Compute the volume of the body defined by the inequalities $$x^2+y^2 \leq 4x, \, |z| \leq x^2+y^2 \\$$ I write the first inequality as $(x-2)^2+y^2 \leq 4$ so it is a disk with radius $2$ and centrum in $(2,0)$. The second inequality is a paraboloid but I don't see how the absolute value of $z$ affects the figure. If $z^+ \leq x^2+y^2$ then $z^- \leq z^2+y^2$ so I don't see the difference it makes. Also, what method is preferable when computing this type of body? As a difference between two double integrals?
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The absolute value affects it, without it all negative values of $z$ would be okay – AnalysisStudent0414 Jun 03 '15 at 12:48
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The inequality containing absolute value can be transformed to $$-(x^2+y^2)\leq z\leq x^2+y^2$$
Because of symmetry, the volume is twice of the object formed between $z=0$ and $z=x^2+y^2$.
Use polar coordinate because the integration region is a circle.
KittyL
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