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compute-the-double-integral-y-0-1-x-0-2-x-2-dxdy-and-y-0-1-x-0-2-y-2-dxdy-

Question Number 226366 by klipto last updated on 26/Nov/25 $$\boldsymbol{\mathrm{compute}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{double}}\:\boldsymbol{\mathrm{integral}} \\ $$$$\int_{\boldsymbol{\mathrm{y}}=\mathrm{0}} ^{\mathrm{1}} \int_{\boldsymbol{\mathrm{x}}=\mathrm{0}} ^{\mathrm{2}} \boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{dxdy}}\:\boldsymbol{\mathrm{and}}\:\:\int_{\boldsymbol{\mathrm{y}}=\mathrm{0}} ^{\mathrm{1}} \int_{\boldsymbol{\mathrm{x}}=\mathrm{0}} ^{\mathrm{2}} \boldsymbol{\mathrm{y}}^{\mathrm{2}} \boldsymbol{\mathrm{dxdy}} \\ $$$$ \\…

Question-226386

Question Number 226386 by fantastic2 last updated on 26/Nov/25 Commented by fantastic2 last updated on 27/Nov/25 $${l}=\mathrm{5}\pi{R} \\ $$$${a}\:{force}\:{F}\:{is}\:{given}\:{horizontally} \\ $$$${for}\:\mathrm{1}{sec}\:{such}\:{that}\:{the}\:{speed}\:{of}\:{the}\:{ball} \\ $$$${becomes}\:\mathrm{0}\:{just}\:{before}\:{touching}\:{the}\:{cylinder} \\ $$$${find}\:{the}\:{time}\:{taken}\:{to}\:{touch}\:{and}\:{the}\:{force}…

Prove-Mo-bious-String-is-Not-a-Orientated-Surface-u-1-u-sin-1-2-cos-1-u-sin-1-2-sin-u-cos-1-2-1-2-u-1-2-0-2pi-

Question Number 226351 by Lara2440 last updated on 26/Nov/25 $$\mathrm{Prove}\:\mathrm{M}\ddot {\mathrm{o}bious}\:\mathrm{String}\:\mathrm{is}\:\mathrm{Not}\:\mathrm{a}\:\mathrm{Orientated}\:\mathrm{Surface}. \\ $$$$\sigma\left({u},\theta\right)=\begin{cases}{\left(\mathrm{1}−{u}\centerdot\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)\right)\mathrm{cos}\left(\theta\right)}\\{\left(\mathrm{1}−{u}\centerdot\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)\right)\mathrm{sin}\left(\theta\right)}\\{{u}\centerdot\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)}\end{cases}\:\:,\:−\frac{\mathrm{1}}{\mathrm{2}}\leq{u}\leq\frac{\mathrm{1}}{\mathrm{2}}\:,\:\mathrm{0}\leq\theta\leq\mathrm{2}\pi \\ $$ Answered by Lara2440 last updated on 26/Nov/25 $$\: \\ $$$$\mathrm{To}\:\mathrm{show}\:\mathrm{that}\:\mathrm{this}\:\mathrm{Surface}\:\mathrm{is}\:\mathrm{Oriented},…

Question-226291

Question Number 226291 by fantastic2 last updated on 25/Nov/25 Answered by mr W last updated on 25/Nov/25 $${m}\omega^{\mathrm{2}} \left({r}+{l}\:\mathrm{sin}\:\theta\right)={mg}\:\mathrm{tan}\:\theta \\ $$$$\Rightarrow\omega=\sqrt{\frac{{g}\:\mathrm{tan}\:\theta}{{r}+{l}\:\mathrm{sin}\:\theta}} \\ $$ Commented by…