Question #283406

Evaluate the following line integrals along with the curve C.

(a) ∫

C (x2 − 2y2 ) ds; C is the line segment parametrized by r(t) = < t / √ 2 , t / √ 2 >, for 0 ≤ t ≤ 4.


1
Expert's answer
2021-12-29T13:00:19-0500

C(x22y2)ds\int_C(x^2 − 2y^2 ) ds

r(t)=<t2,t2>,0t4.r(t) = < \dfrac{t}{\sqrt{2}} , \dfrac{t}{\sqrt{2}} >, 0 ≤ t ≤ 4.


x22y2=t22t2=t22x^2 − 2y^2=\dfrac{t^2}{2}-t^2=-\dfrac{t^2}{2}

dxdt=12,dydt=12\dfrac{dx}{dt}=\dfrac{1}{\sqrt{2}}, \dfrac{dy}{dt}=\dfrac{1}{\sqrt{2}}

(dxdt)2+(dydt)2=(12)2+(12)2=1\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2}=\sqrt{(\dfrac{1}{\sqrt{2}})^2+(\dfrac{1}{\sqrt{2}})^2}=1

C(x22y2)ds=04(t22)(1)dt\int_C(x^2 − 2y^2 ) ds=\displaystyle\int_{0}^{4}(-\dfrac{t^2}{2})(1)dt

=[t36]40=323=\big[-\dfrac{t^3}{6}\big]\begin{matrix} 4 \\ 0 \end{matrix}=-\dfrac{32}{3}


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