Use the fundamental theorem of line integrals to calculate r c ~f ·d~r exactly, if ~f = 2x 1/3~i+e y/7 ~j, and c is the quarter of the unit circle in the first quadrant, traced counterclockwise from (1,0) to (0,1).

Assuming the vector field is [tex]\mathbf f=2x^{1/3}\,\mathbf i+e^{y/7}\,\mathbf j[/tex], we want to find a scalar function [tex]f[/tex] such that [tex]\nabla f=\mathbf f[/tex]. We require

[tex]\dfrac{\partial f}{\partial x}=2x^{1/3}[/tex]
[tex]\dfrac{\partial f}{\partial y}=e^{y/7}[/tex]

Integrating both sides of the first PDE with respect to [tex]x[/tex] gives

[tex]f(x,y)=\dfrac32x^{4/3}+g(y)[/tex]

Differentiating with respect to [tex]y[/tex] gives

[tex]\dfrac{\partial f}{\partial y}=e^{y/7}=\dfrac{\mathrm dg}{\mathrm dy}[/tex]
[tex]\implies g(y)=7e^{y/7}+C[/tex]

[tex]f(x,y)=\dfrac32x^{4/3}+7e^{y/7}+C[/tex]

Such a scalar function exists, so the fundamental theorem holds and any integral along any path is determined exactly by its endpoints. In this case,

[tex]\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(0,1)-f(1,0)=7e^{1/7}-\frac{17}2[/tex]