Fundamental Theorem for Line Integrals — Question 8

Question 8 – Solution

Strategy. The spatial geometry of $C$ is unnecessary because the integrand is the gradient of the supplied scalar function.

Step 1: Initial value $$f(1,-1,0)=(1)(-1)(0)+\frac 12(1+1+0)=1.$$

Step 2: Terminal value $$f(2,1,3)=(2)(1)(3)+\frac 12(4+1+9)=6+7=13.$$ Thus $$\boxed{\int_C\nabla f\cdot d\mathbf r=f(B)-f(A)=13-1=12}.$$

Step 3: Display the field Differentiating componentwise gives $$\nabla f=\langle yz+x,\,xz+y,\,xy+z\rangle.$$ This confirms that the line integral uses all three coordinate components, even though no parametrization is needed.

Verification The quadratic contribution changes from $1$ to $7$, a gain of $6$, and the product contribution changes from $0$ to $6$, another gain of $6$. The total change is $12$.