MvCalc50

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The curve of the intersection of the two paraboloids is given by x^{2}+y^{2}=6-{\frac  {x^{2}+y^{2}}{2}}\, or x^{2}+y^{2}=4\,

Hence,the volume V is given by V=\iint _{R}\int _{{x^{2}+y^{2}}}^{{6-{\frac  {x^{2}+y^{2}}{2}}}}dzdydx\, where R is the region in the (x,y) plane bounded by x^{2}+y^{2}=4\,

Therefore, V=\int _{{-2}}^{{2}}\int _{{-{\sqrt  {4-x^{2}}}}}^{{{\sqrt  {4-x^{2}}}}}[6-{\frac  {1}{2}}(x^{2}+y^{2})-(x^{2}+y^{2})]dydx\,

=6\int _{0}^{2}\int _{{0}}^{{{\sqrt  {4-x^{2}}}}}[4-(x^{2}+y^{2})]dydx\,

=6\int _{{0}}^{{{\frac  {\pi }{2}}}}\int _{0}^{2}(4-r^{2})rdrd\theta \,

=12\pi \,

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