Answer to Question #283404 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

$\begin{aligned} & \int_{c}\left(x^{2}-2 y^{2}\right) d s, r(t)=\left<\frac{1}{\sqrt{2}}, \frac{t}{\sqrt{2}}\right> \text { for } 0 \leq t \leq 4\\ &F(x, y)=x^{2}-2 y^{2}\\ &F(x+t))=\left(\frac{t}{\sqrt{2}}\right)^{2}-2\left(\frac{t}{\sqrt{2}}\right)^{2}\\ &=\frac{t^{2}}{2}-t^{2}=-\frac{t^{2}}{2}\\ &r^{\prime}(t)=\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\\ &\int_{0}^{4}\left(\frac{-t^{2}}{2}\right) \cdot\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \cdot d t\\ &\int_{0}^{4} 2\times \frac{1}{\sqrt{2}}\left(-\frac{t^{2}}{2}\right) d t\\ &\left.\frac{1}{\sqrt{2}} \int \frac{-t^{3}}{3}\right|_{0} ^{4}=\frac{1}{\sqrt{2}} \frac{(-4)^{3}}{3}=-\frac{64}{\sqrt{2} \times 3}\\ &=\frac{-64 \sqrt{2}}{6}=\frac{-32}{3} \sqrt{2} \end{aligned}$