## Thomas' Calculus 13th Edition

$\approx 0.1732$
As we know that, $u=\dfrac{v}{|v|}$ Now, $u=\dfrac{\lt 1,1,1 \gt}{\sqrt{1^2+1^2+(1)^2}}=\lt \dfrac{1}{\sqrt 3},\dfrac{1}{\sqrt 3},\dfrac{1}{\sqrt 3} \gt$ Here, $dh=(\nabla h \cdot u) ds$ Thus, $dh=[\sqrt 3](0.1)\approx 0.1732$