MATH1011 - Lecture 16

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MATH1011 - Lecture 15
Wk8 Skeleton Slides
Wk9 Skeleton Slides

Mass

• Consider a body in ${\mathbb{R}}^3$, let $\rho (x,y,z)$ be a function that gives the mass density of the body at point $(x,y,z)$ $M=\underset{R}{\iiint }\rho (x,y,z)dxdydz$
  • $M$ $-$ mass of the body

Moments of Inertia

• Moments of inertia measure a body’s resistance to change by a rotation around an axis
  • The further away a mass is to the axis, more it resists the change
  • Apparently something about this is wrong in lecture but right in unit reader? Wasn’t elaborated
• Moments of inertia are given by these formulas: M_{yz} = \iiint\limits_{R} x\rho(x,y,z)\ dx\ dy\ dz$$$$M_{xz} = \iiint\limits_{R} y\rho(x,y,z)\ dx\ dy\ dz $M_{xy}=\underset{R}{\iiint }z\rho (x,y,z)dxdydz$
• We need this information to find the centre of mass

Centre of Mass

$(C_x,C_y,C_z)=(\frac{M_{yz}}{M},\frac{M_{xz}}{M},\frac{M_{xy}}{M})$

• If the body is a lamina (that is, a very thin flat plate) we might want to drop the third dimension

Change of Coordinates

• i.e. integrate $x$ over $[a,b]$ instead of integrating $u$ over $[c,d]$

Change of Coordinates Using Double Integrals

${\int }_a^{b}f(x)dx={\int }_c^{d}f(g(u))g^{\prime }(u)du$

• $x=g(u)$
• $dx=g^{\prime }(u)du$
  where if $g$ has an inverse function, the bounds change as follows:$\begin{array}{r}x=a⟹u=g^{-1}(a)=c\\ x=b⟹u=g^{-1}(b)=d\end{array}$
• In many cases a double integral can be simplified by a change of coordinates
  • $(x,y)=g(u,v)$
• We need to figure out:
  • How $du$ $dv$ relates to $dx$ $dy$
  • How to adjust the bounds of integration
• Watch lecture recording for worked solution

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MATH1011 - Lecture 17
Annotated Slides