Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 216350 by issac last updated on 05/Feb/25

$$\mathcal{S}\mathrm{is}\mathrm{the}\mathrm{boundary}\mathrm{surface}\mathrm{of}\mathrm{the}$$$$\mathrm{surrounded}\mathrm{by}\mathrm{the}\mathrm{cylinder}{x}^2+y^2=9$$$$\mathrm{and}\mathrm{plane}z=0,z=2\mathrm{and}$$$$\mathrm{and}\mathrm{vector}\mathrm{Field}\overrightarrow{\mathbf{F}};{\mathbb{R}}^3\to {\mathbb{R}}^3$$$$\overrightarrow{\mathbf{F}}(x,y,z)=3y{\overrightarrow{\mathbf{e}}}_1+yz{\overrightarrow{\mathbf{e}}}_2-{xyz}^5{\overrightarrow{\mathbf{e}}}_3$$$$\underset{\mathcal{S}}{\int \int }\overrightarrow{\mathbf{F}}\cdot \mathrm{d}\overrightarrow{\mathbf{S}}=?$$

Answered by MrGaster last updated on 06/Feb/25

$$\int {\int }_{\mathcal{S}}\overrightarrow{\mathbf{F}}\cdot d\overrightarrow{\mathit{S}}=\int \int {\int }_V▽\cdot \overrightarrow{\mathbf{F}}dV$$$$▽\cdot \overrightarrow{\mathit{F}}=\frac{\partial }{\partial x}\left(3y\right)+\frac{\partial }{\partial y}\left(yz\right)+\frac{\partial }{\partial z}(-{xyz}^5)=0+z-5{xyz}^4$$$$\int \int {\int }_V(z-5{xyz}^4)dV={\int }_0^2{\int }_0^{2\pi }{\int }_0^3(r\cos \theta -5r^2\cos \theta z^4)rdrd\theta dz$$$$={\int }_0^2{\int }_0^{2\pi }{[\frac{r^3}{3}\cos \theta -\frac{5r^4}{4}\cos \theta z^4]}_0^{3}d\theta dz$$$$={\int }_0^2{\int }_0^{2\pi }(9\cos \theta -\frac{405}{4}\cos \theta z^4)d\theta dz$$$$={\int }_0^2{[9\sin \theta -\frac{405}{4}\sin \theta z^4]}_0^{2\pi }dz=0$$$$\int {\int }_{\mathcal{S}}\overrightarrow{\mathbf{F}}\cdot d\overrightarrow{\mathit{S}}=0$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com