∫ C ( x 2 − 2 y 2 ) d s \int_C(x^2 − 2y^2 ) ds ∫ C ( x 2 − 2 y 2 ) d s r ( t ) = < t 2 , t 2 > , 0 ≤ t ≤ 4. r(t) = < \dfrac{t}{\sqrt{2}} , \dfrac{t}{\sqrt{2}} >, 0 ≤ t ≤ 4. r ( t ) =< 2 t , 2 t > , 0 ≤ t ≤ 4.
x 2 − 2 y 2 = t 2 2 − t 2 = − t 2 2 x^2 − 2y^2=\dfrac{t^2}{2}-t^2=-\dfrac{t^2}{2} x 2 − 2 y 2 = 2 t 2 − t 2 = − 2 t 2
d x d t = 1 2 , d y d t = 1 2 \dfrac{dx}{dt}=\dfrac{1}{\sqrt{2}}, \dfrac{dy}{dt}=\dfrac{1}{\sqrt{2}} d t d x = 2 1 , d t d y = 2 1
( d x d t ) 2 + ( d y d t ) 2 = ( 1 2 ) 2 + ( 1 2 ) 2 = 1 \sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2}=\sqrt{(\dfrac{1}{\sqrt{2}})^2+(\dfrac{1}{\sqrt{2}})^2}=1 ( d t d x ) 2 + ( d t d y ) 2 = ( 2 1 ) 2 + ( 2 1 ) 2 = 1
∫ C ( x 2 − 2 y 2 ) d s = ∫ 0 4 ( − t 2 2 ) ( 1 ) d t \int_C(x^2 − 2y^2 ) ds=\displaystyle\int_{0}^{4}(-\dfrac{t^2}{2})(1)dt ∫ C ( x 2 − 2 y 2 ) d s = ∫ 0 4 ( − 2 t 2 ) ( 1 ) d t
= [ − t 3 6 ] 4 0 = − 32 3 =\big[-\dfrac{t^3}{6}\big]\begin{matrix}
4 \\
0
\end{matrix}=-\dfrac{32}{3} = [ − 6 t 3 ] 4 0 = − 3 32
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