Question

Use Exercise 22 to find the centroid of the triangle with vertices \((0,0),(a,0)\), and \((a,b)\), where \(a > 0\) and \(b > 0\).

Short answer

The centroid of triangle with vertices \((0,0),(a,0),(a,b)\) is \(\left( {\frac{2}{3}a,\frac{1}{3}b} \right)\).

Step-by-step solution

given information

Vertices of triangle are \((0,0),(a,0),(a,b)\).

expression for coordinates of centroid and for area of triangle

Write the expression for coordinates of centroid\((\bar x,\bar y)\).

\(\begin{array}{l}\bar x = \frac{1}{{2A}}\oint_C {{x^2}} dy.........(1)\\\bar y = - \frac{1}{{2A}}\oint_C {{y^2}} dx.......(2)\end{array}\)

Here,

\(A\)is the area.

Write the expression for area of triangle \((A)\).

\(A = \frac{1}{2}ab\)

Here,

\(a\) and \(b\) are sides of triangle

parametric equations of curve \({C_1},0 \le t \le a\) and differentiate equation (3) & (4) with respect to \(t\) 

The curve \(C\) is divided into three sub-curves \({C_1},{C_2}\) and \({C_3}\).

Write the parametric equations of curve \({C_1},0 \le t \le a\).

\(\begin{array}{l}x = t.........(3)\\y = 0.........(4)\end{array}\)

Differentiate equation (3) with respect to \(t\).

\(\begin{align} & \frac{d}{dt}(x)=\frac{d}{dt}(t) \\ & \frac{dx}{dt}=1\,\,\,\,\,\,\,\,\left\{ \because \frac{d}{dt}(t)=1 \right\} \\ & dx=dt \end{align}\)

Differentiate equation (4) with respect to \(t\).

\(\begin{align} & \frac{d}{dt}(y)=\frac{d}{dt}(0) \\ & \frac{dy}{dt}=0\,\,\,\,\,\left\{ \because \frac{d}{dt}(0)=0 \right\} \\ & dy=0dt \end{align}\)

parametric equation of curve \({C_2},0 \le t \le b\) and differentiate equation (5) & (6) with respect to \(t\)

Write the parametric equation of curve \({C_2},0 \le t \le b\).

\(\begin{array}{l}x = a.......(5)\\y = t......(6)\end{array}\)

Differentiate equation (5) with respect to \(t\).

\(\begin{align} & \frac{d}{dt}(x)=\frac{d}{dt}(a) \\ & \frac{dx}{dt}=0\,\,\,\,\,\,\,\left\{ \because \frac{d}{dt}(k)=0 \right\} \\ & dx=0dt \end{align}\)

Differentiate equation (6) with respect to \(t\).\(t\)

\(\begin{align} & \frac{d}{dt}(y)=\frac{d}{dt}(t) \\ & \frac{dy}{dt}=1\,\,\,\,\,\,\,\,\left\{ \because \frac{d}{dt}(t)=1 \right\} \\ & dy=dt \end{align}\)

Write the parametric equations of curve \({C_3},t = a\) to \(t = 0\) and differentiate equation (7) & (8) with respect to a

Write the parametric equations of curve \({C_3},t = a\) to \(t = 0\).

\(\begin{array}{l}x = t........(7)\\y = \frac{b}{a}t..........(8)\end{array}\)

Differentiate equation (7) with respect to \(t\).

\(\begin{align} & \frac{d}{dt}(x)=\frac{d}{dt}(t) \\ & \frac{dx}{dt}=1\,\,\,\,\,\,\,\,\,\left\{ \because \frac{d}{dt}(t)=1 \right\} \\ & dx=dt \end{align}\)

Differentiate equation \((8)\) with respect to \(t\).

\(\begin{align} & \frac{d}{dt}(y)=\frac{d}{dt}\left( \frac{b}{a}t \right) \\ & \frac{dy}{dt}=\frac{b}{a}(1)\,\,\,\,\,\,\,\,\,\left\{ \because \frac{d}{dt}(t)=1 \right\} \\ & dy=\frac{b}{a}dt \end{align}\)

value of \(\oint_C {{x^2}} dy\)

Find the value of \(\oint_C {{x^2}} dy\).

\(\begin{align} & \oint_{C}{{{x}^{2}}}dy=\int_{{{C}_{1}}}{{{x}^{2}}}dy+\int_{{{C}_{2}}}{{{x}^{2}}}dy+\int_{{{C}_{3}}}{{{x}^{2}}}dy \\ & =\int_{0}^{a}{{{t}^{2}}}(0dt)+\int_{0}^{b}{{{a}^{2}}}dt+\int_{a}^{0}{{{t}^{2}}}\left( \frac{b}{a}dt \right) \\ & =0+{{a}^{2}}\int_{0}^{b}{d}t+\frac{b}{a}\int_{a}^{0}{{{t}^{2}}}dt\,\,\,\,\,\,\,\,\,\left\{ \because \int{0}dt=0 \right\} \\ & ={{a}^{2}}[t]_{0}^{b}+\frac{b}{a}\left[ \frac{{{t}^{3}}}{3} \right]_{a}^{0}\,\,\,\,\,\,\,\,\,\,\,\left\{ \because \int{d}t=t,\int{{{t}^{n}}}dt=\frac{{{t}^{n+1}}}{n+1} \right\} \end{align}\)

Apply limits and simplify the equation.

\(\begin{aligned}\oint_C {{x^2}} dy & = {a^2}(b - 0) + \frac{b}{a}\left( {\frac{{{0^3}}}{3} - \frac{{{a^3}}}{3}} \right)\\ & = {a^2}b - \frac{b}{a}\left( {\frac{{{a^3}}}{3}} \right)\\ & = {a^2}b - \frac{1}{3}{a^2}b\\ & = \frac{2}{3}{a^2}b\end{aligned}\)

Substitute \(\frac{2}{3}{a^2}b\) for \(\oint_C {{x^2}} dy\) and \(\frac{1}{2}ab\) for \(A\) in equation (1),

\(\left( \begin{aligned}\bar x & = \frac{1}{{2\left( {\frac{1}{2}ab} \right)}}\left( {\frac{2}{3}{a^2}b} \right)\\ & = \frac{1}{{ab}}\left( {\frac{2}{3}{a^2}b} \right)\\ & = \frac{2}{3}a\end{aligned} \right)\)

value of \(\oint_C {{y^2}} dx\). 

Find the value of \(\oint_C {{y^2}} dx\).

\(\begin{aligned}\oint_C {{y^2}} dx & = \int_{{C_1}} {{y^2}} dx + \int_{{C_2}} {{y^2}} dx + \int_{{C_3}} {{y^2}} dx\\ & = \int_0^a 0 dt + \int_0^b {{t^2}} (0dt) + \int_a^0 {{{\left( {\frac{b}{a}t} \right)}^2}} dt\\ & = 0 + 0 + \frac{{{b^2}}}{{{a^2}}}\int_a^0 {{t^2}} dt\\ & = \frac{{{b^2}}}{{{a^2}}}\left( {\frac{{{t^3}}}{3}} \right)_a^0\end{aligned}\)

Apply limits and simplify the equation.

\(\begin{aligned}\oint_C {{y^2}} dx & = \frac{{{b^2}}}{{{a^2}}}\left( {\frac{{{0^3}}}{3} - \frac{{{a^3}}}{3}} \right)\\ & = \frac{{{b^2}}}{{{a^2}}}\left( { - \frac{{{a^3}}}{3}} \right)\\ & = - \frac{1}{3}a{b^2}\end{aligned}\)

Substitute \( - \frac{1}{3}a{b^2}\,\)for \(\oint_C {{y^2}} dx\)and \(\frac{1}{2}ab\)for \(A\) in equation (2),

\(\begin{aligned}\bar y & = - \frac{1}{{2\left( {\frac{1}{2}ab} \right)}}\left( { - \frac{1}{3}a{b^2}} \right)\\ & = \frac{1}{{ab}}\left( {\frac{1}{3}a{b^2}} \right)\\ & = \frac{1}{3}b\end{aligned}\)

Thus, the centroid of triangle with vertices a\(\left( {0,0} \right),\left( {a,{\rm{ }}0} \right),(a,{\rm{ }}b)\) is \(\left( {\frac{2}{3}a,\frac{1}{3}b} \right).\)