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.

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