Answer
$\int_C\sqrt {xy}dx+e^{y}dy+xzdz=e-\frac{9}{70}$
Work Step by Step
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, $\int_C\sqrt {xy}dx+e^{y}dy+xzdz=e-\frac{9}{70}$
