Question Number 220950 by Nicholas666 last updated on 21/May/25
$$ \\ $$$$\:\:\:\:\int_{\:\mathrm{0}} ^{\:\pi} \int_{\:\mathrm{0}} ^{\:\mathrm{1}} \int_{\:\mathrm{0}} ^{\:\:\pi} \:\mathrm{sin}^{\:\mathrm{2}} \:{x}\:+\:{y}\:\mathrm{sin}\:{z}\:{dxdydz}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\pi\:\left(\mathrm{2}\:+\:\pi\right)\:\:\:\:\:\: \\ $$$$ \\ $$
Answered by SdC355 last updated on 21/May/25
![∫_0 ^( π) (sin^2 (r)+y∙sin(z))dr=∫_0 ^( π) sin^2 (x)dx+πysin(z) ∫_0 ^( π) ((1/2)−(1/2)cos(2x))dx=(1/2)π+[(1/4)sin(2x)]_(x=0) ^(x=π) = (π/2)+πysin(z) ∫_0 ^( 1) ((π/2)−πysin(z))dy=(π/2)−[(π/2)y^2 sin(z)]_(y=0) ^(y=1) = (π/2)(1−sin(z)) (π/2)∫_0 ^( π) (1−sin(z))dz=[(π/2)(z+cos(z))]_(z=0) ^(z=π) (π/2)(π+2)](https://www.tinkutara.com/question/Q220960.png)
$$\int_{\mathrm{0}} ^{\:\pi} \:\left(\mathrm{sin}^{\mathrm{2}} \left({r}\right)+{y}\centerdot\mathrm{sin}\left({z}\right)\right)\mathrm{d}{r}=\int_{\mathrm{0}} ^{\:\pi} \:\mathrm{sin}^{\mathrm{2}} \left({x}\right)\mathrm{d}{x}+\pi{y}\mathrm{sin}\left({z}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \:\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{2}{x}\right)\right)\mathrm{d}{x}=\frac{\mathrm{1}}{\mathrm{2}}\pi+\left[\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{2}{x}\right)\right]_{{x}=\mathrm{0}} ^{{x}=\pi} = \\ $$$$\frac{\pi}{\mathrm{2}}+\pi{y}\mathrm{sin}\left({z}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\left(\frac{\pi}{\mathrm{2}}−\pi{y}\mathrm{sin}\left({z}\right)\right)\mathrm{d}{y}=\frac{\pi}{\mathrm{2}}−\left[\frac{\pi}{\mathrm{2}}{y}^{\mathrm{2}} \mathrm{sin}\left({z}\right)\right]_{{y}=\mathrm{0}} ^{{y}=\mathrm{1}} = \\ $$$$\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\mathrm{sin}\left({z}\right)\right) \\ $$$$\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\pi} \:\left(\mathrm{1}−\mathrm{sin}\left({z}\right)\right)\mathrm{d}{z}=\left[\frac{\pi}{\mathrm{2}}\left({z}+\mathrm{cos}\left({z}\right)\right)\right]_{{z}=\mathrm{0}} ^{{z}=\pi} \\ $$$$\frac{\pi}{\mathrm{2}}\left(\pi+\mathrm{2}\right) \\ $$