V=∫022πx∗(x2+x)dx=V = \int_{0}^{2} 2\pi x*(x^2 +x) dx =V=∫022πx∗(x2+x)dx=
=2π∫02x3+x2dx=2π[x44+x33]∣02== 2\pi \int_{0}^{2} x^3 +x^2 dx = 2\pi [\frac{x^4}{4} + \frac{x^3}{3}] \Big\vert_{0}^{2} ==2π∫02x3+x2dx=2π[4x4+3x3]∣∣02=
=2π[4+83]=2π[203]=40π3= 2\pi [4+ \frac{8}{3}] = 2\pi [\frac{20}{3}] = \frac{40 \pi}{3}=2π[4+38]=2π[320]=340π
Answer
V=40π3V = \frac{40\pi}{3}V=340π
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