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Question Number 216351 by issac last updated on 05/Feb/25

$$\mathrm{Vector}\mathrm{field}\overrightarrow{\mathbf{F}};{\mathbb{R}}^3\to {\mathbb{R}}^3$$$$\overrightarrow{\mathbf{F}}(x,y,z)=xy{\overrightarrow{\mathbf{e}}}_1-5y{\overrightarrow{\mathbf{e}}}_2-3yz{\overrightarrow{\mathbf{e}}}_3$$$$\underset{\mathcal{S};x^2+y^2+z^2=r^2}{\int \int }\overrightarrow{\mathbf{F}}\cdot \mathrm{d}\overrightarrow{\mathbf{S}}=?$$

Answered by MrGaster last updated on 06/Feb/25

$$▽\cdot \overrightarrow{\mathbf{F}}=\frac{\partial }{\partial x}\left(\partial y\right)+\frac{\partial }{\partial y}(-5y)+\frac{\partial }{\partial z}(-3yz)$$$$=y-5-3y=-2y-5$$$$\int \int {\int }_V(-2y-5)dV$$$$x=r\sin \theta \cos \varphi ,y=r\sin \theta \sin \varphi ,z=r\cos \theta$$$$dV=r^2\sin \theta drd\theta d\varphi$$$$0\le r\le r,0\le \theta \le \pi ,0\le \varphi \le 2\pi$$$$\int \int {\int }_V(-2y-5)dV={\int }_0^r{\int }_0^{\pi }{\int }_0^{2\pi }(-2r\sin \theta \varphi -5)r^2\sin \theta d\varphi d\theta dr$$$$={\int }_0^r{\int }_0^{\pi }[-2r^3{\sin }^2\theta {\int }_0^{2\pi }\sin \varphi d\varphi -{5\mathrm{r}}^2\sin \theta {\int }_0^{2\pi }d\varphi ]d\theta dr$$$${\int }_0^r{\int }_0^{\pi }[0-10\pi r^2\sin \theta ]d\theta dr$$$$=-10\pi {\int }_0^r{\int }_0^{\pi }{r}^2\sin \theta d\theta dr$$$$=-10\pi {\int }_0^r{\int }_0^{\pi }{[-r^2\cos \theta ]}_0^{\pi }dr$$$$=-10\pi {\int }_0^{r}2r^{2}dr$$$$=-20\pi {\left[\frac{r^3}{3}\right]}_0^r$$$$=\overline{)\begin{array}{c}-\frac{20\pi r^3}{3}\end{array}}$$

Commented by issac last updated on 06/Feb/25

$$\mathrm{Thx}!\mathrm{Mr}\mathrm{Gaster}\mathrm{you}\mathrm{r}\mathrm{my}\mathrm{Hero}:⟩$$

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