Question Number 89804 by peter frank last updated on 19/Apr/20
Answered by jagoll last updated on 19/Apr/20
Commented by jagoll last updated on 19/Apr/20
$$\mathrm{vol}\:=\:\pi\:\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\:\left(\sqrt{\mathrm{x}}\:\sqrt[{\mathrm{4}\:\:}]{\mathrm{x}−\mathrm{1}}\right)^{\mathrm{2}} \:\mathrm{dx} \\ $$$$=\:\pi\:\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\:\left(\mathrm{x}\sqrt{\mathrm{x}−\mathrm{1}}\right)\:\mathrm{dx} \\ $$$$=\:\pi\:\left[\:\frac{\mathrm{2x}\left(\mathrm{x}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}}−\:\frac{\mathrm{4}\left(\mathrm{x}−\mathrm{1}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{15}}\:\underset{\:\:\mathrm{1}} {\right]}^{\mathrm{3}} \\ $$$$=\:\pi\:\left[\:\mathrm{4}\sqrt{\mathrm{2}}\:−\:\frac{\mathrm{16}\sqrt{\mathrm{2}}}{\mathrm{15}}\:\right]\:=\:\frac{\mathrm{44}\pi\sqrt{\mathrm{2}}}{\mathrm{15}} \\ $$$$ \\ $$
Commented by peter frank last updated on 19/Apr/20
$${thank}\:{you} \\ $$
Commented by peter frank last updated on 19/Apr/20
$${why}\:{y}=\mathrm{0}\:{excluded} \\ $$
Commented by jagoll last updated on 19/Apr/20
$$\mathrm{y}\:=\:\mathrm{0}\:\mathrm{same}\:\mathrm{to}\:\mathrm{x}−\mathrm{axis}\: \\ $$