Consider the line integral Z C (sin x dx + cos y dy), where C consists of the top part of the circle x 2 + y 2 = 1 from (1, 0) to (−1, 0), followed by the line segment from (−1, 0) to (2, −π). Evaluate this line integral in two ways:

Direct computation:Parameterize the top part of the circle [tex]x^2+y^2=1[/tex] by[tex]\vec r(t)=(x(t),y(t))=(\cos t,\sin t)[/tex]with [tex]0\le t\le\pi[/tex], and the line segment by[tex]\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)[/tex]with [tex]0\le t\le1[/tex]. Then[tex]\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)[/tex][tex]=\displaystyle\int_0^\pi(-\sin t\sin(\cos t)+\cos t\cos(\sin t)\,\mathrm dt+\int_0^1(3\sin(3t-1)-\pi\cos(-\pi t))\,\mathrm dt[/tex][tex]=0+(\cos1-\cos2)=\boxed{\cos1-\cos2}[/tex]Using the fundamental theorem of calculus:The integral can be written as[tex]\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=\int_C\underbrace{(\sin x,\cos y)}_{\vec F}\cdot\underbrace{(\mathrm dx,\mathrm dy)}_{\vec r}[/tex]If there happens to be a scalar function [tex]f[/tex] such that [tex]\vec F=\nabla f[/tex], then [tex]\vec F[/tex] is conservative and the integral is path-independent, so we only need to worry about the value of [tex]f[/tex] at the path's endpoints.This requires[tex]\dfrac{\partial f}{\partial x}=\sin x\implies f(x,y)=-\cos x+g(y)[/tex][tex]\dfrac{\partial f}{\partial y}=\cos y=\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=\sin y+C[/tex]So we have[tex]f(x,y)=-\cos x+\sin y+C[/tex]which means [tex]\vec F[/tex] is indeed conservative. By the fundamental theorem, we have[tex]\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=f(2,-\pi)-f(1,0)=-\cos2-(-\cos1)=\boxed{\cos1-\cos2}[/tex]