Question #343907

use the method of cylinders to determine the volume of the solid by rotating the region bounded by y=-x^2-10x+6 and y=2x+26 about the

a. line x=2 b. line x=-1 c. line x=-14


1
Expert's answer
2022-05-24T10:26:44-0400
x210x+6=2x+26-x^2-10x+6=2x+26

x2+12x+20=0x^2+12x+20=0

x1=10,x2=2x_1=-10, x_2=-2

a.


V=1022π(2x)(x210x+6(2x+26)dxV=\displaystyle\int_{-10}^{-2}2\pi(2-x)(-x^{2}-10x+6-(2x+26)dx=2π102(x3+10x24x40)dx=2\pi\displaystyle\int_{-10}^{-2}(x^3+10x^2-4x-40)dx

=2π[x44+10x332x240x]210=2\pi[\dfrac{x^4}{4}+\dfrac{10x^3}{3}-2x^2-40x]\begin{matrix} -2\\ -10 \end{matrix}

=4096π3(units3)=\dfrac{4096\pi}{3}({units}^3)

b.


V=1022π(1x)(x210x+6(2x+26)dxV=\displaystyle\int_{-10}^{-2}2\pi(-1-x)(-x^{2}-10x+6-(2x+26)dx=2π102(x3+10x24x40)dx=2\pi\displaystyle\int_{-10}^{-2}(x^3+10x^2-4x-40)dx

=2π[x44+13x33+16x2+20x]210=2\pi[\dfrac{x^4}{4}+\dfrac{13x^3}{3}+16x^2+20x]\begin{matrix} -2\\ -10 \end{matrix}

=2560π3(units3)=\dfrac{2560\pi}{3}({units}^3)

c.


V=1022π(x+14)(x210x+6(2x+26)dxV=\displaystyle\int_{-10}^{-2}2\pi(x+14)(-x^{2}-10x+6-(2x+26)dx=2π102(x3+10x24x40)dx=2\pi\displaystyle\int_{-10}^{-2}(x^3+10x^2-4x-40)dx

=2π[x4426x3394x2280x]210=2\pi[\dfrac{x^4}{4}-\dfrac{26x^3}{3}-94x^2-280x]\begin{matrix} -2\\ -10 \end{matrix}

=4096π3(units3)=\dfrac{4096\pi}{3}({units}^3)


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