Question

Use a computer algebra system to find the mass, center of mass, and moments of inertia of the lamina that occupies the region \(D\) and has the given density function.

\(D\)is enclosed by the right loop of the four-leaved rose \(r = \cos 2\theta ;\quad \rho (x,y) = {x^2} + {y^2}\)

Short answer

The mass is of given lamina is \(m = \frac{{3\pi }}{{64}}\)

The centre of mass of given lamina:

\(\begin{aligned}{l}\bar x \approx 0.7095133\\\bar y = 0\end{aligned}\)

The moments of inertia are following:

\(\begin{aligned}{l}{I_x} \approx 0.00281\\{I_y} \approx 0.079\\{I_o} \approx 0.081812\end{aligned}\)

Step-by-step solution

Calculate the mass of the given lamina

Total mass of lamina is obtained as:

We're given \(\rho (x,y) = {x^2} + {y^2} = {r^2}\),

so we can state more explicitly:

The region \({\rm{D}}\) over which we are integrating will be bounded by the graph of:

\(r = \cos (2\theta )\)

So limits will be:

\(\begin{aligned}{l}r:(0,\cos (2\theta ))\\\theta :\left( { - \frac{\pi }{4},\frac{\pi }{4}} \right)\end{aligned}\)

Which makes our integration for the total mass:

\(\begin{aligned}{c}M = \int_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\int_0^{\cos (2\theta )} {{r^3}} } drd\theta \\ = \frac{{3\pi }}{{64}}\\ \approx .147262\end{aligned}\)

Which is computed using Wolframalpha.com

Calculate the centre of mass of the given lamina

By symmetry we know that the center of mass will lie along the\(x\)-axis, and thus we need only calculate the\(x\)-displacement of mass.

From the definition of center of mass, assuming\(\rho \)is continuous for our\(x\):

\(\begin{aligned}{c}M \cdot {x_{cm}} = \sum\limits_{i = 1}^n {{x_i}} {m_i}\\ \Rightarrow \int x \cdot dm = \int x \cdot \rho \cdot dA\end{aligned}\)

We can derive the equation which will give the x-coordinate of our center of mass:\({x_{cm}} = \frac{1}{M}\int_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\int_0^{\cos (2\theta )} x } \cdot \rho \cdot dA\)

We know \(x = r\cos (\theta )\) across the entire region,

and\(\rho (x,y) \equiv \rho (r,\theta ) = {r^2}\)

So, plugging that into our equation, we get:

\(\begin{aligned}{c}{x_{cm}} = \frac{1}{M}\int_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\int_0^{\cos (2\theta )} r } \cdot \cos (\theta ) \cdot {r^2} \cdot rdrd\theta \\ = \frac{1}{M}\int_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\int_0^{\cos (2\theta )} {{r^4}} } \cos (\theta )drd\theta \end{aligned}\)

From Wolframalpha.com, computing the integral, and multiplyingby \(1/{\rm{M}}\), we get:

\({x_{cm}} \approx .7095133 \ldots \)

Calculate the moment of inertia of the given lamina

Moment of inertia is calculated as mass element times squared distance from the axis of rotation:

\(I = \sum {{m_i}} r_i^2\)

Which translates for a continuous distribution as

\(\int {r_\alpha ^2} dm = \int {r_\alpha ^2} \cdot \rho \cdot dA\)

Using: \({r_\alpha }\) to be the radius around a specified axis

The region is flat, so our radial distance around the \(x\)-axis is simply \(y\), and distance around the \(y\)-axis is simply \(x\)

Using: \(y = r\sin (\theta )\), and: \(x = r\cos (\theta )\)

We get our centres of mass to be:

\(\begin{aligned}{c}{I_x} = \int_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\int_0^{\cos (2\theta )} {{r^5}} } {\sin ^2}(\theta )drd\theta \\ = \frac{{5\pi }}{{384}} - \frac{4}{{105}}\\ \approx .00281\end{aligned}\)

and

\(\begin{aligned}{c}{I_y} = \int_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\int_0^{\cos (2\theta )} {{r^5}} } {\cos ^2}(\theta )drd\theta \\ = \frac{{5\pi }}{{384}} + \frac{4}{{105}}\\ \approx .079\end{aligned}\)

And the polar moment of inertia:

\(\begin{aligned}{c}{I_0} = \int_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\int_0^{\cos (2\theta )} {{r^5}} } drd\theta \\ = \frac{{5\pi }}{{192}}\\ \approx 0.081812\end{aligned}\)

Which is computed directly from Wolframalpha.com, as directed in the question.

Thus, the mass is of given lamina is \(m = \frac{{3\pi }}{{64}}\)

The centre of mass of given lamina:

\(\begin{aligned}{l}\bar x \approx 0.7095133\\\bar y = 0\end{aligned}\)

The moments of inertia are following:

\(\begin{aligned}{l}{I_x} \approx 0.00281\\{I_y} \approx 0.079\\{I_o} \approx 0.081812\end{aligned}\)