This is question 10 from Exercise 8b from A Second Course in Mathematical Analysis (J. C. Burkill, H. Burkill).
A wire is shaped in the form of a twisted cubic given by $$\mathbf{r}(t) = (3t - t^3, 3t^2, 3t+ t^3), 0 \leq t \leq \sqrt{2}.$$ Show that the length of the wire between the origin and the point $(x,y,z)$ is $\sqrt{2}z$ and hence that the total length of the wire is $10$.
I could do this part. One useful thing I obtained here is that $ds = 3 \sqrt{2} (t^2+1) dt$.
Find the moment of inertia of the wire about the $y$-axis, if its mass per unit length varies uniformly with arc length, from 0 at the origin to 1 at the other end. (If a wire is shaped as a curve $\gamma$ in $\mathbb{R}^3$, its moment of inertia about a straight line $\lambda$ is $\int_{\gamma} \mu \rho^2 ds$, where $\mu(x,y,z)$ is the mass per unit length at $(x,y,z)$ and $\rho(x,y,z)$ is the distance of $(x,y,z)$ from $\lambda$.)
To tackle this, I use the definitions given. I find that $$\rho = y = 3t^2$$ because $\rho$ is the distance of the curve from the $y$-axis. I also get $$\mu = \dfrac{\sqrt{2} z}{10} = \dfrac{\sqrt{2} (3t+t^3)}{10}.$$
Putting this together and using $ds = 3 \sqrt{2} (t^2+1) dt$ from the earlier part, I obtain the integral as $$\dfrac{342 \sqrt{2}}{25}.$$ This is unfortunately wrong and the printed answer is $$\dfrac{3514}{25}.$$
Please could someoe help me onbtain the right answer? Thank you!
The printed answer is correct. The distance from $y$-axis is $\sqrt{x^2+z^2}$ (not $y$) and the density is $$\mu(t)=\frac{\int_0^t ds}{\int_0^{\sqrt{2}} ds}=\frac{\sqrt{2}t(t^2+3)}{10}.$$ Therefore the moment of inertia with respect to the $y$-axis of the given curve is $$\begin{align} \int_{\gamma} (x^2+z^2) \,\mu ds&=\int_0^{\sqrt{2}} ((3t-t^3)^2+(3t+t^3)^2)\, \mu(t)\, 3 \sqrt{2} (t^2+1) dt\\ &=\frac{6}{5}\int_0^{\sqrt{2}}t^3(t^4+9)(t^2+3)(t^2+1)\,dt\\ &=\frac{3}{5}\int_0^{2}u(u^2+9)(u+3)(u+1)\,du =\frac{3514}{25} \end{align}$$ which is confirmed by WA.