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baherus [9]
3 years ago
12

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) t

o (−1, 0), followed by the line segment from (−1, 0) to (2, −π). Evaluate this line integral in two ways:
Mathematics
1 answer:
pashok25 [27]3 years ago
3 0

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=\nabla 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

\dfrac{\partial f}{\partial x}=\sin x\implies f(x,y)=-\cos x+g(y)

\dfrac{\partial f}{\partial y}=\cos y=\dfrac{\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}

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