# Problem of the Week #66 - July 1st, 2013

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#### Chris L T521

##### Well-known member
Staff member
Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Background Info: The tendency of a lamina to resist a change in rotational motion about an axis is measured by its moment of inertia about that axis. If a lamina occupies a region $R$ of the $xy$-plane, and if its density function $\delta(x,y)$ is continuous on $R$, then the moments of inertia about the $x$-axis, $y$-axis, and $z$-axis are denoted by $I_x$, $I_y$, and $I_z$ respectively, and are defined by
$I_x= \iint\limits_R y^2\delta(x,y)\,dA,\qquad I_y = \iint\limits_R x^2\delta(x,y)\,dA$
$I_z= \iint\limits_R (x^2+y^2)\delta(x,y)\,dA$

Problem: Consider the circular lamina that occupies the region described by the inequalities $0\leq x^2+y^2\leq a^2$. Assuming that the lamina has constant density $\delta$, show that
$I_x=I_y=\frac{\delta\pi a^4}{4},\qquad I_z=\frac{\delta\pi a^4}{2}.$

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#### Chris L T521

##### Well-known member
Staff member
This week's problem was correctly answered by Ackbach and Sudharaka. You can find Sudharaka's solution below.

Let $$x=r\cos\theta$$ and $$y=r\sin\theta$$ where $$r\geq 0$$ and $$0\leq\theta<2\pi$$. Then,
\begin{eqnarray}
I_x &=& \delta\int^{2\pi}_{\theta=0} \int^{a}_{r=0}(r^2 \sin^{2} \theta) r \,dr\,d\theta\\
&=&\frac{\delta a^4}{4}\int_{0}^{2\pi}\sin^{2}\theta\,d\theta\\
&=&\frac{\delta\pi a^4}{4}
\end{eqnarray}
Similarly,
\begin{eqnarray}
I_y &=& \delta\int^{2\pi}_{\theta=0} \int^{a}_{r=0}(r^2 \cos^{2}\theta) r \,dr\,d\theta\\
&=&\frac{\delta a^4}{4}\int_{0}^{2\pi}\cos^{2}\theta\,d\theta\\
&=&\frac{\delta\pi a^4}{4}
\end{eqnarray}

\begin{eqnarray}
I_z &=& \delta\int^{2\pi}_{\theta=0} \int^{a}_{r=0}(r^2\sin^{2}\theta+r^2 \cos^{2}\theta) r \,dr\,d\theta\\
&=& \delta\int^{2\pi}_{\theta=0} \int^{a}_{r=0}r^3 \,dr\,d\theta\\
&=&\frac{\delta a^4}{4}\int_{0}^{2\pi}\,d\theta\\
&=&\frac{\delta\pi a^4}{2}
\end{eqnarray}

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