Question Number 211749 by universe last updated on 19/Sep/24
$${volume}\:{bounded}\:{by}\:{the}\:{curve} \\ $$$$\:{z}\:=\:\sqrt{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} }\:\:\:{and}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\:\mathrm{6}^{\mathrm{2}} \\ $$