Hello I am getting stuck on this question:
- The total work needed to move an object along a path of straight lines from $(-1,0)$ to $(0, a)$ and from $(0, a)$ to $(1,0)$, $a>0$, $a\in \mathbb R$, under the force field $\vec F = (4y^3-3xy^2, 3x+12y^2x-3x^2y)$ is equal $3$, if the value of constant $a$ is:
(A) $1$;
(B) $2$;
(C) $3$;
(D) $4$;
(E) none of the above.
I am unsure of whether or not I am supposed to use the fundamental theorem for line integrals to evaluate this, which would result in the $a$ being irrelevant, or if there is another way to do this problem. Any help is appreciated, thanks!
Note that $\bf F$ is the sum of two vector fields: ${\bf F}_1=(0,3x)$ (which is not conservative) and ${\bf F}_2=(4y^3-3xy^2,12y^2x-3x^2y)$ (which is conservative with potential function $U(x,y)=4xy^3-\frac{3x^2y^2}{2}$). The work for ${\bf F}_2$ does not depend on $a$ and it is equal to $U(1,0)-U(-1,0)=0$. On the other hand, the work for ${\bf F}_1$ does depend on $a$ and it can be evaluated by computing the (easy) line integral along the straight lines from $(−1,0)$ to $(0,a)$ and from $(0,a)$ to $(1,0)$.