Find the volume of the solid that is enclosed by the cone z=(x2+y2)12 and the sphere x2+y2+z2=2

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Yusuf Keller · Accepted Answer

Convert the equation od the cone to cylindrical coordinates  z=x2+y2=r2=r  Do the same for the equation of the sphere.  x2+y2+z2=2→z2=2−r2,z=2−r2  Find the intersection of the sphere the cone by setting the equations equal to each other and solving for r. This value is the maximum value for r in our integral.  r=2−r2→2r2=2,r=1  Set up the integral. Since the region is spherical of the x/y axis, we integrate from 0 to 2 pi. The equation of the sphere defines the top of the region, and the equation of be cone defines the bottom  Integrate with respect to z  ∫02π∫01r⋅z∣z=rz=2−r2drdθ  ∫02π∫01r2−r2−r2drdθ  Break the inner integral into separate pieces so that we can perform a u-substitution on the left portion.  ∫02π[∫01r2−r2dr−∫01r2dr]dθ  u=2−r2,du=−2rdr  Perform the substitution and simplify.  ∫02π[∫21−12udu−∫01r2dr]dθ  ∫02π[∫1212udu−∫01r2dr]dθ  Integrate with respect to u and r, and simplifly  ∫02π[∫21−12udu−∫01r2dr]dθ  ∫02π[∫1212udu−∫01r2dr]dθ  ∫02π23(2−1)dθ  Integrate with respect to theta  23(2−1)⋅θ∣θ=0θ=2π=4π3(2

Find the volume of the solid that is enclosed by the cone z=(x^2+y^2)^{1/2}ZS

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