Question

Consider a wire of density \(\rho(x, y)\) given by the space curve \(C: \mathbf{r}(t)=x(t) \mathbf{i}+y(t) \mathbf{j}, \quad a \leq t \leq b\) The moments of inertia about the \(x\) - and \(y\) -axes are given by \(I_{x}=\int_{C} y^{2} \rho(x, y) d s\) and \(I_{y}=\int_{C} x^{2} \rho(x, y) d s\) In Exercises 63 and \(64,\) find the moments of inertia for the wire of density \(\boldsymbol{\rho}\). A wire lies along \(\mathbf{r}(t)=a \cos t \mathbf{i}+a \sin t \mathbf{j}, 0 \leq t \leq 2 \pi\) and \(a>0,\) with density \(\rho(x, y)=1\).

Short answer

The moments of inertia for the wire of density \(\rho\) about the \(x\)-axis and \(y\)-axis are both \(\frac{a^3 \pi}{2}\).

Step-by-step solution

Expressing moments of inertia in terms of given parametrized curve

Since \(\mathbf{r}(t)=a \cos t \mathbf{i}+a \sin t \mathbf{j}, 0 \leq t \leq 2 \pi\), we can write \(x=a \cos t\) and \(y=a \sin t\). For the curve \(C:\mathbf{r}(t)=x(t) \mathbf{i}+y(t) \mathbf{j}, \quad a \leq t \leq b\), we can rewrite the moments of inertia as: \[I_{x}=\int_{0}^{2 \pi} (a \sin t)^{2} \rho ds , \quad I_{y}=\int_{0}^{2 \pi} (a \cos t)^{2} \rho ds\].

Calculating \(ds\)

The line element \(ds\) in the integral is given by \(ds = \sqrt{(dx/dt)^2 + (dy/dt)^2} dt\), where \(dx/dt\) and \(dy/dt\) are the derivatives of the parametric function with respect to \(t\). For \(\mathbf{r}(t)=a \cos t \mathbf{i}+a \sin t \mathbf{j}\), \(dx/dt=-a \sin t\) and \(dy/dt=a \cos t\). Therefore, \(ds = \sqrt{(-a \sin t)^2 + (a \cos t)^2} dt = a dt\).

Substituting \(ds\) and \(\rho\) values

Since \(\rho(x, y)=1\) and \(ds=a dt\), the moments of inertia become: \[I_{x}=\int_{0}^{2 \pi} (a \sin t)^{2} a dt = a^3 \int_{0}^{2 \pi} \sin^2 t dt , \ I_{y}=\int_{0}^{2 \pi} (a \cos t)^{2} a dt = a^3 \int_{0}^{2 \pi} \cos^2 t dt.\].

Solve the integrals

The definite integrals can be solved using the power-reduction identity \(\sin^2 t = (1-\cos 2t)/2\) and \(\cos^2 t = (1+\cos 2t)/2\), yielding: \[I_{x}= a^3 \int_{0}^{2 \pi} (1-\cos 2t)/2 dt = \frac{a^3 \pi}{2}=\frac{a^3 \pi}{2} , \ I_{y}=a^3 \int_{0}^{2 \pi} (1+\cos 2t)/2 dt= \frac{a^3 \pi}{2}.\]