Question

a) Draw the region \(R\).

b) Find the coordinates of the centroid of region \(R\).

c) Find the values of \(m\) and \(n\) such that the centroid lies outside R.

Short answer

a)For \(0 < x < 1\), the one with the greater exponent will be below the other.

b) The coordinates of the centroid \((\bar x,\bar y)\) of the region \(R\)is\(\left( {\frac{{(n + 1)(m + 1)}}{{(n + 2)(m + 2)}},\frac{{(n + 1)(m + 1)}}{{(2n + 1)(2m + 1)}}} \right)\).

c) The values of \(m\) and \(n\) is \(\left( {\frac{2}{3},\frac{{20}}{{63}}} \right)\).

Step-by-step solution

Calculate the equations.

(a)

The equations are \(y = {x^m},y = {x^n}\)

The region lies between 0 to 1.

The values \(m\) and \(n\) are integers.

\(\begin{aligned}{}y = {x^m} - - - - - eq(1)\\y = {x^n} - - - - - eq(2)\end{aligned}\)

Plot a graph for the equations \(y = {x^m}\) and \(y = {x^n}\) to find the value of \(x\)and\(y\).

Consider \(m = 1\) and \(n = 1\).

Calculate value of y using Equation (1)

Substitute 0 for \(x\) and 1 for \(m\) in Equation (1).

\(\begin{aligned}{}y = {0^1}\\y = 0\end{aligned}\)

The co-ordinate of is \((0,0)\).

Calculate \(x\) value using Equation (2).

Substitute 1 for \(x\) and 1 for \(n\) in Equation (2).

\(\begin{aligned}{}y = {1^1}\\y = 1\end{aligned}\)

the co-ordinate of (x, y) is \((1,1)\).

Similarly calculate the coordinate values up to bound the region in the graph.

Draw the region as shown in Figure 1.

Calculate the area of region R.

(b) Formula used

\(A = \int_0^1 {\left| {{x^n} - {x^m}} \right|} dx = \int_0^1 {\left( {{x^n} - {x^m}} \right)} dx\)

\( = \left( {\frac{{{x^{n + 1}}}}{{n + 1}} - \frac{{{x^{m + 1}}}}{{m + 1}}} \right)_0^1\)

\( = \frac{{m - n}}{{(m + 1)(n + 1)}}\)

Find the coordinates of the centroid of \(R\).

\(\begin{aligned}\bar x &= \frac{1}{A}\int_0^1 x \left( {{x^n} - {x^m}} \right)dx\\\bar x &= \frac{1}{A}\left( {\frac{{{x^{n + 2}}}}{{n + 2}} - \frac{{{x^{m + 2}}}}{{m + 2}}} \right)_0^1\\\bar x &= \frac{1}{A}\frac{{m - n}}{{(m + 2)(n + 2)}}\\\bar x &= \frac{{(n + 1)(m + 1)}}{{(n + 2)(m + 2)}}\end{aligned}\)

Find the coordinates of the centroid of \(R\).

\(\begin{aligned}\bar y &= \frac{1}{A}\int_0^1 {\frac{1}{2}} \left( {{x^{2n}} - {x^{2m}}} \right)dx\\ &= \frac{1}{{2A}}\left( {\frac{{{x^{2n + 1}}}}{{2n + 1}} - \frac{{{x^{2m + 1}}}}{{2m + 1}}} \right)_0^1\\ &= \frac{1}{{2A}}\frac{{2m - 2n}}{{(2n + 1)(2m + 1)}}\\ &= \frac{{(n + 1)(m + 1)}}{{(2n + 1)(2m + 1)}}\end{aligned}\)

Hence, \((\bar x,\bar y) = \left( {\frac{{(n + 1)(m + 1)}}{{(n + 2)(m + 2)}},\frac{{(n + 1)(m + 1)}}{{(2n + 1)(2m + 1)}}} \right)\)

Calculate the values of \(m\) and \(n\) such that the centroid \((\bar x,\bar y)\) lies outside\(R\)

(c)

Consider \(m = 4\) and \(n = 3\).

\((\bar x,\bar y) = \left( {\frac{{(m + 1)(n + 1)}}{{(m + 2)(n + 2)}},\frac{{(m + 1)(n + 1)}}{{(2m + 1)(2n + 1)}}} \right)\)

Substitute 4 for \(m\) and 3 for \(n\).

\(\begin{aligned}(\bar x,\bar y) &= \left\{{\frac{{(4 + 1)(3 + 1)}}{{(4 + 2)(3 + 2)}},\frac{{(4 + 1)(3 + 1)}}{{|2(4) + 1|(2(3) + 1\mid }}} \right\}\\(\bar x,\bar y) &= \left( {\left( {\frac{{5 \times 4}}{{6 \times 5}}} \right),\left( {\frac{{5 \times 4}}{{9 \times 7}}} \right)} \right)\\(\bar x,\bar y) &= \left( {\frac{2}{3},\frac{{20}}{{63}}} \right)\end{aligned}\)

Hence, the values of \(m\) and \(n\) such that the centroid lies outside \(R\) is \(\left( {\frac{2}{3},\frac{{20}}{{63}}} \right)\).