Question Number 50689 by Tawa1 last updated on 18/Dec/18
$$\mathrm{If}\:\:\mathrm{c}\:\mathrm{is}\:\mathrm{the}\:\mathrm{line}\:\mathrm{segement}\:\mathrm{from}\:\:\left(\mathrm{0},\:\mathrm{0},\:\mathrm{0}\right)\:\mathrm{to}\:\left(\mathrm{1},\:\mathrm{2},\:\mathrm{3}\right) \\ $$$$\mathrm{find}\:\:\:\int\:\mathrm{x}\:\mathrm{e}^{\mathrm{yz}} \:\:\mathrm{ds} \\ $$
Answered by mr W last updated on 19/Dec/18
![((x−1)/1)=((y−2)/2)=((z−3)/3) ⇒y=2x ⇒z=3x ds=(√((dx)^2 +(dy)^2 +(dz)^2 ))=(√(1+2^2 +3^2 ))dx=(√(14))dx ∫xe^(yz) ds=∫_0 ^1 xe^(6x^2 ) (√(14)) dx =((√(14))/(12))∫_0 ^1 e^(6x^2 ) d(6x^2 ) =((√(14))/(12))[e^(6x^2 ) ]_0 ^1 =((√(14))/(12))(e^6 −1)](Q50704.png)
$$\frac{{x}−\mathrm{1}}{\mathrm{1}}=\frac{{y}−\mathrm{2}}{\mathrm{2}}=\frac{{z}−\mathrm{3}}{\mathrm{3}} \\ $$$$\Rightarrow{y}=\mathrm{2}{x} \\ $$$$\Rightarrow{z}=\mathrm{3}{x} \\ $$$${ds}=\sqrt{\left({dx}\right)^{\mathrm{2}} +\left({dy}\right)^{\mathrm{2}} +\left({dz}\right)^{\mathrm{2}} }=\sqrt{\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }{dx}=\sqrt{\mathrm{14}}{dx} \\ $$$$\int{xe}^{{yz}} {ds}=\int_{\mathrm{0}} ^{\mathrm{1}} {xe}^{\mathrm{6}{x}^{\mathrm{2}} } \sqrt{\mathrm{14}}\:{dx} \\ $$$$=\frac{\sqrt{\mathrm{14}}}{\mathrm{12}}\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{\mathrm{6}{x}^{\mathrm{2}} } \:{d}\left(\mathrm{6}{x}^{\mathrm{2}} \right) \\ $$$$=\frac{\sqrt{\mathrm{14}}}{\mathrm{12}}\left[{e}^{\mathrm{6}{x}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\sqrt{\mathrm{14}}}{\mathrm{12}}\left({e}^{\mathrm{6}} −\mathrm{1}\right) \\ $$
Commented by Tawa1 last updated on 19/Dec/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$