## Calculus 8th Edition

$\int_C\sqrt {xy}dx+e^{y}dy+xzdz=e-\frac{9}{70}$
Given: $C: r(t)=t^{4}i+t^{2}j+t^{3}k$ Therefore, $x=t^{4},y=t^{2},z=t^{3}$ and $dx=4t^{3}dt,y=2tdt,z=3t^{2}dt$ $\int_C\sqrt {xy}dx+e^{y}dy+xzdz=\int_{0}^{1}\sqrt {t^{4}t^{2}}(4t^{3}dt)+e^{t^{2}}(2tdt)+t^{4}.t^{3}(3t^{2}dt)$ $=\int_{0}^{1}4t^{6}+3t^{9}dt+\int_{0}^{1}2te^{t^{2}}dt$ $=A+B$ $A=\int_{0}^{1}4t^{6}+3t^{9}dt=[\frac{4t^{7}}{7}+\frac{3t^{10}}{10}]_{0}^{1}$ $A=\frac{61}{70}$ Now, $B=\int_{0}^{1}2te^{t^{2}}dt$ Take $t^{2}=u$ and $2tdt=du$ Limits of integration remains unchanged. $B=\int_{0}^{1}e^{u}du$ $=e-1$ The given line integral = A+B $\int_C\sqrt {xydx+e^{y}dy+xzdz}=\frac{61}{70}+e-1$ Hence, $\int_C\sqrt {xy}dx+e^{y}dy+xzdz=e-\frac{9}{70}$