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 $V=\iint_R \int_{x^2+y^2}^{6-\frac{x^2+y^2}{2}} dz dy dx\,$ where R is the region in the (x,y) plane bounded by $x^2+y^2=4\,$

Therefore, $V=\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}[6-\frac{1}{2}(x^2+y^2)-(x^2+y^2)]dy dx\,$

= $6\int_0^2\int_{0}^{\sqrt{4-x^2}}[4-(x^2+y^2)]dy dx\,$

= $6\int_{0}^{\frac{\pi}{2}}\int_0^2(4-r^2)r dr d\theta\,$

= $12\pi\,$

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