The question in its entirety: Determine for which constants A & B the vector field $$\mathbb{F} = (Axln(z))\mathbb{i} + (By^2z)\mathbb{j} + ((\frac{x^2}{z})+y^3)\mathbb{j}$$ is conservative. If $\Bbb{C}$ is the straight line from $(1,1,1)$ to $(2,1,2)$ find $$\int_c2xln(z)\,dx+2y^2z\,dy + y^3\,dz$$
Now the trick here is supposed to be to observe that for A = 2 and B = 3 the field is conservative, and very similiar to the integral that we have to evaluate, so we can split it up into two parts (one conservative and a non-conservetive). However, as an exercise I wanted to see if I could do it without the "trick". I expressed the line parametrically as $$\Bbb{C(t)} = [t,1,t]$$ for $1\leq t\leq 2$. Inserting this into the integral we get $$\int_1^22tln(t)+2t + 1\,dt$$ which evaluates to $$4\,ln(2)+\frac{5}{2}$$ and the answer is supposed to be $$4\,ln(2)-\frac{1}{2}$$ What is it that I'm doing wrong? Is there something wrong with my understanding of work integrals, or have I just missed something minor?
UPDATE: I have experienced a similar problem on the next couple of questions that have similar structure, so I am assuming that there is some catch to the integral that it might not be as easily parametriziable as I thought. Where exactly did I go wrong in my line of thinking?
I think I have it! The dx,dy & dz terms in the integral represent the derivatives of the x, y and & components of the parametric position vector! With $\frac{d}{dt}(t,1,t) = (1,0,1)$ the integral simplifies to $$\int_1^2 2t\,ln(t)+ 1$$ which has solution $4\,ln(2) - \frac{1}{2}$! I thought I would post it here, in case someone else has a similar question!