Question Number 155919 by horpy4 last updated on 05/Oct/21
Answered by physicstutes last updated on 05/Oct/21
![F×V = determinant ((i,j,k),((3u),u^2 ,(u+2)),((2u),(−3u),(u−2)))=i(u^3 −2u^2 +3u^2 −6u) − j(3u^2 −6u−2u^2 −4u) + k(−9u^2 −2u^3 ) F×V= (u^3 +u^2 −6u)i+(−u^2 +10u)j+(−9u^2 −2u^3 )k ∫_0 ^2 F×V du = [((u^4 /4)+(u^2 /2)−3u^2 )i + (−(u^3 /3)+5u^2 )j+(−3u^3 −(1/2)u^4 )k]_0 ^2](Q155922.png)
$$\boldsymbol{\mathrm{F}}×\boldsymbol{\mathrm{V}}\:=\:\begin{vmatrix}{{i}}&{{j}}&{{k}}\\{\mathrm{3}{u}}&{{u}^{\mathrm{2}} }&{{u}+\mathrm{2}}\\{\mathrm{2}{u}}&{−\mathrm{3}{u}}&{{u}−\mathrm{2}}\end{vmatrix}={i}\left({u}^{\mathrm{3}} −\mathrm{2}{u}^{\mathrm{2}} +\mathrm{3}{u}^{\mathrm{2}} −\mathrm{6}{u}\right) \\ $$$$−\:{j}\left(\mathrm{3}{u}^{\mathrm{2}} −\mathrm{6}{u}−\mathrm{2}{u}^{\mathrm{2}} −\mathrm{4}{u}\right)\:+\:{k}\left(−\mathrm{9}{u}^{\mathrm{2}} −\mathrm{2}{u}^{\mathrm{3}} \right) \\ $$$$\boldsymbol{\mathrm{F}}×\boldsymbol{\mathrm{V}}=\:\left({u}^{\mathrm{3}} +{u}^{\mathrm{2}} −\mathrm{6}{u}\right){i}+\left(−{u}^{\mathrm{2}} +\mathrm{10}{u}\right){j}+\left(−\mathrm{9}{u}^{\mathrm{2}} −\mathrm{2}{u}^{\mathrm{3}} \right){k} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\boldsymbol{\mathrm{F}}×\boldsymbol{\mathrm{V}}\:{du}\:=\:\left[\left(\frac{{u}^{\mathrm{4}} }{\mathrm{4}}+\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{3}{u}^{\mathrm{2}} \right){i}\:+\:\left(−\frac{{u}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{5}{u}^{\mathrm{2}} \right){j}+\left(−\mathrm{3}{u}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{4}} \right){k}\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$
Commented by horpy4 last updated on 05/Oct/21
$${it}\:{seems}\:{there}\:{is}\:{error} \\ $$