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$$\boldsymbol{\mathrm{compute}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{double}}\:\boldsymbol{\mathrm{integral}} \\ $$$$\int_{\boldsymbol{\mathrm{y}}=\mathrm{0}} ^{\mathrm{1}} \int_{\boldsymbol{\mathrm{x}}=\mathrm{0}} ^{\mathrm{2}} \boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{dxdy}}\:\boldsymbol{\mathrm{and}}\:\:\int_{\boldsymbol{\mathrm{y}}=\mathrm{0}} ^{\mathrm{1}} \int_{\boldsymbol{\mathrm{x}}=\mathrm{0}} ^{\mathrm{2}} \boldsymbol{\mathrm{y}}^{\mathrm{2}} \boldsymbol{\mathrm{dxdy}} \\ $$$$ \\ $$
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$${calculate}\:{the}\:{volume}\:{of}\:{a}\:{sphere} \\ $$$${using}\:{double}\:{integral} \\ $$
Question Number 226351 Answers: 1 Comments: 0
$$\mathrm{Prove}\:\mathrm{M}\ddot {\mathrm{o}bious}\:\mathrm{String}\:\mathrm{is}\:\mathrm{Not}\:\mathrm{a}\:\mathrm{Orientated}\:\mathrm{Surface}. \\ $$$$\sigma\left({u},\theta\right)=\begin{cases}{\left(\mathrm{1}−{u}\centerdot\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)\right)\mathrm{cos}\left(\theta\right)}\\{\left(\mathrm{1}−{u}\centerdot\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)\right)\mathrm{sin}\left(\theta\right)}\\{{u}\centerdot\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right)}\end{cases}\:\:,\:−\frac{\mathrm{1}}{\mathrm{2}}\leq{u}\leq\frac{\mathrm{1}}{\mathrm{2}}\:,\:\mathrm{0}\leq\theta\leq\mathrm{2}\pi \\ $$
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Question Number 226293 Answers: 2 Comments: 1
$${A}\:{hemispherical}\:{bowl}\:{of}\:{radius}\:{R} \\ $$$$\:{with}\:{maimum}\:{water}\:{in}\:{it}\:{without} \\ $$$${needing}\:{to}\:{spill}\:{is}\:{spinning}\:{with}\:{the} \\ $$$${content}\:{at}\:{constant}\:\omega.\:{Find}\:{volume} \\ $$$${of}\:{water}\:{in}\:{bowl}. \\ $$$$ \\ $$
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