Answer to Question #343907 in Calculus for fhu

Question

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

Answer 1

$$-x^2-10x+6=2x+26$$

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

$$x_1=-10, x_2=-2$$

a.

$$V=\displaystyle\int_{-10}^{-2}2\pi(2-x)(-x^{2}-10x+6-(2x+26)dx$$ $$=2\pi\displaystyle\int_{-10}^{-2}(x^3+10x^2-4x-40)dx$$

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

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

b.

$$V=\displaystyle\int_{-10}^{-2}2\pi(-1-x)(-x^{2}-10x+6-(2x+26)dx$$ $$=2\pi\displaystyle\int_{-10}^{-2}(x^3+10x^2-4x-40)dx$$

$$=2\pi[\dfrac{x^4}{4}+\dfrac{13x^3}{3}+16x^2+20x]\begin{matrix} -2\\ -10 \end{matrix}$$

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

c.

$$V=\displaystyle\int_{-10}^{-2}2\pi(x+14)(-x^{2}-10x+6-(2x+26)dx$$ $$=2\pi\displaystyle\int_{-10}^{-2}(x^3+10x^2-4x-40)dx$$

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

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