Evaluate the line integral ∮ C 2 d s \oint_C2ds ∮ C 2 d s where C C C is the unit circle.
∮ C f ( x , y ) d s = ∫ a b f ( x ( t ) , y ( t ) ) ( d x d t ) 2 + ( d y d t ) 2 d t \oint_Cf(x, y)ds=\displaystyle\int_{a}^bf(x(t),y(t))\sqrt{\big({dx \over dt}\big)^2+\big({dy \over dt}\big)^2}dt ∮ C f ( x , y ) d s = ∫ a b f ( x ( t ) , y ( t )) ( d t d x ) 2 + ( d t d y ) 2 d t We first need parametric equations to represent C . C. C . The unit circle can be parametrized by means of the equations
x = cos t , y = sin t , 0 ≤ t ≤ 2 π . x=\cos t, y=\sin t, \ 0\leq t\leq2\pi. x = cos t , y = sin t , 0 ≤ t ≤ 2 π .
d x d t = − sin t , d y d t = cos t {dx \over dt}=-\sin t,\ {dy \over dt}=\cos t d t d x = − sin t , d t d y = cos t
( d x d t ) 2 + ( d y d t ) 2 = ( − sin t ) 2 + ( cos t ) 2 = 1 \sqrt{\big({dx \over dt}\big)^2+\big({dy \over dt}\big)^2}=\sqrt{\big(-\sin t\big)^2+\big(\cos t\big)^2}=1 ( d t d x ) 2 + ( d t d y ) 2 = ( − sin t ) 2 + ( cos t ) 2 = 1
∮ C 2 d s = ∫ 0 2 π 2 ( 1 ) d t = 2 [ t ] 2 π 0 = 2 ( 2 π − 0 ) = 4 π \oint_C2ds=\displaystyle\int_{0}^{2\pi}2(1)dt=2\big[t\big]\begin{matrix}
2\pi \\
0
\end{matrix}=2(2\pi-0)=4\pi ∮ C 2 d s = ∫ 0 2 π 2 ( 1 ) d t = 2 [ t ] 2 π 0 = 2 ( 2 π − 0 ) = 4 π
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