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Question Number 182367 by universe last updated on 08/Dec/22

find volume of region bounded above by z = 1+(√(1−x^2 −y^2 )) and below by z = (√(x^2 +y^2 ))

$$ \\ $$$$\:\:\:\:\mathrm{find}\:\mathrm{volume}\:\mathrm{of}\:\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{above}\: \\ $$$$\:\:\:\mathrm{by}\:\mathrm{z}\:=\:\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:}\:\:\mathrm{and}\:\mathrm{below} \\ $$$$\:\:\:\:\mathrm{by}\:\:\:\mathrm{z}\:=\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:}\: \\ $$

Answered by Tokugami last updated on 08/Dec/22

top half is a semisphere with radius r=1: V_1 =(1/2)((4/3)π(1)^3 )=(4/6)π=((2π)/3) bottom half is a cone with radius r=1 and h=1: V_2 =(1/3)π(1)^2 (1)=(π/3) V=V_1 +V_2 =((2π)/3)+(π/3)=π

$$\mathrm{top}\:\mathrm{half}\:\mathrm{is}\:\mathrm{a}\:\mathrm{semisphere}\:\mathrm{with}\:\mathrm{radius}\:{r}=\mathrm{1}: \\ $$$${V}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{4}}{\mathrm{3}}\pi\left(\mathrm{1}\right)^{\mathrm{3}} \right)=\frac{\mathrm{4}}{\mathrm{6}}\pi=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\mathrm{bottom}\:\mathrm{half}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cone}\:\mathrm{with}\:\mathrm{radius}\:{r}=\mathrm{1}\:\mathrm{and}\:{h}=\mathrm{1}: \\ $$$${V}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}\pi\left(\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}\right)=\frac{\pi}{\mathrm{3}} \\ $$$${V}={V}_{\mathrm{1}} +{V}_{\mathrm{2}} =\frac{\mathrm{2}\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{3}}=\pi \\ $$$$ \\ $$

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