{d}x\int_{0}^{\sqrt{2ax-x^{2}}}\mathrm{d}y\,\frac{x\left(x^{2}+y^{2}\right)\phi^{\prime}{\left(y\right)}}{\sqrt{4a^{2}x^{2}-\left(x^{2}+y^{2}\right)^{2}}}.\tag{1}$$


Given $a\in\mathbb{R}_{>0}\land\phi\in C^{1}{\left[0,a\right]}$, we find by changing the order of integration

$$\begin{align} \mathcal{F}{\left[\phi\right]}{\left(a\right)} &=\int_{0}^{2a}\mathrm{d}x\int_{0}^{\sqrt{2ax-x^{2}}}\mathrm{d}y\,\frac{x\left(x^{2}+y^{2}\right)\phi^{\prime}{\left(y\right)}}{\sqrt{4a^{2}x^{2}-\left(x^{2}+y^{2}\right)^{2}}}\\ &=\int_{0}^{a}\mathrm{d}y\int_{a-\sqrt{a^{2}-y^{2}}}^{a+\sqrt{a^{2}-y^{2}}}\mathrm{d}x\,\frac{x\left(x^{2}+y^{2}\right)\phi^{\prime}{\left(y\right)}}{\sqrt{4a^{2}x^{2}-\left(x^{2}+y^{2}\right)^{2}}}\\ &=\int_{0}^{a}\mathrm{d}y\,\phi^{\prime}{\left(y\right)}\int_{a-a\sqrt{1-\left(\frac{y}{a}\right)^{2}}}^{a+a\sqrt{1-\left(\frac{y}{a}\right)^{2}}}\mathrm{d}x\,\frac{x\left(x^{2}+y^{2}\right)}{\sqrt{4a^{2}x^{2}-\left(x^{2}+y^{2}\right)^{2}}}.\tag{2a}\\ \end{align}$$

Rescaling the integration variables by $a$, we then find

$$\begin{align} \mathcal{F}{\left[\phi\right]}{\left(a\right)} &=\int_{0}^{a}\mathrm{d}y\,\phi^{\prime}{\left(y\right)}\int_{a-a\sqrt{1-\left(\frac{y}{a}\right)^{2}}}^{a+a\sqrt{1-\left(\frac{y}{a}\right)^{2}}}\mathrm{d}x\,\frac{x\left(x^{2}+y^{2}\right)}{\sqrt{4a^{2}x^{2}-\left(x^{2}+y^{2}\right)^{2}}}\\ &=\int_{0}^{1}\mathrm{d}t\,\phi^{\prime}{\left(at\right)}\int_{1-\sqrt{1-t^{2}}}^{1+\sqrt{1-t^{2}}}\mathrm{d}u\,\frac{a^{3}u\left(u^{2}+t^{2}\right)}{\sqrt{4u^{2}-\left(u^{2}+t^{2}\right)^{2}}};~~~\small{\left[\left(x,y\right)\mapsto\left(au,at\right)\right]}\\ &=a^{3}\int_{0}^{1}\mathrm{d}t\,\phi^{\prime}{\left(at\right)}\int_{1-\sqrt{1-t^{2}}}^{1+\sqrt{1-t^{2}}}\mathrm{d}u\,\frac{u\left(u^{2}+t^{2}\right)}{\sqrt{4u^{2}-\left(u^{2}+t^{2}\right)^{2}}}\\ &=a^{3}\lim_{b\to1^{-}}\int_{0}^{b}\mathrm{d}t\,\phi^{\prime}{\left(at\right)}\int_{1-\sqrt{1-t^{2}}}^{1+\sqrt{1-t^{2}}}\mathrm{d}u\,\frac{u\left(u^{2}+t^{2}\right)}{\sqrt{4u^{2}-\left(u^{2}+t^{2}\right)^{2}}}.\tag{2b}\\ \end{align}$$

Now, it's obvious from the last line above that the integration over $u$ is independent of the parameter $a$, but it also happens to be independent of the other variable $t$, as we shall see shortly.


Define the auxiliary function $J:\left[0,1\right)\rightarrow\mathbb{R}$ via the definite integral,

$$J{\left(t\right)}:=\int_{1-\sqrt{1-t^{2}}}^{1+\sqrt{1-t^{2}}}\mathrm{d}u\,\frac{u\left(u^{2}+t^{2}\right)}{\sqrt{4u^{2}-\left(u^{2}+t^{2}\right)^{2}}}.\tag{3}$$

Supposing $t\in\left[0,1\right)$ and defining the auxiliary parameter $\sqrt{1-t^{2}}=:r\in\left(0,1\right]$, we obtain

$$\begin{align} J{\left(t\right)} &=\int_{1-\sqrt{1-t^{2}}}^{1+\sqrt{1-t^{2}}}\mathrm{d}u\,\frac{u\left(u^{2}+t^{2}\right)}{\sqrt{4u^{2}-\left(u^{2}+t^{2}\right)^{2}}}\\ &=\int_{1-r}^{1+r}\mathrm{d}u\,\frac{u\left(u^{2}+1-r^{2}\right)}{\sqrt{4u^{2}-\left(u^{2}+1-r^{2}\right)^{2}}};~~~\small{\left[\sqrt{1-t^{2}}=:r\right]}\\ &=\int_{1-r}^{1+r}\mathrm{d}u\,\frac{2u\left(u^{2}+1-r^{2}\right)}{2\sqrt{4r^{2}-\left(1+r^{2}-u^{2}\right)^{2}}}\\ &=\frac12\int_{1-r}^{1+r}\mathrm{d}u\,\frac{2u\left(u^{2}+1-r^{2}\right)}{\sqrt{\left[2r+\left(1+r^{2}-u^{2}\right)\right]\left[2r-\left(1+r^{2}-u^{2}\right)\right]}}\\ &=\frac12\int_{1-r}^{1+r}\mathrm{d}u\,\frac{2u\left(u^{2}+1-r^{2}\right)}{\sqrt{\left[\left(1+r\right)^{2}-u^{2}\right]\left[u^{2}-\left(1-r\right)^{2}\right]}}\\ &=\frac12\int_{\left(1-r\right)^{2}}^{\left(1+r\right)^{2}}\mathrm{d}v\,\frac{\left(v+1-r^{2}\right)}{\sqrt{\left[\left(1+r\right)^{2}-v\right]\left[v-\left(1-r\right)^{2}\right]}};~~~\small{\left[u^{2}=v\right]}\\ &=\frac12\int_{0}^{1}\mathrm{d}w\,4r\,\frac{2\left(2rw-r+1\right)}{\sqrt{\left[4r\left(1-w\right)\right]\left(4rw\right)}};~~~\small{\left[v=4rw+\left(1-r\right)^{2}\right]}\\ &=\int_{0}^{1}\mathrm{d}w\,\frac{2rw-r+1}{\sqrt{w\left(1-w\right)}}\\ &=\int_{0}^{1}\mathrm{d}w\,\frac{1-r}{\sqrt{w\left(1-w\right)}}+\int_{0}^{1}\mathrm{d}w\,\frac{2rw}{\sqrt{w\left(1-w\right)}}\\ &=\int_{0}^{1}\mathrm{d}w\,\frac{1-r}{\sqrt{w\left(1-w\right)}}+\left[2r\sqrt{w\left(1-w\right)}\right]_{w=0}^{w=1}+\int_{0}^{1}\mathrm{d}w\,\frac{r}{\sqrt{w\left(1-w\right)}};~~~\small{I.B.P.s}\\ &=\int_{0}^{1}\frac{\mathrm{d}w}{\sqrt{w\left(1-w\right)}}\\ &=\pi.\tag{4}\\ \end{align}$$


Finally, continuing with the main calculation from where we left off in the last line of $(2b)$ above, we obtain using result $(4)$ that

$$\begin{align} \mathcal{F}{\left[\phi\right]}{\left(a\right)} &=a^{3}\lim_{b\to1^{-}}\int_{0}^{b}\mathrm{d}t\,\phi^{\prime}{\left(at\right)}\int_{1-\sqrt{1-t^{2}}}^{1+\sqrt{1-t^{2}}}\mathrm{d}u\,\frac{u\left(u^{2}+t^{2}\right)}{\sqrt{4u^{2}-\left(u^{2}+t^{2}\right)^{2}}}\\ &=a^{3}\lim_{b\to1^{-}}\int_{0}^{b}\mathrm{d}t\,\phi^{\prime}{\left(at\right)}\,J{\left(t\right)}\\ &=a^{3}\lim_{b\to1^{-}}\int_{0}^{b}\mathrm{d}t\,\phi^{\prime}{\left(at\right)}\,\pi\\ &=\pi\,a^{3}\int_{0}^{1}\mathrm{d}t\,\phi^{\prime}{\left(at\right)}\\ &=\pi\,a^{2}\int_{0}^{a}\mathrm{d}y\,\phi^{\prime}{\left(y\right)};~~~\small{\left[at=y\right]}\\ &=\pi\,a^{2}\left[\phi{\left(a\right)}-\phi{\left(0\right)}\right].\blacksquare\\ \end{align}$$

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  • $\begingroup$ Woah! I am stumped. I have been at it since two days. As the number of substitutions increased, I used to stop midway, thinking the calculation would get complicated. I guess I should've kept on. Thank you so much! $\endgroup$ – user467745 Aug 8 '17 at 11:02

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