Evaluate $\int z ds $ over the curve $x=y=z^2$ from $(1,1,1)$ to $(4,4,2)$

Question

Let r denote the path $x=y=z^2$ from $(1,1,1)$ to $(4,4,2)$. Evaluate $$\int_rz ds.$$

I am not sure where to even begin on this one. Any help would be much appreciated.

Answer 1

Such an integral is called the line integral of a scalar field. You can see here for some introduction. Here $ds$ is the infinitesimal of the arc length. In the 3-dimensional space we have $ds = dz\sqrt{(\frac{dx}{dz})^2+(\frac{dy}{dz})^2+1}$. By the relationship of $x,y,z$ in this problem $\frac{dy}{dz} = \frac{dx}{dz} = 2z$, so we have $ds = dz\sqrt{8z^2 + 1}$. Therefore the original integral is $$ \int_r zds = \int_1^2 z\sqrt{8z^2 + 1}dz \overset{t=z^2}{=} \frac{1}{2}\int_1^4\sqrt{8t+1}dt = \frac{1}{24}[(8t+1)^{3/2}]^4_1 = \frac{1}{24}(33^{3/2}-27). $$