Question

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:

Answer 1

Direct computation:

Parameterize the top part of the circle
`x^2+y^2=1` by

`\vec r(t)=(x(t),y(t))=(\cos t,\sin t)`

with
`0\le t\le\pi`, and the line segment by

`\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)`

with
`0\le t\le1`. Then

`\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)`

`=\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`

`=0+(\cos1-\cos2)=\boxed{\cos1-\cos2}`

Using the fundamental theorem of calculus:

The integral can be written as

`\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)`

If there happens to be a scalar function
`f` such that
`\vec F=\\abla f`, then
`\vec F` is conservative and the integral is path-independent, so we only need to worry about the value of
`f` at the path's endpoints.

This requires

`(\partial f)/(\partial x)=\sin x\implies f(x,y)=-\cos x+g(y)`

`(\partial f)/(\partial y)=\cos y=(\mathrm dg)/(\mathrm dy)\implies g(y)=\sin y+C`

So we have

`f(x,y)=-\cos x+\sin y+C`

which means
`\vec F` is indeed conservative. By the fundamental theorem, we have

`\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=f(2,-\pi)-f(1,0)=-\cos2-(-\cos1)=\boxed{\cos1-\cos2}`