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Question Number 204522 by universe last updated on 20/Feb/24

Commented by witcher3 last updated on 20/Feb/24

$${\varphi }^{\prime }\left(\mathrm{y}\right)\mathrm{not}\mathrm{mor}\mathrm{information}\mathrm{about}\varphi \mathrm{finction}?$$$$$$

Commented by universe last updated on 21/Feb/24

$$sir\varphi \left(y\right)isaanydifferentiablefunctionof$$$$y$$$$andanswerofthisquestionis\pi a^2[\varphi \left(a\right)-\varphi \left(0\right)]$$$$ithinkchangingorderthenintegrate$$

Commented by witcher3 last updated on 21/Feb/24

$$\mathrm{i}\mathrm{will}\mathrm{tchek}\mathrm{i}\mathrm{found}\mathrm{a}\mathrm{factor}\mathrm{of}1/2$$

Answered by witcher3 last updated on 21/Feb/24

$$\mathrm{x}\to \mathrm{at}$$$$\mathrm{y}\to \mathrm{as}$$$${\int }_0^2{\int }_0^{\sqrt{2\mathrm{t}-{\mathrm{t}}^2}}\frac{{\phi }^{\prime }\left(\mathrm{as}\right){\mathrm{a}}^5({\mathrm{s}}^2+{\mathrm{t}}^2)\mathrm{tdsdt}}{{\mathrm{a}}^2\sqrt{{4\mathrm{t}}^2-{({\mathrm{s}}^2+{\mathrm{t}}^2)}^2}}$$$$0⩽\mathrm{t}⩽2;0⩽\mathrm{s}⩽\sqrt{2\mathrm{t}-{\mathrm{t}}^2}\Rightarrow {\mathrm{s}}^2+{\mathrm{t}}^2-2\mathrm{t}⩽0$$$$\Rightarrow {(\mathrm{t}-1)}^2⩽1-{\mathrm{s}}^2$$$$1-\sqrt{1-{\mathrm{s}}^2}⩽\mathrm{t}⩽1+\sqrt{1-{\mathrm{s}}^2}$$$$\mathrm{s}\in [0,1]$$$${\int }_0^1{\mathrm{a}}^3{\phi }^{\prime }\left(\mathrm{as}\right){\int }_{1-\sqrt{1-{\mathrm{s}}^2}}^{1+\sqrt{1-{\mathrm{s}}^2}}\frac{({\mathrm{s}}^2+{\mathrm{t}}^2)\mathrm{sdsdt}}{\sqrt{{4\mathrm{t}}^2-{({\mathrm{t}}^2+{\mathrm{s}}^2)}^2}}$$$${\mathrm{t}}^2+{\mathrm{s}}^2=\mathrm{u}\Rightarrow \mathrm{tdt}=\frac{\mathrm{du}}{2}$$$${\int }_0^1{\mathrm{a}}^3{\phi }^{\prime }\left(\mathrm{as}\right){\int }_{2-2\sqrt{1-{\mathrm{s}}^2}}^{2+2\sqrt{-{\mathrm{s}}^2}}\frac{\mathrm{udu}}{2\sqrt{-{\mathrm{u}}^2+4(\mathrm{u}-{\mathrm{s}}^2)}}\mathrm{ds}$$$${\int }_0^1{\mathrm{a}}^3{\phi }^{\prime }\left(\mathrm{as}\right){\int }_{2-2\sqrt{1-{\mathrm{s}}^2}}^{2+2\sqrt{1-{\mathrm{s}}^2}}\frac{\mathrm{udu}}{2\sqrt{-{(\mathrm{u}-2)}^2+4-{4\mathrm{s}}^2}}\mathrm{ds}$$$$\mathrm{s}=\sin \left(\mathrm{r}\right)\Rightarrow \mathrm{ds}=\cos \left(\mathrm{r}\right)\mathrm{dr}$$$${\int }_0^{\frac{\pi }{2}}{\mathrm{a}}^3{\phi }^{\prime }\left(\mathrm{asin}\left(\mathrm{r}\right)\right)\cos \left(\mathrm{r}\right){\int }_{2-2\cos \left(\mathrm{r}\right)}^{2+2\cos \left(\mathrm{r}\right)}\frac{\mathrm{udu}}{2\sqrt{{4\cos }^2\left(\mathrm{r}\right)-{(\mathrm{u}-2)}^2}}$$$$={\mathrm{a}}^3{\int }_0^{\frac{\pi }{2}}\frac{{\varphi }^{\prime }\left(\mathrm{asin}\left(\mathrm{r}\right)\right)\cos \left(\mathrm{r}\right)}{2}{\int }_{2-2\cos \left(\mathrm{r}\right)}^{2+2\cos \left(\mathrm{r}\right)}\frac{\mathrm{dudr}}{2\cos \left(\mathrm{r}\right)\sqrt{1-{\left(\frac{\mathrm{u}-2}{2\cos \left(\mathrm{r}\right)}\right)}^2}}$$$$=\frac{{\mathrm{a}}^3}{2}{\int }_0^{\frac{\pi }{2}}{\varphi }^{\prime }\left(\mathrm{asin}\left(\mathrm{r}\right)\right)\cos \left(\mathrm{r}\right)[-{\cos }^{-1}{\left(\frac{\mathrm{u}-2}{2\cos \left(\mathrm{r}\right)}\right]}_{2-2\cos \left(\mathrm{r}\right)}^{2+2\cos \left(\mathrm{r}\right)}]\mathrm{dr}$$$$=\frac{{\mathrm{a}}^3}{2}{\int }_0^{\frac{\pi }{2}}{\phi }^{\prime }\left(\mathrm{asin}\left(\mathrm{r}\right)\right)\cos \left(\mathrm{r}\right)[-{\cos }^{-}\left(1\right)+{\cos }^{-}(-1)]$$$$=\frac{\pi {\mathrm{a}}^2}{2}{\int }_0^{\frac{\pi }{2}}{\phi }^{\prime }\left(\mathrm{asin}\left(\mathrm{r}\right)\right)\mathrm{d}\left(\mathrm{asin}\left(\mathrm{r}\right)\right)=\frac{\pi {\mathrm{a}}^2}{2}{\left[\phi \left(\mathrm{asin}\left(\mathrm{r}\right)\right)\right]}_0^{\frac{\pi }{2}}$$$$=\frac{\pi {\mathrm{a}}^2}{2}[\phi \left(\mathrm{a}\right)-\phi \left(0\right)]$$$$$$$$$$

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