Question Number 204522 by universe last updated on 20/Feb/24
Commented by witcher3 last updated on 20/Feb/24
$$\phi'\left(\mathrm{y}\right)\:\:\mathrm{not}\:\mathrm{mor}\:\mathrm{information}\:\mathrm{about}\:\phi\:\mathrm{finction}? \\ $$$$ \\ $$
Commented by universe last updated on 21/Feb/24
![sir φ(y) is a any differentiable function of y and answer of this question is πa^2 [φ(a)−φ(0)] i think changing order then integrate](https://www.tinkutara.com/question/Q204538.png)
$$\:{sir}\:\phi\left({y}\right)\:{is}\:{a}\:{any}\:{differentiable}\:{function}\:{of} \\ $$$${y}\: \\ $$$${and}\:{answer}\:{of}\:{this}\:{question}\:{is}\:\:\pi{a}^{\mathrm{2}} \left[\phi\left({a}\right)−\phi\left(\mathrm{0}\right)\right] \\ $$$${i}\:{think}\:{changing}\:{order}\:{then}\:{integrate} \\ $$
Commented by witcher3 last updated on 21/Feb/24
$$\mathrm{i}\:\mathrm{will}\:\mathrm{tchek}\:\mathrm{i}\:\mathrm{found}\:\mathrm{a}\:\mathrm{factor}\:\mathrm{of}\:\mathrm{1}/\mathrm{2} \\ $$
Answered by witcher3 last updated on 21/Feb/24
![x→at y→as ∫_0 ^2 ∫_0 ^(√(2t−t^2 )) ((ϕ′(as)a^5 (s^2 +t^2 )tdsdt)/(a^2 (√(4t^2 −(s^2 +t^2 )^2 )))) 0≤t≤2; 0≤s≤(√(2t−t^2 ))⇒ s^2 +t^2 −2t≤0 ⇒(t−1)^2 ≤1−s^2 1−(√(1−s^2 ))≤t≤1+(√(1−s^2 )) s∈[0,1] ∫_0 ^1 a^3 ϕ′(as)∫_(1−(√(1−s^2 ))) ^(1+(√(1−s^2 ))) (((s^2 +t^2 )sdsdt)/( (√(4t^2 −(t^2 +s^2 )^2 )))) t^2 +s^2 =u⇒tdt=(du/2) ∫_0 ^1 a^3 ϕ′(as)∫_(2−2(√(1−s^2 ))) ^(2+2(√(−s^2 ))) ((udu)/(2(√(−u^2 +4(u−s^2 )))))ds ∫_0 ^1 a^3 ϕ′(as)∫_(2−2(√(1−s^2 ))) ^(2+2(√(1−s^2 ))) ((udu)/(2(√(−(u−2)^2 +4−4s^2 ))))ds s=sin(r)⇒ds=cos(r)dr ∫_0 ^(π/2) a^3 ϕ′(asin(r))cos(r)∫_(2−2cos(r)) ^(2+2cos(r)) ((udu)/(2(√(4cos^2 (r)−(u−2)^2 )))) =a^3 ∫_0 ^(π/2) ((φ′(asin(r))cos(r))/2)∫_(2−2cos(r)) ^(2+2cos(r)) ((dudr)/(2cos(r)(√(1−(((u−2)/(2cos(r))))^2 )))) =(a^3 /2)∫_0 ^(π/2) φ′(asin(r))cos(r)[−cos^(−1) (((u−2)/(2cos(r)))]_(2−2cos(r)) ^(2+2cos(r)) ]dr =(a^3 /2)∫_0 ^(π/2) ϕ′(asin(r))cos(r)[−cos^− (1)+cos^− (−1)] =((πa^2 )/2)∫_0 ^(π/2) ϕ′(asin(r))d(asin(r))=((πa^2 )/2)[ϕ(asin(r))]_0 ^(π/2) =((πa^2 )/2)[ϕ(a)−ϕ(0)]](https://www.tinkutara.com/question/Q204539.png)
$$\mathrm{x}\rightarrow\mathrm{at} \\ $$$$\mathrm{y}\rightarrow\mathrm{as} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} \int_{\mathrm{0}} ^{\sqrt{\mathrm{2t}−\mathrm{t}^{\mathrm{2}} }} \frac{\varphi'\left(\mathrm{as}\right)\mathrm{a}^{\mathrm{5}} \left(\mathrm{s}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} \right)\mathrm{tdsdt}}{\mathrm{a}^{\mathrm{2}} \sqrt{\mathrm{4t}^{\mathrm{2}} −\left(\mathrm{s}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$$\mathrm{0}\leqslant\mathrm{t}\leqslant\mathrm{2};\:\:\:\:\:\mathrm{0}\leqslant\mathrm{s}\leqslant\sqrt{\mathrm{2t}−\mathrm{t}^{\mathrm{2}} }\Rightarrow\:\:\:\mathrm{s}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} −\mathrm{2t}\leqslant\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} \leqslant\mathrm{1}−\mathrm{s}^{\mathrm{2}} \\ $$$$\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{s}^{\mathrm{2}} }\leqslant\mathrm{t}\leqslant\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{s}^{\mathrm{2}} } \\ $$$$\mathrm{s}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{a}^{\mathrm{3}} \varphi'\left(\mathrm{as}\right)\int_{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{s}^{\mathrm{2}} }} ^{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{s}^{\mathrm{2}} }} \frac{\left(\mathrm{s}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} \right)\mathrm{sdsdt}}{\:\sqrt{\mathrm{4t}^{\mathrm{2}} −\left(\mathrm{t}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$$\mathrm{t}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} =\mathrm{u}\Rightarrow\mathrm{tdt}=\frac{\mathrm{du}}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{a}^{\mathrm{3}} \varphi'\left(\mathrm{as}\right)\int_{\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}−\mathrm{s}^{\mathrm{2}} }} ^{\mathrm{2}+\mathrm{2}\sqrt{−\mathrm{s}^{\mathrm{2}} }} \frac{\mathrm{udu}}{\mathrm{2}\sqrt{−\mathrm{u}^{\mathrm{2}} +\mathrm{4}\left(\mathrm{u}−\mathrm{s}^{\mathrm{2}} \right)}}\mathrm{ds} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{a}^{\mathrm{3}} \varphi'\left(\mathrm{as}\right)\int_{\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}−\mathrm{s}^{\mathrm{2}} }} ^{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{1}−\mathrm{s}^{\mathrm{2}} }} \frac{\mathrm{udu}}{\mathrm{2}\sqrt{−\left(\mathrm{u}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{4}−\mathrm{4s}^{\mathrm{2}} }}\mathrm{ds} \\ $$$$\mathrm{s}=\mathrm{sin}\left(\mathrm{r}\right)\Rightarrow\mathrm{ds}=\mathrm{cos}\left(\mathrm{r}\right)\mathrm{dr} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{a}^{\mathrm{3}} \varphi'\left(\mathrm{asin}\left(\mathrm{r}\right)\right)\mathrm{cos}\left(\mathrm{r}\right)\int_{\mathrm{2}−\mathrm{2cos}\left(\mathrm{r}\right)} ^{\mathrm{2}+\mathrm{2cos}\left(\mathrm{r}\right)} \frac{\mathrm{udu}}{\mathrm{2}\sqrt{\mathrm{4cos}^{\mathrm{2}} \left(\mathrm{r}\right)−\left(\mathrm{u}−\mathrm{2}\right)^{\mathrm{2}} }} \\ $$$$=\mathrm{a}^{\mathrm{3}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\phi'\left(\mathrm{asin}\left(\mathrm{r}\right)\right)\mathrm{cos}\left(\mathrm{r}\right)}{\mathrm{2}}\int_{\mathrm{2}−\mathrm{2cos}\left(\mathrm{r}\right)} ^{\mathrm{2}+\mathrm{2cos}\left(\mathrm{r}\right)} \frac{\mathrm{dudr}}{\mathrm{2cos}\left(\mathrm{r}\right)\sqrt{\mathrm{1}−\left(\frac{\mathrm{u}−\mathrm{2}}{\mathrm{2cos}\left(\mathrm{r}\right)}\right)^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{a}^{\mathrm{3}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \phi'\left(\mathrm{asin}\left(\mathrm{r}\right)\right)\mathrm{cos}\left(\mathrm{r}\right)\left[−\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{u}−\mathrm{2}}{\mathrm{2cos}\left(\mathrm{r}\right)}\right]_{\mathrm{2}−\mathrm{2cos}\left(\mathrm{r}\right)} ^{\mathrm{2}+\mathrm{2cos}\left(\mathrm{r}\right)} \right]\mathrm{dr} \\ $$$$=\frac{\mathrm{a}^{\mathrm{3}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \varphi'\left(\mathrm{asin}\left(\mathrm{r}\right)\right)\mathrm{cos}\left(\mathrm{r}\right)\left[−\mathrm{cos}^{−} \left(\mathrm{1}\right)+\mathrm{cos}^{−} \left(−\mathrm{1}\right)\right] \\ $$$$=\frac{\pi\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \varphi'\left(\mathrm{asin}\left(\mathrm{r}\right)\right)\mathrm{d}\left(\mathrm{asin}\left(\mathrm{r}\right)\right)=\frac{\pi\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\left[\varphi\left(\mathrm{asin}\left(\mathrm{r}\right)\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\pi\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\left[\varphi\left(\mathrm{a}\right)−\varphi\left(\mathrm{0}\right)\right] \\ $$$$ \\ $$$$ \\ $$
Commented by universe last updated on 21/Feb/24
![and ans is πa^2 [φ(a)−φ(0)] sir can you share your telegram i′d ?? beacouse i added a good math talegram group](https://www.tinkutara.com/question/Q204557.png)
$${and}\:{ans}\:{is}\:\pi{a}^{\mathrm{2}} \left[\phi\left({a}\right)−\phi\left(\mathrm{0}\right)\right] \\ $$$${sir}\:{can}\:{you}\:{share}\:{your}\:{telegram}\:{i}'{d}\:?? \\ $$$${beacouse}\:{i}\:{added}\:{a}\:{good}\:{math}\:{talegram} \\ $$$${group} \\ $$