Volume of solids of revolution

The Volume $V_{1} = \pi * \int_{-2}^{\infty} (1 - y)^2 \: dx =$ $100\pi * \int_{-2}^{\infty} \dfrac{x^2 + 2x + 1}{(x + 3)^2 (x + 4)^2} \:$ $$

Using partial fractions we obtain: $$

$\dfrac{x^2 + 2x + 1}{(x + 3)^2 (x + 4)^2} = \dfrac{A}{x + 3} + \dfrac{B}{(x + 3)^2} + \dfrac{C}{(x + 4)} + \dfrac{D}{(x + 4)^2}$

$\implies$ $$

$x^2 + 2x + 1 = (A + C) x^3 + (11A + B + 10C + D) x^2 + (40A + 8B + 33C + 6D) +$ $(48A + 16B + 36C + 9D)$ $$

$\implies$

$A = -C \implies$ $$

$B - C + D = 1$ $$

$8B - 7C + 6D = 2$ $$

$16B - 12C + 9D = 1$ $$

$\implies$

$2C - 3D = -3$ $$

$C - 2D = -6$ $$

$\implies D = 9, C = 12 \implies A = -12 \implies B = 4$ $$

$\implies$ $$

$V_{1} = 100\pi * \int_{-2}^{\infty} (\dfrac{-12}{x + 3} + \dfrac{4}{(x + 3)^2} + \dfrac{12}{x + 4} + \dfrac{9}{(x + 4)^2}) \: dx =$ $$

$100\pi * (12 * \ln(1 + \dfrac{1}{x + 3}) - \dfrac{4}{x + 3} - \dfrac{9}{x +4})|_{-2}^{\infty} =$ $$

$100\pi * (\dfrac{17}{2} - 12 * \ln(2))$ $$

Similarly, for $V_{2}$ we have: $$

$V_{2} = \pi * \int_{-\infty}^{-5} (y - 1)^2 \: dx =$ $$

$100\pi * \int_{-\infty}^{-5} \dfrac{x^2 + 2x + 1}{(x + 3)^2 (x + 4)^2} \: dx$ $$

$100\pi * (12 * \ln(1 + \dfrac{1}{x + 3}) - \dfrac{4}{x + 3} - \dfrac{9}{x +4})|_{-\infty}^{-5} =$ $$

$100\pi * (12 * \ln(\dfrac{1}{2}) + 11)$ $$

$\implies$

$V_{2} - V_{1} = 250\pi = a\pi \implies a = 250.$