Question

Use Exercise 22 to find the centroid of a quarter-circular region of radius \(a\).

Short answer

The centroid of the quarter-circular region of radius \(a\) is \(\left( {\frac{{4a}}{{3\pi }},\frac{{4a}}{{3\pi }}} \right)\).

Step-by-step solution

Given information

Radius is \(a\).

Formula used for centroid

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

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

Step 3:Sketch the region to get a better understanding

find the centroid of a semi-circular region of radius \(a\). From Exercise 22 we have

\(\bar x = \frac{1}{{2A}}\int_C {{x^2}} dy,\quad \bar y = - \frac{1}{{2A}}\int_C {{y^2}} dx\)

Sketch the region to get a better understanding of the situation.

Area of the semicircle is \(\frac{1}{2}{r^2}\pi \), in this case \( \Rightarrow \frac{1}{2}{a^2}\pi \)

Step 4:applying Green's Theorem

Plug \(A\) in and find \(\bar x\) and \(\bar y\), change to polar coordinates after applying Green's Theorem here

\(\begin{aligned}{l}x & = r\cos \theta ,\;\;\;y = r\sin \theta ,\;\;\;0 \le r \le a,\;\;\;0 \le \theta \le \pi \\\bar x & = \frac{1}{{\pi {a^2}}}\int_C {{x^2}} dy\\ & = \frac{1}{{\pi {a^2}}}\int_D {\int_D {\left( {\frac{\partial }{{\partial x}}\left( {{x^2}} \right) - \frac{\partial }{{\partial y}}(0)} \right)} } dA\\ & = \frac{1}{{\pi {a^2}}}\int_0^\pi {\int_0^a 2 } r\cos \theta rdrd\theta \end{aligned}\)

Solve further

\(\begin{aligned}{l} & = \frac{1}{{\pi {a^2}}}\int_0^\pi {\cos } \theta d\theta \int_0^a 2 {r^2}dr\\ & = \left. {\frac{1}{{\pi {a^2}}} \cdot (\sin \theta )_0^\pi \cdot \frac{2}{3}{r^3}} \right|_0^a\\ & = 0\end{aligned}\)

solve further

because \(\sin \pi - \sin 0 = 0\)

\(\)\(\begin{aligned}\bar y & = - \frac{1}{{\pi {a^2}}}\int_C {{y^2}} dx\\ & = - \frac{1}{{\pi {a^2}}}\int_D {\int_D {\left( {\frac{\partial }{{\partial x}}(0) - \frac{\partial }{{\partial y}}\left( {{y^2}} \right)} \right)} } dA\\& = \frac{1}{{\pi {a^2}}}\int_0^\pi {\sin } \theta d\theta \int_0^a 2 {r^2}dr\\ & = \left. {\frac{1}{{\pi {a^2}}} \cdot ( - \cos \theta )_0^\pi \cdot \frac{2}{3}{r^3}} \right|_0^a\end{aligned}\)

Solve further

\(\begin{aligned} & = \frac{1}{{\pi {a^2}}}(1 + 1) \cdot \frac{2}{3}{a^3}\\ & = \frac{{4a}}{{3\pi }}(\bar x,\bar y)\\ & = \left( {0,\frac{{4a}}{{3\pi }}} \right)\end{aligned}\)