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Calculus - Multivariate≻Integration≻Center of Mass≻Polar

Table 6.5.5(a)   Calculation of center of mass via task template

Obtain $m$, the total mass in region $R$ 

${\int }_0^{\mathrm{\π}}{\int }_0^1{r}^2 \mathit{\ⅆ}r \mathit{\ⅆ}\mathrm{\θ}$ = $\frac{1}{3}\mathrm{\π}$$\stackrel{\text{assign to a name}}{\to }$$m$

Obtain $M_x$, the total moments about the $x$-axis

${\int }_0^{\mathrm{\pi }}{\int }_0^1{r}^3 \sin \left(\mathrm{\θ}\right) \mathit{\ⅆ}r \mathit{\ⅆ}\mathrm{\θ}$ = $\frac{1}{2}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Mx}$

Obtain $M_y$, the total moments about the $y$-axis

${\int }_0^{\mathrm{\pi }}{\int }_0^1{r}^3 \cos \left(\mathrm{\theta }\right) \mathit{\ⅆ}r \mathit{\ⅆ}\mathrm{\θ}$ = $0$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{My}$

Obtain $\stackrel{\&conjugate0;}{x}\=M_y\/m$ 

$\mathrm{My}\/m$ = $0$$$

Obtain $\stackrel{\&conjugate0;}{y}\=M_x\/m$ 

$\mathrm{Mx}\/m$ = $\frac{3}{2\mathrm{\π}}$$$

Table 6.5.5(b)   Calculation of the centroid from first principles

Total mass

$q≔\mathrm{Int}\left(r^2\,\left[r\=0..1\,\mathrm{\θ}\=0..\mathrm{\π}\right]\right)$

${\∫}_0^{\mathrm{\π}}{\∫}_0^1{r}^2\ⅆr\ⅆ\mathrm{\θ}$

$m≔\mathrm{value}\left(q\right)$

$\frac{1}{3}\mathrm{\π}$

First Moments

$q≔\mathrm{Int}\left(r^3 \sin \left(\mathrm{\θ}\right)\,\left[r\=0..1\,\mathrm{\θ}\=0.. \mathrm{\π}\right]\right)$

${\∫}_0^{\mathrm{\π}}{\∫}_0^1{r}^3\sin \left(\mathrm{\θ}\right)\ⅆr\ⅆ\mathrm{\θ}$

$M_x≔\mathrm{value}\left(q\right)$

$\frac{1}{2}$

$q≔\mathrm{Int}\left(r^3 \cos \left(\mathrm{\theta }\right)\,\left[r\=0..1\,\mathrm{\theta }\=0.. \mathrm{\pi }\right]\right)$

${\∫}_0^{\mathrm{\π}}{\∫}_0^1{r}^3\cos \left(\mathrm{\θ}\right)\ⅆr\ⅆ\mathrm{\θ}$

$M_y≔\mathrm{value}\left(q\right)$

$0$

Coordinates of centroid $\left(\stackrel{\&conjugate0;}{x}\,\stackrel{\&conjugate0;}{y}\right)$

$\left[\frac{M_y}{m}\,\frac{M_x}{m}\right]$ = $\left[0\,\frac{3}{2\mathrm{\π}}\right]$$$