Taking the red line

$$ We have $y=\sqrt{2}x^2$ for $0\le x\le 1$ and $$ $\begin{aligned} dS&=\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\\ &=\sqrt{1+(2x\sqrt{2})^2}\,dx\\ &=\sqrt{1+8x^2}\,dx. \end{aligned}$ $$ Therefore $\begin{aligned} N &= \int_{0}^{1} 2x\sqrt{1+8x^2}\,dx\\ \end{aligned}.$ $$ Let $u=1+8x^2$ , then $\,du=16x dx$ and $\,1\le u\le 9$ . Thus $$ $\begin{aligned} N &= \frac{1}{8}\int_{1}^{9} u^\frac{1}{2}\,du\\ &= \frac{1}{12} \left.u^\frac{3}{2}\right|_1^9\\ &= \frac{1}{12}(27-1)\\ &= \frac{13}{6}\\ &=\boxed{2.167} \end{aligned}$ $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

$\int_{c}2xdS=\int_{0}^{1}2xdx\sqrt{1+(\frac{dy}{dx})^{2}}=\int_{0}^{1}\sqrt{1+8t}dt=13/6$