How do you find the volume of the solid with base region bounded by the curves #y=1-x^2# and #y=x^2-9# if cross sections perpendicular to the #x#-axis are squares?

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Answer 1 · 2014-11-03T05:18:02

Since the length of a square cross-section can be expressed by

#1-x^2-(x^2-9)=10-2x^2=2(5-x^2)#,

the area of the cross-section can be found by

#A(x)=[2(5-x^2)]^2=4(25-10x^2+x^4)#.

Let us find the #x#-coordinates of the intersections of the two parabola.

#2(5-x^2)=0 => x^2=5 => x=pm sqrt{5}#,

Since the base of the solid spans from #-sqrt{5}# to #sqrt{5}#, the volume #V# of the solid can be found by

#V=4int_{-sqrt{5}}^{sqrt{5}}(25-10x^2+x^4)dx#

by using the symmetry (even function),

#=8int_0^{sqrt{5}}(25-10x^2+x^4)dx#

#=8[25x-10/3x^3+1/5x^5]_0^{sqrt{5}}#

#=8(25sqrt{5}-50/3sqrt{5}+5sqrt{5})={320}/3sqrt{5}#.

I hope that this was helpful.