problem with Line integral of vector field

Question

Taking the xyz-coordinate system with $i,j,k$ are the unit vector of each axis, there is a Vector Field $F = {5x+y, 3y-2xz, z} = (5x+y)i + (3y-2xz)j + zk$ I want to find the integral of F on the line connect 2 point $A(1,2,0)$ and $B=(3,-4,1)$

$$\int \vec{F}d\vec{h}$$

for

$$dh= idx + jdy + zdk$$

Then

$$\int \vec{F}d\vec{h} = \int((5x+y)i + (3y-2xz)j + zk)(idx + jdy + zdk) = \int (5x+y)dx + (3y-2xy)dy + zdz$$

I got stuck at this point because this doesn't look like multi-variable integral since there is no product $dxdydz$ like I have learned from my math book

Answer 1

You have to parametrize $[A,B]$. Thus $$[A,B]=\{\gamma(t)=(1+2t, 2-6t,t)\mid t\in[0,1]\}$$ and thus $\dot\gamma(t)=\frac{d\vec \gamma(t)}{dt}=(2,-6,1)$ Therefore

$$\int_{[A,B]}\vec F \cdot d\vec h=\int_0^1 F(\gamma(t))\cdot \dot\gamma(t)dt=\int_0^1 F(1+2t, 2-6t,t)\cdot \frac{d\gamma(t)}{dt}dt=\int_0^1\begin{pmatrix}5(1+2t)+(2-5t)\\3(2-6t)-2t(1+2t)\\t\end{pmatrix}\cdot \begin{pmatrix}2\\-6\\ 1\end{pmatrix}dt=...$$

Answer 2

Continuing idm's calculation \begin{align} I &= \int\limits_0^1 \left( \begin{matrix} 5(1+2t)+(2-6t) \\ 3(2-6t)-2(1+2t)t \\ t \end{matrix} \right) \cdot \left( \begin{matrix} 2 \\ -6 \\ 1 \end{matrix} \right) \, dt \\ &= \int\limits_0^1 24 t^2+129 t-22 \, dt \\ &= \left[8 t^3 + \frac{129}{2} t^2 - 22 t \right]_0^1 \\ &= 8 + \frac{129}{2} - 22 \\ &= \frac{101}{2} \end{align}