Question

(a) Find the value of Z C (x + y) ds, where C is parameterized by x = t, y = t and 0 ≤ t ≤ 1

Answer 1

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

$$\int_C(x+y)ds=\displaystyle\int_{0}^{1}(t+t)\sqrt{(1)^2+(1)^2}dt$$

$$=\sqrt{2}[t^2]\begin{matrix} 1\\ 0 \end{matrix}=\sqrt{2}$$

$$\int_C(x+y)ds=\sqrt{2}$$