Question

Find the total mass of two turns of a spring with density \(\rho\) in the shape of the circular helix \(\mathbf{r}(t)=3 \cos t \mathbf{i}+3 \sin t \mathbf{j}+2 t \mathbf{k}\) \(\rho(x, y, z)=z\)

Short answer

The total mass of the spring is \(16\pi^2 \sqrt{22}\).

Step-by-step solution

Convert Density Function

We start by converting the density function \(\rho(x, y, z)= z\) into a function of \(t\) using the given parametrization. Given \(\mathbf{r}(t)=3 \cos t \mathbf{i}+3 \sin t \mathbf{j}+2 t \mathbf{k}\), the z component is \(z=2t\). Thus we get the density as a function of \(t\) as: \(\rho(t)= 2t\).

Parameterize the Mass Element

The infinitesimal mass element \(dm\) of a small piece of wire is the density times the length of that piece. The length element \(ds\) is given by \(\|r'(t)\| dt\), where \(r'(t)\) is the derivative of \(r(t)\). We compute the derivative as: \(r'(t) = -3 \sin t \mathbf{i} + 3 \cos t \mathbf{j} + 2 \mathbf{k}\). Its magnitude is \(\|r'(t)\| = \sqrt{(-3 \sin t)^2+(3 \cos t)^2+2^2} = \sqrt{18+4} = \sqrt{22}\). Thus, the mass element is \(dm = \rho ds = \rho \|r'(t)\| dt = 2t \sqrt{22} dt\).

Setup and Evaluate the Integral

The total mass is the integration of the mass element over all \(t\) which parameterize the helix. The helix makes two turns, so \(t\) ranges from 0 to \(4\pi\). The integral becomes: \(M = \int dm = \int_{0}^{4\pi} 2t \sqrt{22} dt\). The antiderivative is \(t^2 \sqrt{22}\), thus we have: \(M = [t^2 \sqrt{22}]_{0}^{4\pi} = (4\pi)^2 \sqrt{22} = 16\pi^2 \sqrt{22}\).