Answer to Question #283406 in Calculus for Haider

Question

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.

Answer 1

$$\int_C(x^2 − 2y^2 ) ds$$

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

$$x^2 − 2y^2=\dfrac{t^2}{2}-t^2=-\dfrac{t^2}{2}$$

$$\dfrac{dx}{dt}=\dfrac{1}{\sqrt{2}}, \dfrac{dy}{dt}=\dfrac{1}{\sqrt{2}}$$

$$\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2}=\sqrt{(\dfrac{1}{\sqrt{2}})^2+(\dfrac{1}{\sqrt{2}})^2}=1$$

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

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