Find the volume of the region in 3-space with x > 0, y > 0 and z > 0 given by z^2/2 < x + y < z? Hint: First find the area of the horizontal cross-sections

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Answer 1 · 2017-08-19T12:14:57

#8/15#

#z^2/2 < x+y < z# we can conclude

#z^2/2-z < x+y-z < 0# then

#z^2/2-z < 0 rArr z(z/2-1) < 0# but #z > 0# then #z/2-1 < 0# and #z < 2#

so we have

#0 < z < 2#

Analyzing now for a fixed #z = z_0# we have

#z_0^2/2 < x + y# and #x + y < z_0# so the delimited region is the trapezoid with vertices

#(z_0^2/2,0), (z_0,0),(0,z_0),(0,z_0^2/2)# with area

#S_(z_0) = 1/2(sqrt2z_0^2/2+sqrt2 z_0)(1/sqrt2(z_0-z_0^2/2)) = 1/2(z_0^2-(z_0^2/2)^2)#

Now integrating this transversal area we have

#S = int_(z=0)^(z=2)S_z dz=int_(z=0)^(z=2)1/2(z^2-(z^2/2)^2)dz = 8/15#

Attached a plot showing the trapezoidal cross section.

enter image source here