Answer
$\dfrac94$
Work Step by Step
$F(x,y,z)=(xy,xz,yz)$
$C:~~r(t)=(t,t^2,2t),~t\in[0,1]$
$\Rightarrow r'(t)=(1,2t,2),~t\in[0,1]$
We are now ready to calculate the line integral:
\begin{eqnarray*}\int_C F\cdot d r&=&\int_{0}^{1}(F\circ r)(t)\cdot r'(t)\,dt\\
&=&\int_{0}^{1}(x(t)y(t)(1)+x(t)z(t)(2t)+yz(2))\,dt\\
&=&\int_{0}^{1}(t^3+4t^3+4t^3)\,dt\\
&=&9\int_{0}^{1}t^3\,dt~~=~~9\left[\dfrac{t^4}{4}\right]_{0}^{1}\\
&=&\dfrac94.
\end{eqnarray*}
