Step 1

We are given the curve equation as $y = a\cos \frac{{\pi x}}{L}$.

We are asked to locate the centroid of the shaded area.

 
Step 2

We can consider a differential element inside the shaded area which can be represented as:

Here, $dx$ is the width of the differential element, $x'$ is the x coordinate of the centroid of the elemental area and $y'$ is the y co-ordinate of the centroid of the elemental area.

The curve equation is given as:

\[y = a\cos \frac{{\pi x}}{L}\] …. (1)

Here, a is a constant.

To calculate the y co-ordinate of the centroid of the shaded area, we have:

\[\bar y = \frac{{\int\limits_A {y'dA} }}{{\int\limits_A {dA} }}\] …. (2)

The area of the differential element $\left( {dA} \right)$ can be given as:

\[dA = ydx\]

Substitute the values in the above expression, we get:

\[dA = \left( {a\cos \frac{{\pi x}}{L}} \right)dx\] … (3)
 
Step 3

Now, to calculate the y co-ordinate of the centroid of the area located about the y-axis, we have:

\[y' = \frac{y}{2}\]

Substitute the value of x in the above expression, we get:

\[\begin{array}{c} y' = \left( {\frac{{a\cos \frac{{\pi x}}{L}}}{2}} \right)\\ y' = \frac{a}{2}\cos \left( {\frac{{\pi x}}{L}} \right) \end{array}\]

To calculate the integral of $dA$, we need to integrate equation (3) from limit $\left( { - \frac{L}{2}} \right)$ to $\left( {\frac{L}{2}} \right)$ as:

\begin{array}{c} \int\limits_A {dA} = \int\limits_{ - \frac{L}{2}}^{\frac{L}{2}} {\left( {a\cos \frac{{\pi x}}{L}} \right)} \cdot dx\\ \int\limits_A {dA} = \left( {\frac{{aL}}{\pi }} \right)\left[ {\sin \left( {\frac{{\pi x}}{L}} \right)} \right]_{ - \frac{L}{2}}^{\frac{L}{2}} \end{array}

Substitute the limits in x, we get:

\begin{array}{c} \int\limits_A {dA} = \left( {\frac{{aL}}{\pi }} \right)\left[ {\sin \left( {\frac{{\pi \times \left( {\frac{L}{2}} \right)}}{L}} \right) - \sin \left( {\frac{{\pi \times \left( { - \frac{L}{2}} \right)}}{L}} \right)} \right]_{ - \frac{L}{2}}^{\frac{L}{2}}\\ \int\limits_A {dA} = \left( {\frac{{aL}}{\pi }} \right) \times \left( {1 + 1} \right)\\ \int\limits_A {dA} = \frac{{2aL}}{\pi } \end{array}

Substitute the values in equation (2), we get:

\[\begin{array}{c} \bar y = \frac{{\int\limits_{ - \frac{L}{2}}^{\frac{L}{2}} {\left( {\frac{a}{2}\cos \left( {\frac{{\pi x}}{L}} \right)} \right)\left( {\left( {a\cos \frac{{\pi x}}{L}} \right)dx} \right)} }}{{\left( {\frac{{2aL}}{\pi }} \right)}}\\ \bar y = \frac{{\int\limits_{ - \frac{L}{2}}^{\frac{L}{2}} {\left( {\frac{{{a^2}}}{4}} \right)\left( {\cos \frac{{2\pi x}}{L} + 1} \right) \cdot dx} }}{{\left( {\frac{{2aL}}{\pi }} \right)}}\\ \bar y = \frac{{\left( {\frac{{{a^2}}}{4}} \right)\left( {\frac{L}{{2\pi }}\sin \frac{{2\pi }}{L}x + x} \right)_{ - \frac{L}{2}}^{\frac{L}{2}}}}{{\left( {\frac{{2aL}}{\pi }} \right)}} \end{array}\]

Substitute the limits in x, we get:

\[\begin{array}{c} \bar y = \frac{{\left[ {\left( {\frac{{{a^2}}}{4}} \right)\left( {\frac{L}{{2\pi }}\sin \frac{{2\pi }}{L} \times \left( {\frac{L}{2}} \right) + \left( {\frac{L}{2}} \right)} \right)} \right] - \left[ {\left( {\frac{{{a^2}}}{4}} \right)\left( {\frac{L}{{2\pi }}\sin \frac{{2\pi }}{L} \times \left( { - \frac{L}{2}} \right) + \left( { - \frac{L}{2}} \right)} \right)} \right]}}{{\left( {\frac{{2aL}}{\pi }} \right)}}\\ \bar y = \frac{{\left( {\frac{{{a^2}}}{4}} \right)\left[ {\left( {\frac{L}{2}} \right) + \left( {\frac{L}{2}} \right)} \right]}}{{\left( {\frac{{2aL}}{\pi }} \right)}}\\ \bar y = \frac{{\left( {\frac{{{a^2}L}}{4}} \right)}}{{\left( {\frac{{2aL}}{\pi }} \right)}}\\ \bar y = \frac{\pi }{8}a \end{array}\]

Hence, the centroid of the $\left( {\bar x,\;\bar y} \right)$ of the shaded area will be $\left( {0,\;\frac{\pi }{8}a} \right)$.