Step 1

We are given the following data:

The curve function is $y = {\left( {4 - {x^{\frac{1}{2}}}} \right)^2}$.

We are asked to locate the centroid $\bar x$ of the area.

 
Step 2

The free-body diagram of the curve is given below.

Here, consider an arbitrary element of thickness $dx$, $dA$ represent the area of the element and $\left( {\tilde x,\tilde y} \right)$ represent the coordinate of the centroid of the curve.

The area of the element can be represented as,

\[dA = ydx\]

The centroid will located at distance of,

\[\tilde x = x\]
 
Step 3

The integration limit will be ${x_1} = 0$ to ${x_2} = 4\;{\rm{ft}}$.

The formula to locate centriod $\bar x$ of the area is given by,

\[\begin{array}{c} \bar x = \frac{{\int_{\rm{A}} {\tilde xdA} }}{{\int_{\rm{A}} {dA} }}\\ = \frac{{\int_{{x_1}}^{{x_2}} {\left( {xydx} \right)} }}{{\int_{{x_1}}^{{x_2}} {\left( {ydx} \right)} }} \end{array}\]

Substitute all the known values in the above formula.

\[\begin{array}{c} \bar x = \frac{{\int_0^{4\;{\rm{ft}}} {\left[ {\left( x \right){{\left( {4 - {x^{\frac{1}{2}}}} \right)}^2}dx} \right]} }}{{\int_0^{4\;{\rm{ft}}} {\left[ {{{\left( {4 - {x^{\frac{1}{2}}}} \right)}^2}dx} \right]} }}\\ = \frac{{\int_0^{4\;{\rm{ft}}} {\left[ {x\left( {x - 8{{\left( x \right)}^{\frac{1}{2}}} + 16} \right)dx} \right]} }}{{\int_0^{4\;{\rm{ft}}} {\left( {x - 8{{\left( x \right)}^{\frac{1}{2}}} + 16} \right)dx} }}\\ = \frac{{\left[ {\frac{{{x^3}}}{3} - \frac{{16}}{5}{x^{\left( {\frac{5}{2}} \right)}} + 8{x^2}} \right]_0^{4\;{\rm{ft}}}}}{{\left[ {\frac{{{x^2}}}{2} - \frac{{16}}{3}{x^{\left( {\frac{3}{2}} \right)}} + 16x} \right]_0^{4\;{\rm{ft}}}}}\\ = \frac{{\left[ {\frac{{{{\left( 4 \right)}^3}}}{3} - \frac{{16}}{5}{{\left( 4 \right)}^{\left( {\frac{5}{2}} \right)}} + 8{{\left( 4 \right)}^2} - 0 + 0 - 0} \right]}}{{\left[ {\frac{{{{\left( 4 \right)}^2}}}{2} - \frac{{16}}{3}{{\left( 4 \right)}^{\left( {\frac{3}{2}} \right)}} + 16\left( 4 \right) - 0 + 0 - 0} \right]}} \end{array}\]

On further solving the above equation,

\[\begin{array}{c} \bar x = \left[ {\frac{{\left( {46.93} \right)}}{{\left( {29.33} \right)}}} \right]\;{\rm{ft}}\\ \approx 1.6\;{\rm{ft}} \end{array}\]