Calculating line integral $\int_{\gamma}K(x) dx$

Question

$G=\{(x,y,z)^t \in \mathbb{R}^3: y>0,z>0\}$

$K\Big((x,y,z)^t\Big)= \Big(\ln(zy),\frac{x}{y},\beta \frac{x}{z}\Big)$

I want to calculate the line integral $$\int_{\gamma}K(x) dx$$ for $\beta=1$ and $\beta=2$ for $\gamma$ going from $(0,1,1)^t$ directly to $(x_0,y_0,z_0)^t$ in $G$.

For $\beta=1$ it's pretty easy because I can use the indefinite integral $F:=x \ln(zy)$ so $\int_{\gamma}K(x) dx = F\Big((x_0,y_0,z_0)^t\Big)-F\Big((0,1,1)^t\Big)=x_0\ln(z_0y_0)$.

Now, for $\beta=2$ there is no such $F$. But apparently one can caluclate the integral using:

$\mathrm {integral(\beta=1)}+\mathrm{what's\ missing}$

Can someone explain to me how this is done? So how do I calculate the line integral for $\beta=2$?

Thanks!

Answer 1

I will write all vectors as row vectors. Let $$ K_1(x)=\Bigl(\ln(z\,y),\frac{x}{y},\frac{x}{z}\Bigr),\quad K_2(x)=\Bigl(0,0,(\beta-1)\frac{x}{z}\Bigr). $$ Then $$ \int_\gamma K(x)\,dx=\int_\gamma K_1(x)\,dx+\int_\gamma K_2(x)\,dx. $$ The first one you know how to do. The second one can be computed explicitly using the definition of line integral