MATH 251 Lecture 23

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Applications of Integrals

Given mass density $\rho (x,y)$ of ideal laminar region (perfectly flat)

mass

$\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}=\underset{R}{\iint }\rho (x,y)\mathrm{d}A$

center of mass

$\overline{x}=\frac{1}{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}\underset{R}{\iint }x\rho (x,y)\mathrm{d}A$

$\overline{y}=\frac{1}{\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}\underset{R}{\iint }y\rho (x,y)\mathrm{d}A$

moment of inertia

first moment about $y$ axis: $\iint x\rho (x,y)\mathrm{d}A$

first moment about $x$ axis: $\iint y\rho (x,y)\mathrm{d}A$

second moment about $y$ axis: $\iint x^2\rho (x,y)\mathrm{d}A$

second moment about $x$ axis $\iint y^2\rho (x,y)\mathrm{d}A$

third moment about $y$ axis: $\iint x^3\rho (x,y)\mathrm{d}A$

third moment about $x$ axis: $\iint y^3\rho (x,y)\mathrm{d}A$

etc.

Example

Find the 2nd moments of:

• $\rho (x,y)=1+x+y$
• $y=1-\frac{x}{3}$
• $y=\frac{x}{3}-1$
• $x\in [0,3]$

Second moment about $y$ axis $\underset{0}{\overset{3}{\int }}\underset{\frac{x}{3}-1}{\overset{1-\frac{x}{3}}{\int }}{x}^2(1+x+y)\mathrm{d}y\mathrm{d}x=45-\frac{162}{5}$

Second moment about $x$ axis $\underset{0}{\overset{3}{\int }}\underset{\frac{x}{3}-1}{\overset{1-\frac{x}{3}}{\int }}{y}^2(1+x+y)\mathrm{d}y\mathrm{d}x=\dots$

Gaussian Curve

Find $I=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-\frac{x^2}{2}}\mathrm{d}x$

Instead, we're going to find $I^2=I\cdot I$, which can be simplified to

$\underset{{\mathbb{ℝ}}^2}{\iint }{\mathrm{e}}^{-\frac{x^2+y^2}{2}}\mathrm{d}A$

Converting this to polar coordinates gives

$\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}r{\mathrm{e}}^{-\frac{r^2}{2}}\mathrm{d}r\mathrm{d}\theta$

Which evaluates to $2\pi$ by u-substitution. Therefore, $I=\sqrt{2\pi }$, so the normalized probabylity density function (pdf) $N(0,1)$ is given by

$p(x)=\frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{-\frac{x^2}{2}}$