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Author: Tinku Tara

F-x-y-z-x-x-2-y-2-z-2-e-1-y-x-2-y-2-z-2-e-2-z-x-2-y-2-z-2-e-3-x-2-y-2-z-2-R-2-S-F-dS-

Question Number 219340 by SdC355 last updated on 23/Apr/25 $$\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\left({x},{y},{z}\right)=−\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} −\frac{{y}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} −\frac{{z}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}}…

dx-1-sin-3-x-cos-3-x-

Question Number 219341 by alcohol last updated on 23/Apr/25 $$\int\frac{{dx}}{\mathrm{1}\:+\:{sin}^{\mathrm{3}} {x}\:+\:{cos}^{\mathrm{3}} {x}} \\ $$ Commented by Ghisom last updated on 23/Apr/25 $$\mathrm{simply}\:\mathrm{use}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{decompose} \\ $$$$\mathrm{which}\:\mathrm{leads}\:\mathrm{to} \\…

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…