Step 1

We are given the equation of curve of the shaded area is $y = {x^3}$.

We are asked to determine the centroid $\left( {\bar x,\;\bar y} \right)$ of the shaded area.

 
Step 2

Consider a small vertical strip of thickness $dx$ in the shaded region,

Here, $x$ is the distance of centroid of the strip from the y axis, $y$ is the height of the small strip, $\frac{y}{2}$ is the distance of centroid of the strip from the x axis,

The equation of the curve is given as:

\[y = {x^3}\]

To calculate the area of the small strip, we have:

\[dA = ydx\]

Substitute the known value in the above expression, we get:

\[dA = {x^3}dx\]
 
Step 3

To calculate the centroid of the shaded area along the x axis, we have:

\[\bar x = \frac{{\int\limits_0^1 {xdA} }}{{\int\limits_0^1 {dA} }}\]

Here, x is the x-coordinate of centroid of the small strip.

Substitute all the known values in the above expression, we get:

\[\begin{array}{c} \bar x = \frac{{\int\limits_0^1 {x\left( {{x^3}} \right)dx} }}{{\int\limits_0^1 {{x^3}dx} }}\\ \bar x = \frac{{\left( {\frac{{{x^5}}}{5}} \right)_0^1}}{{\left( {\frac{{{x^4}}}{4}} \right)_0^1}}\\ \bar x = \frac{{\left[ {\frac{{{{\left( 1 \right)}^5}}}{5} - 0} \right]}}{{\left[ {\frac{{{{\left( 1 \right)}^4}}}{4} - 0} \right]}}\\ \bar x = 0.8 \end{array}\]
 
Step 4

To calculate the centroid of the shaded area along the y axis, we have:

\[\bar y = \frac{{\int\limits_0^1 {\left( {\frac{y}{2}} \right)dA} }}{{\int\limits_0^1 {dA} }}\]

Substitute all the known values in the above expression,

\[\bar y = \frac{{\int\limits_0^1 {\left( {\frac{y}{2}} \right){x^3}dx} }}{{\int\limits_0^1 {{x^3}dx} }}\]

Substitute all the known values in the above expression,

\[\begin{array}{l} \bar y = \frac{{\int\limits_0^1 {\left( {\frac{{{x^3}}}{2}} \right){x^3}dx} }}{{\int\limits_0^1 {{x^3}dx} }}\\ \bar y = \frac{{\left( {\frac{1}{2}} \right)\int\limits_0^1 {{x^6}dx} }}{{\int\limits_0^1 {{x^3}dx} }}\\ \bar y = \frac{{\left( {\frac{1}{2}} \right) \times \left( {\frac{{{x^7}}}{7}} \right)_0^1}}{{\left[ {\frac{{{x^4}}}{4}} \right]_0^1}}\\ \bar y = \frac{{\left( {\frac{1}{2}} \right) \times \left( {\frac{{{{\left( 1 \right)}^7}}}{7} - \frac{{{{\left( 0 \right)}^7}}}{7}} \right)}}{{\left( {\frac{{{{\left( 1 \right)}^4}}}{4} - \frac{{{{\left( 0 \right)}^4}}}{4}} \right)}} \end{array}\]

On further solving the above equation, we get:

\begin{array}{l} \bar y = \frac{{\left( {\frac{1}{2}} \right) \times \left( {\frac{1}{7}} \right)}}{{\left( {\frac{1}{4}} \right)}}\\ \bar y = \frac{{\left( {\frac{1}{{14}}} \right)}}{{\left( {\frac{1}{4}} \right)}}\\ \bar y = \left( {\frac{1}{{14}}} \right) \times \left( {\frac{4}{1}} \right)\\ \bar y = 0.286\;{\rm{m}} \end{array}