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Author: Tinku Tara

F-xe-1-ye-2-ze-3-S-u-v-2-3sin-u-cos-v-2-3sin-v-sin-u-3cos-u-u-0-2pi-v-0-2pi-S-F-dS-

Question Number 219338 by SdC355 last updated on 23/Apr/25 $$\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}=−{x}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} −{y}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} −{z}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} \\ $$$$\overset{\rightarrow} {\boldsymbol{\mathcal{S}}}\left({u},{v}\right)=\begin{cases}{\left(\mathrm{2}+\mathrm{3sin}\left({u}\right)\right)\mathrm{cos}\left({v}\right)}\\{\left(\mathrm{2}+\mathrm{3sin}\left({v}\right)\right)\mathrm{sin}\left({u}\right)}\\{\mathrm{3cos}\left({u}\right)}\end{cases} \\ $$$${u}\in\left[\mathrm{0},\mathrm{2}\pi\right]\:,\:{v}\in\left[\mathrm{0},\mathrm{2}\pi\right] \\ $$$$\int\int_{\:\boldsymbol{\mathcal{S}}} \:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\centerdot\mathrm{d}\overset{\rightarrow}…

Question-219339

Question Number 219339 by Spillover last updated on 23/Apr/25 Answered by mr W last updated on 23/Apr/25 $$\alpha=\mathrm{45}°+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{3} \\ $$ Answered by…

F-x-y-z-xye-1-yze-2-xye-3-S-u-v-2-v-cos-u-sin-2piv-v-cos-u-2-v-cos-u-cos-2piv-2v-2-u-pi-pi-v-0-pi-2-S-F-dS-

Question Number 219333 by SdC355 last updated on 23/Apr/25 $$\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\left({x},{y},{z}\right)=−{xy}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} +{yz}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} −{xy}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} \\ $$$$\overset{\rightarrow} {\boldsymbol{\mathcal{S}}}\left({u},{v}\right)\begin{cases}{\left(\mathrm{2}+{v}\centerdot\mathrm{cos}\left({u}\right)\right)\mathrm{sin}\left(\mathrm{2}\pi{v}\right)}\\{{v}\centerdot\mathrm{cos}\left({u}\right)}\\{\left(\mathrm{2}+{v}\centerdot\mathrm{cos}\left({u}\right)\right)\mathrm{cos}\left(\mathrm{2}\pi{v}\right)+\mathrm{2}{v}−\mathrm{2}}\end{cases} \\ $$$${u}\in\left[−\pi,\pi\right]\:,\:{v}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right] \\ $$$$\int\int_{\:\boldsymbol{\mathcal{S}}} \:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\centerdot\mathrm{d}\overset{\rightarrow}…

Question-219281

Question Number 219281 by Rojarani last updated on 22/Apr/25 Answered by Ghisom last updated on 22/Apr/25 $${P}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{1} \\ $$$${P}\:'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3} \\ $$$${P}\:''\left({x}\right)=\mathrm{6}{x} \\ $$$$…