Question Number 167588 by AlbertEinstein last updated on 20/Mar/22
$$\mathrm{1}+\mathrm{1}=¿ \\ $$
Answered by chhaythean last updated on 20/Mar/22
![=(1/3)∫_0 ^(2π) ∫_0 ^(π/4) sinϕcos^3 θdϕdθ =(1/3)∫_0 ^(2π) cos^3 θ×[−cosϕ]_0 ^(π/4) dθ =(1/3)×[−(((√2)/2)−1)](∫_0 ^(2π) cosθdθ−∫_0 ^(2π) sin^2 θd(sinθ)) =(1/3)×[−(((√2)/2)−1)]×0 So determinant (((∫_0 ^(2π) ∫_0 ^(π/4) ∫_0 ^(cosθ) r^2 sinϕdrdϕdθ=0)))](https://www.tinkutara.com/question/Q167590.png)
$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{sin}\varphi\mathrm{cos}^{\mathrm{3}} \theta\mathrm{d}\varphi\mathrm{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{cos}^{\mathrm{3}} \theta×\left[−\mathrm{cos}\varphi\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×\left[−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\mathrm{1}\right)\right]\left(\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{cos}\theta\mathrm{d}\theta−\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{sin}^{\mathrm{2}} \theta\mathrm{d}\left(\mathrm{sin}\theta\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×\left[−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\mathrm{1}\right)\right]×\mathrm{0} \\ $$$$\mathrm{So}\:\begin{array}{|c|}{\int_{\mathrm{0}} ^{\mathrm{2}\pi} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \int_{\mathrm{0}} ^{\mathrm{cos}\theta} \mathrm{r}^{\mathrm{2}} \mathrm{sin}\varphi\mathrm{drd}\varphi\mathrm{d}\theta=\mathrm{0}}\\\hline\end{array} \\ $$