Question Number 224915 by fantastic last updated on 11/Oct/25
$${evaluate}\: \\ $$$$\int_{{C}} \left(\mathrm{3}{y}^{\mathrm{2}} +\mathrm{2}{z}^{\mathrm{2}} \right){dx}+\left(\mathrm{6}{x}−\mathrm{10}{z}\right){y}\:{dy}\:+\left(\mathrm{4}{xz}−\mathrm{5}{y}^{\mathrm{2}} \right){dz} \\ $$$${along}\:{the}\:{portion}\:{from}\:\left(\mathrm{1},\mathrm{0},\mathrm{1}\right)\:{to}\:\left(\mathrm{3},\mathrm{4},\mathrm{5}\right)\:{of} \\ $$$${the}\:{curve}\:{C}, \\ $$$${which}\:{is}\:{the}\:{intersection}\:{of}\:{the} \\ $$$${two}\:{surfaces}\:{z}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:{and}\:{z}={y}+\mathrm{1} \\ $$
Answered by ajfour last updated on 13/Oct/25
$${Intersection}\:{curve} \\ $$$$\mathrm{2}{y}={x}^{\mathrm{2}} −\mathrm{1} \\ $$$${let}\:{y}+\frac{\mathrm{1}}{\mathrm{2}}={t}^{\mathrm{2}} \:\:\:\:\:,\:{x}={t}\sqrt{\mathrm{2}} \\ $$$${for}\:{x}=\mathrm{1},\:{y}=\mathrm{0},\:{z}=\mathrm{1},\:{t}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$${for}\:{x}=\mathrm{3},\:{y}=\mathrm{4},\:{z}=\mathrm{5},\:{t}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}} \\ $$$${dx}=\sqrt{\mathrm{2}}{dt} \\ $$$${dy}=\mathrm{2}{tdt} \\ $$$$\overset{\:\:\mathrm{3}/\sqrt{\mathrm{2}}} {\int}_{\mathrm{1}/\sqrt{\mathrm{2}}} \left\{\mathrm{3}\left({t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{2}\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \right\}{dt} \\ $$$$=\int\left(\mathrm{5}{t}^{\mathrm{4}} −{t}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{4}}\right)=\left({t}^{\mathrm{5}} −\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{5}{t}}{\mathrm{4}}\underset{\mathrm{1}/\sqrt{\mathrm{2}}} {\right)}^{\mathrm{3}/\sqrt{\mathrm{2}}} \\ $$$${similarly}\:{the}\:{rest}… \\ $$
Commented by ajfour last updated on 13/Oct/25
https://youtu.be/drf__zRLz4M?si=AJwQMGb8yZAk0wth