Question Number 224907 by fantastic last updated on 11/Oct/25
$${Let}\:{B}=\left\{\left({x},{y},{z}\right)\in\mathbb{R}^{\mathrm{3}} :{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \leqslant\mathrm{1}\right\} \\ $$$${and}\:{define}\:{u}\left({x},{y},{z}\right)=\mathrm{sin}\:\left(\left(\mathrm{1}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} −{z}^{\mathrm{2}} \right)^{\mathrm{2}} \right) \\ $$$${for}\:\left({x},{y},{z}\right)\in{B}. \\ $$$${Then}\:{the}\:{value}\:{of} \\ $$$$\underset{{B}} {\int\int\int}\left(\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {u}}{\partial{z}^{\mathrm{2}} }\right){dxdydz}\:{is}\:………. \\ $$
Answered by fkwow344 last updated on 12/Oct/25
![x=rsin(θ)cos(ρ) y=rsin(θ)sin(ρ) z=rcos(θ) ▽^2 u=(1/(r^2 sin(θ))) ((∂ )/∂r)(r∙(∂u/∂r))+(1/(r^2 sin(θ))) ((∂ )/∂θ)(sin(θ)(∂u/∂θ))+(1/(r^2 sin^2 (θ))) (∂^2 u/∂ρ^2 ) dV=r^2 sin(θ)dr∧dθ∧dρ ∫∫∫_( B) [ ((∂ )/∂r)(r∙(∂u/∂r))+((∂ )/∂θ)(sin(θ)(∂u/∂θ))+(1/(sin(θ))) (∂^2 u/∂ρ^2 ) ]dr∧dθ∧dρ u_B (r,θ,ρ)=sin((1−r^2 )^2 ) ∫∫∫_( B) ((∂ )/∂r)(r∙((∂u(r,θ,ρ))/∂r)) dr∧dθ∧dρ ∫∫∫_( B) ▽^2 u_B (r,θ,ρ) dr∧dθ∧dρ=0 because ∫_0 ^( 1) ((d )/dr)(r∙((du_B (r,θ,ρ))/dr)) dr=0 ∫_0 ^( 1) ((d )/dr)(r∙(du_B /dr)) dr=[r∙((du_B (r,θ,ρ))/dr)]_(r=0) ^(r=1) =[((du_B (r,θ,ρ))/dr)]_(r=1) [(du/dr)]_(r=1) =[−4r(1−r^2 )cos((r^2 −1)^2 )]_(r=1) =0. −](https://www.tinkutara.com/question/Q224912.png)
$${x}={r}\mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\rho\right) \\ $$$${y}={r}\mathrm{sin}\left(\theta\right)\mathrm{sin}\left(\rho\right) \\ $$$${z}={r}\mathrm{cos}\left(\theta\right) \\ $$$$\bigtriangledown^{\mathrm{2}} {u}=\frac{\mathrm{1}}{{r}^{\mathrm{2}} \mathrm{sin}\left(\theta\right)}\:\frac{\partial\:\:}{\partial{r}}\left({r}\centerdot\frac{\partial{u}}{\partial{r}}\right)+\frac{\mathrm{1}}{{r}^{\mathrm{2}} \mathrm{sin}\left(\theta\right)}\:\frac{\partial\:\:}{\partial\theta}\left(\mathrm{sin}\left(\theta\right)\frac{\partial{u}}{\partial\theta}\right)+\frac{\mathrm{1}}{{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right)}\:\frac{\partial^{\mathrm{2}} {u}}{\partial\rho^{\mathrm{2}} } \\ $$$$\mathrm{d}{V}={r}^{\mathrm{2}} \mathrm{sin}\left(\theta\right)\mathrm{d}{r}\wedge\mathrm{d}\theta\wedge\mathrm{d}\rho \\ $$$$\int\int\int_{\:\mathcal{B}} \left[\:\frac{\partial\:\:}{\partial{r}}\left({r}\centerdot\frac{\partial{u}}{\partial{r}}\right)+\frac{\partial\:\:}{\partial\theta}\left(\mathrm{sin}\left(\theta\right)\frac{\partial{u}}{\partial\theta}\right)+\frac{\mathrm{1}}{\mathrm{sin}\left(\theta\right)}\:\frac{\partial^{\mathrm{2}} {u}}{\partial\rho^{\mathrm{2}} }\:\right]\mathrm{d}{r}\wedge\mathrm{d}\theta\wedge\mathrm{d}\rho \\ $$$${u}_{\mathcal{B}} \left({r},\theta,\rho\right)=\mathrm{sin}\left(\left(\mathrm{1}−{r}^{\mathrm{2}} \right)^{\mathrm{2}} \right) \\ $$$$\int\int\int_{\:\mathcal{B}} \:\frac{\partial\:\:}{\partial{r}}\left({r}\centerdot\frac{\partial{u}\left({r},\theta,\rho\right)}{\partial{r}}\right)\:\mathrm{d}{r}\wedge\mathrm{d}\theta\wedge\mathrm{d}\rho \\ $$$$\int\int\int_{\:\mathcal{B}} \:\:\bigtriangledown^{\mathrm{2}} {u}_{\mathcal{B}} \left({r},\theta,\rho\right)\:\mathrm{d}{r}\wedge\mathrm{d}\theta\wedge\mathrm{d}\rho=\mathrm{0} \\ $$$$\mathrm{because}\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{\mathrm{d}\:\:}{\mathrm{d}{r}}\left({r}\centerdot\frac{\mathrm{d}{u}_{\mathcal{B}} \left({r},\theta,\rho\right)}{\mathrm{d}{r}}\right)\:\mathrm{d}{r}=\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{\mathrm{d}\:\:}{\mathrm{d}{r}}\left({r}\centerdot\frac{\mathrm{d}{u}_{\mathcal{B}} }{\mathrm{d}{r}}\right)\:\mathrm{d}{r}=\left[{r}\centerdot\frac{\mathrm{d}{u}_{\mathcal{B}} \left({r},\theta,\rho\right)}{\mathrm{d}{r}}\right]_{{r}=\mathrm{0}} ^{{r}=\mathrm{1}} =\left[\frac{\mathrm{d}{u}_{\mathcal{B}} \left({r},\theta,\rho\right)}{\mathrm{d}{r}}\right]_{{r}=\mathrm{1}} \\ $$$$\left[\frac{\mathrm{d}{u}}{\mathrm{d}{r}}\right]_{{r}=\mathrm{1}} =\left[−\mathrm{4}{r}\left(\mathrm{1}−{r}^{\mathrm{2}} \right)\mathrm{cos}\left(\left({r}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \right)\right]_{{r}=\mathrm{1}} =\mathrm{0}. \\ $$$$\:\:\:\:\:\:\:− \\ $$