Find $\int_C F.dr $ where C is the smooth curve joining $(1,0,0) $ to $(0,0,2)$.

Question

Suppose the vector field is the gradient of the function $f(x,y,z)=-\frac{1}{x^2+y^2+z^2}.$

Find $\int_C F\cdot dr $ where $C$ is the smooth curve joining $(1,0,0) $ to $(0,0,2)$.

I found $\operatorname{grad} f=\left(\dfrac{2x}{(x^2+y^2+z^2)^2},\dfrac{2y}{(x^2+y^2+z^2)^2},\dfrac{2z}{(x^2+y^2+z^2)^2}\right)$

Then parametrizing the curve C we get $r=((1-t),0,2t)$

Then I computed $dr=(-1,0,2)$

So $F(r)\cdot dr=\dfrac{10t-2}{((1-t)^2+4t^2)^2}$

But the answer is not coming Can anyone please help where it is going wrong

I am trying to recollect my vector calculus since I am not in touch for long

Answer 1

Fundamental theorem of calculus: You know the vector field is the gradient of a known scalar function $f$, so the line integral is path-independent and

$$\int_C\nabla f\cdot\mathrm d\mathbf r=f(0,0,2)-f(1,0,0)$$

Answer 2

That the vector field is the gradient of a continuous function implies that the field is conservative, which means that the path integral over the vector field only depends on its endpoints, which is fitting because we are only given the endpoints here. $$f(1,0,0)=-\frac1{1^2+0^2+0^2}=-1$$ $$f(0,0,2)=-\frac1{0^2+0^2+2^2}=-\frac14$$ Then $$\int_C\mathbf F\cdot dr=f(0,0,2)-f(1,0,0)=\frac34$$

Answer 3

Please review the Fundamental theorem of line integrals.

You already have the potential function and all you have to do is to evaluate $$f(x,y,z)=-\frac{1}{x^2+y^2+z^2}$$ at the endpoints and subtract.