Simplify.

${\int}_{0}^{2}\ufeff2\pi y((\frac{{y}^{2}}{2})-(\frac{{y}^{4}}{4}-\frac{{y}^{2}}{2}))dy$

${\int}_{0}^{2}\ufeff2\pi y((\frac{{y}^{2}}{2})-(\frac{{y}^{4}}{4}-\frac{{y}^{2}}{2}))dy$

vangstosiis
2022-07-27
Answered

Simplify.

${\int}_{0}^{2}\ufeff2\pi y((\frac{{y}^{2}}{2})-(\frac{{y}^{4}}{4}-\frac{{y}^{2}}{2}))dy$

${\int}_{0}^{2}\ufeff2\pi y((\frac{{y}^{2}}{2})-(\frac{{y}^{4}}{4}-\frac{{y}^{2}}{2}))dy$

You can still ask an expert for help

abortargy

Answered 2022-07-28
Author has **19** answers

${\int}_{0}^{2}2\pi y((\frac{{y}^{2}}{2})-(\frac{{y}^{4}}{4}-\frac{{y}^{2}}{2}))dy$

$={\int}_{0}^{2}2\pi y(\frac{{y}^{2}}{2}-\frac{{y}^{4}}{4}+\frac{{y}^{2}}{2})dy=2\pi {\int}_{0}^{2}y(\frac{2{y}^{2}}{2}-\frac{{y}^{4}}{4})dy$

$=2\pi {\int}_{0}^{2}y({y}^{2}-\frac{1}{4}{y}^{4})dy=2\pi {\int}_{0}^{2}({y}^{3}-\frac{1}{4}{y}^{5})dy=2\pi [\frac{{y}^{4}}{4}-\frac{1}{4}\ast \frac{{y}^{6}}{6}{]}_{0}^{2}$

$=2\pi [(\frac{{2}^{4}}{4}-\frac{{2}^{6}}{24})-(\frac{{0}^{4}}{4}-\frac{{0}^{6}}{24})]=2\pi [4-\frac{8}{3}]=2\pi (\frac{4}{3})=\frac{8\pi}{3}$

$={\int}_{0}^{2}2\pi y(\frac{{y}^{2}}{2}-\frac{{y}^{4}}{4}+\frac{{y}^{2}}{2})dy=2\pi {\int}_{0}^{2}y(\frac{2{y}^{2}}{2}-\frac{{y}^{4}}{4})dy$

$=2\pi {\int}_{0}^{2}y({y}^{2}-\frac{1}{4}{y}^{4})dy=2\pi {\int}_{0}^{2}({y}^{3}-\frac{1}{4}{y}^{5})dy=2\pi [\frac{{y}^{4}}{4}-\frac{1}{4}\ast \frac{{y}^{6}}{6}{]}_{0}^{2}$

$=2\pi [(\frac{{2}^{4}}{4}-\frac{{2}^{6}}{24})-(\frac{{0}^{4}}{4}-\frac{{0}^{6}}{24})]=2\pi [4-\frac{8}{3}]=2\pi (\frac{4}{3})=\frac{8\pi}{3}$

Aleah Booth

Answered 2022-07-29
Author has **5** answers

$2\pi {\int}_{0}^{2}\ufeff(\frac{{y}^{3}}{2}-\frac{{y}^{5}}{4}+\frac{{y}^{3}}{2})dy=2\pi {\int}_{0}^{2}\frac{-{y}^{5}}{4}dy$

$=-2\pi [\frac{{y}^{6}}{24}{]}_{0}^{2}=-2\pi (\frac{64}{24}-0)=\frac{-16\pi}{3}$

$=-2\pi [\frac{{y}^{6}}{24}{]}_{0}^{2}=-2\pi (\frac{64}{24}-0)=\frac{-16\pi}{3}$

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I have hit this in my book and the way I do it I get

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