Question Number 102606 by bemath last updated on 10/Jul/20
$${what}\:{is}\:{the}\:{volume}\:{of}\:{region} \\ $$$${bounded}\:{by}\:{y}\:={x}^{\mathrm{2}} −\mathrm{2}{x}\:{and} \\ $$$${y}={x}\:{that}\:{is}\:{rotated}\:{about} \\ $$$${y}=\mathrm{4}\:? \\ $$
Commented by bemath last updated on 10/Jul/20
Answered by Ar Brandon last updated on 10/Jul/20
$$\mathrm{Volume}\:,\:\mathrm{V}=\pi\int_{\mathrm{0}} ^{\mathrm{3}} \left\{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{4}\right)^{\mathrm{2}} −\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} \right\}\mathrm{dx} \\ $$
Answered by bobhans last updated on 10/Jul/20
$${vol}\:=\:\pi\overset{\mathrm{3}} {\int}_{\mathrm{0}} \left(\mathrm{4}^{\mathrm{2}} −\left({x}^{\mathrm{2}} −\mathrm{2}{x}\right)^{\mathrm{2}} \right)−\left(\mathrm{4}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)\:{dx} \\ $$$$=\:\pi\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\left({x}^{\mathrm{2}} −\left({x}^{\mathrm{2}} −\mathrm{2}{x}\right)^{\mathrm{2}} \:{dx}\:\right. \\ $$$$=\:\pi\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\left({x}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}{x}\right)\left(\mathrm{2}{x}\right){dx} \\ $$$$=\mathrm{4}\pi\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}{x}\left({x}^{\mathrm{2}} −{x}\right){dx}\:=\:\mathrm{4}\pi\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\:\left({x}^{\mathrm{3}} −{x}^{\mathrm{2}} \right)\:{dx} \\ $$$$=\:\mathrm{4}\pi\:\left\{\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} \right\}_{\mathrm{0}} ^{\mathrm{3}} =\mathrm{4}\pi\:\left\{\frac{\mathrm{81}}{\mathrm{4}}−\frac{\mathrm{27}}{\mathrm{3}}\right\}\: \\ $$$$=\mathrm{4}\pi\:\left\{\frac{\mathrm{45}}{\mathrm{4}}\right\}\:=\:\mathrm{45}\pi \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 10/Jul/20
What difference will it make if it was rather rotated about y=0 with respect to your solution ? ��
Commented by bemath last updated on 10/Jul/20
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