$\left(\stackrel{\&conjugate0;}{x}\,\stackrel{\&conjugate0;}{y}\right)\=\left(\frac{1}{3}\sum _{k\=1}^3{x}_k\,\frac{1}{3}\sum _{k\=1}^3{y}_k\right)$

Obtain the equations of the edges of the triangle in Figure 6.5.3(a)

Loading Student:-Precalculus

$\left[0\,0\right]\,\left[b\,c\right]$$\stackrel{\text{equation of line}}{\to }$$y\=\frac{cx}{b}$$$

$\left[a\,0\right]\,\left[b\,c\right]$$\stackrel{\text{equation of line}}{\to }$$y\=-\frac{cx}{-b\+a}\+\frac{ca}{-b\+a}$$\stackrel{\text{simplify}}{\=}$$y\=\frac{c\left(a-x\right)}{-b\+a}$

Total area

${\int }_0^b{\int }_0^{c x\/b}1 \mathit{\ⅆ}y \mathit{\ⅆ}x\+{\int }_b^a{\int }_0^{c \left(x-a\right)\/\left(b-a\right)}1 \mathit{\ⅆ}y \mathit{\ⅆ}x$

$\frac{1}{2}cb\+\frac{1}{2}\frac{c\left(a^2-b^2\right)}{b-a}-\frac{ac\left(-b\+a\right)}{b-a}$

$\stackrel{\text{simplify}}{\=}$

$\frac{1}{2}ac$

$\stackrel{\text{assign to a name}}{\to }$

$A$

First Moments

${\int }_0^b{\int }_0^{c x\/b}y \mathit{\ⅆ}y \mathit{\ⅆ}x\+{\int }_b^a{\int }_0^{c \left(x-a\right)\/\left(b-a\right)}y \mathit{\ⅆ}y \mathit{\ⅆ}x$

$\frac{1}{6}{c}^{2}b\+\frac{1}{6}\frac{c^2\left(a^3-b^3\right)}{{\left(b-a\right)}^2}-\frac{1}{2}\frac{c^{2}a\left(a^2-b^2\right)}{{\left(b-a\right)}^2}\+\frac{1}{2}\frac{c^2{a}^2\left(-b\+a\right)}{{\left(b-a\right)}^2}$

$\stackrel{\text{simplify}}{\=}$

$\frac{1}{6}{c}^{2}a$

$\stackrel{\text{assign to a name}}{\to }$

$\mathrm{Mx}$

${\int }_0^b{\int }_0^{c x\/b}x \mathit{\ⅆ}y \mathit{\ⅆ}x\+{\int }_b^a{\int }_0^{c \left(x-a\right)\/\left(b-a\right)}x \mathit{\ⅆ}y \mathit{\ⅆ}x$

$\frac{1}{3}c{b}^2\+\frac{1}{3}\frac{c\left(a^3-b^3\right)}{b-a}-\frac{1}{2}\frac{ac\left(a^2-b^2\right)}{b-a}$

$\stackrel{\text{simplify}}{\=}$

$\frac{1}{6}\left(a\+b\right)ac$

$\stackrel{\text{assign to a name}}{\to }$

$\mathrm{My}$

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

$\left[\frac{\mathrm{My}}{A}\,\frac{\mathrm{Mx}}{A}\right]$ = $\left[\frac{1}{3}a\+\frac{1}{3}b\,\frac{1}{3}c\right]$$$

Table 6.5.3(a)   Calculations for the centroid of the triangle in Figure 6.5.3(a)

Initialize

Loading Student:-MultivariateCalculus

Obtain the area of the triangle

$$\begin{gathered} \mathrm{MultiInt}\left(1\&comma;\left[x\&comma;y\right]\&equals;\mathrm{Triangle}\left(⟨0\&comma;0⟩\&comma;⟨a\&comma;0⟩\&comma;⟨b\&comma;c⟩\right)\&comma;\mathrm{output}\&equals;\mathrm{integral}\right) assuming b\&gt;0\&comma;b<a\&comma;c\&gt;0\&semi; \\ A≔\mathrm{simplify}\left(\mathrm{MultiInt}\left(1\&comma;\left[x\&comma;y\right]\&equals;\mathrm{Triangle}\left(⟨0\&comma;0⟩\&comma;⟨a\&comma;0⟩\&comma;⟨b\&comma;c⟩\right)\right)\right) assuming b\&gt;0\&comma;b<a\&comma;c\&gt;0 \end{gathered}$$

$\frac{1}{2}ac$

Obtain $M_x$, the first moment about the x-axis

$$\begin{gathered} \mathrm{MultiInt}\left(y\&comma;\left[x\&comma;y\right]\&equals;\mathrm{Triangle}\left(⟨0\&comma;0⟩\&comma;⟨a\&comma;0⟩\&comma;⟨b\&comma;c⟩\right)\&comma;\mathrm{output}\&equals;\mathrm{integral}\right) assuming b\&gt;0\&comma;b<a\&comma;c\&gt;0\&semi; \\ {M}_x≔\mathrm{simplify}\left(\mathrm{MultiInt}\left(y\&comma;\left[x\&comma;y\right]\&equals;\mathrm{Triangle}\left(⟨0\&comma;0⟩\&comma;⟨a\&comma;0⟩\&comma;⟨b\&comma;c⟩\right)\right)\right) assuming b\&gt;0\&comma;b<a\&comma;c\&gt;0 \end{gathered}$$

$\frac{1}{6}{c}^{2}a$

Obtain $M_y$, the first moment about the y-axis

$$\begin{gathered} \mathrm{MultiInt}\left(x\&comma;\left[x\&comma;y\right]\&equals;\mathrm{Triangle}\left(⟨0\&comma;0⟩\&comma;⟨a\&comma;0⟩\&comma;⟨b\&comma;c⟩\right)\&comma;\mathrm{output}\&equals;\mathrm{integral}\right) assuming b\&gt;0\&comma;b<a\&comma;c\&gt;0\&semi; \\ {M}_y≔\mathrm{simplify}\left(\mathrm{MultiInt}\left(x\&comma;\left[x\&comma;y\right]\&equals;\mathrm{Triangle}\left(⟨0\&comma;0⟩\&comma;⟨a\&comma;0⟩\&comma;⟨b\&comma;c⟩\right)\right)\right) assuming b\&gt;0\&comma;b<a\&comma;c\&gt;0 \end{gathered}$$

$\frac{1}{6}\left(a\&plus;b\right)ca$

Obtain $\left(\stackrel{\&conjugate0;}{x}\&comma;\stackrel{\&conjugate0;}{y}\right)$, the coordinates of the centroid

$\left[\frac{M_y}{A}\&comma;\frac{M_x}{A}\right]$ = $\left[\frac{1}{3}a\&plus;\frac{1}{3}b\&comma;\frac{1}{3}c\right]$$$