 # Simplify. int_0^2 2pi y((y^2)/2)) -((y^4)/4- (y^2)/2)dy vangstosiis 2022-07-27 Answered
Simplify.
${\int }_{0}^{2}﻿2\pi y\left(\left(\frac{{y}^{2}}{2}\right)-\left(\frac{{y}^{4}}{4}-\frac{{y}^{2}}{2}\right)\right)dy$
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${\int }_{0}^{2}2\pi y\left(\left(\frac{{y}^{2}}{2}\right)-\left(\frac{{y}^{4}}{4}-\frac{{y}^{2}}{2}\right)\right)dy$
$={\int }_{0}^{2}2\pi y\left(\frac{{y}^{2}}{2}-\frac{{y}^{4}}{4}+\frac{{y}^{2}}{2}\right)dy=2\pi {\int }_{0}^{2}y\left(\frac{2{y}^{2}}{2}-\frac{{y}^{4}}{4}\right)dy$
$=2\pi {\int }_{0}^{2}y\left({y}^{2}-\frac{1}{4}{y}^{4}\right)dy=2\pi {\int }_{0}^{2}\left({y}^{3}-\frac{1}{4}{y}^{5}\right)dy=2\pi \left[\frac{{y}^{4}}{4}-\frac{1}{4}\ast \frac{{y}^{6}}{6}{\right]}_{0}^{2}$
$=2\pi \left[\left(\frac{{2}^{4}}{4}-\frac{{2}^{6}}{24}\right)-\left(\frac{{0}^{4}}{4}-\frac{{0}^{6}}{24}\right)\right]=2\pi \left[4-\frac{8}{3}\right]=2\pi \left(\frac{4}{3}\right)=\frac{8\pi }{3}$
###### Not exactly what you’re looking for? Aleah Booth
$2\pi {\int }_{0}^{2}﻿\left(\frac{{y}^{3}}{2}-\frac{{y}^{5}}{4}+\frac{{y}^{3}}{2}\right)dy=2\pi {\int }_{0}^{2}\frac{-{y}^{5}}{4}dy$
$=-2\pi \left[\frac{{y}^{6}}{24}{\right]}_{0}^{2}=-2\pi \left(\frac{64}{24}-0\right)=\frac{-16\pi }{3}$