Question

Найдите обьем тела вращения ограниченного линиями у=х в квадрате -4

Answer 1

$y=x^2 -4 \\ ---------- \\ x^2 -4=0 \\ (x+2)(x-2)=0 \\ x_1=-2 \\ x_2=2 \\ ---------- \\ V= \pi \int\limits^b_{a} {f^2(x)} \, dx \\ ---------- \\$

$V= \pi \int\limits^2_{-2} {(x^2 -4)^2} \, dx =\pi \int\limits^2_{-2} {(x^4 -8x^2+16)} \, dx = \\ \\ =\pi ( \frac{x^5}{5} -8* \frac{x^3}{3}+16x)|^2_{-2} = \\ \\ =\pi [( \frac{2^5}{5} -8* \frac{2^3}{3}+16*2)-(\frac{(-2)^5}{5} -8* \frac{(-2)^3}{3}+16*(-2))]= \\ \\ = \pi [( \frac{32}{5} -8* \frac{8}{3}+32)-(\frac{-32}{5} -8* \frac{-8}{3}-32)]= \\ \\ = \pi [( \frac{32}{5} -\frac{64}{3}+32)-(\frac{-32}{5} + \frac{64}{3}-32)]=$

$= \pi [( \frac{32}{5} -\frac{64}{3}+32+\frac{32}{5} - \frac{64}{3}+32)]= \\ \\ = \pi [( \frac{64}{5} -\frac{128}{3}+64)]= \\ \\ = \pi ( \frac{64*3-128*5+64*15}{15})= \\ \\ = \pi ( \frac{192-640+960}{15})= \\ \\ = \frac{512*\pi}{15}=107,233$