Question

Find the total mass of the wire with density \(\boldsymbol{\rho}\). \(\mathbf{r}(t)=2 \cos t \mathbf{i}+2 \sin t \mathbf{j}+3 t \mathbf{k}, \quad \rho(x, y, z)=k+z\) \((k>0), \quad 0 \leq t \leq 2 \pi\)

Short answer

The total mass of the wire is \(2k\pi\sqrt{13} + 6\pi^{2}\sqrt{13}\).

Step-by-step solution

Find Line Element

First, calculate the derivative \(\mathbf{r}'(t)\) of r(t) and then the magnitude of this derivative to find ds. \(\mathbf{r}'(t) = -2 \sin t \mathbf{i} + 2 \cos t \mathbf{j} + 3 \mathbf {k}\). Compute the magnitude as \(ds = |\mathbf{r}'(t)| = \sqrt{(-2\sin t)^2 + (2\cos t)^2 + 3^2 } = \sqrt{13}\). This constant value indicates that the wire is uniformly bent.

Substitute into density

We substitute \(x\), \(y\), and \(z\) from the path function into the density function to get: \( \rho(2\cos t, 2\sin t, 3t) = k + 3t\) where \(z = 3t\).

Integration

To find the total mass, we integrate the product of density and the line element over t from 0 to \(2\pi\). So we get \( m = \int_{0}^{2\pi} \rho ds dt = \int_{0}^{2\pi} (k + 3t) \sqrt{13} dt = \sqrt{13} \left. \left( kt + \frac{3}{2} t^2 \right) \right\vert_{0}^{2\pi} = 2k\pi\sqrt{13} + 6\pi^{2}\sqrt{13}\).