4.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c14975b7-6afa-44ce-beab-1cba2e82b249-14_675_528_242_772}
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\caption{Figure 3}
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A uniform lamina \(O A B\) is in the shape of the region \(R\).
Region \(R\) lies in the first quadrant and is bounded by the curve with equation \(\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 36 } = 1\), the \(x\)-axis, and the \(y\)-axis, as shown shaded in Figure 3.
The point \(A\) is the point of intersection of the curve and the \(x\)-axis.
The point \(B\) is the point of intersection of the curve and the \(y\)-axis.
One unit on each axis represents 1 m .
The area of \(R\) is \(6 \pi\)
The centre of mass of \(R\) lies at the point with coordinates \(( \bar { x } , \bar { y } )\)
- Use algebraic integration to show that \(\bar { x } = \frac { 16 } { 3 \pi }\)
- Use algebraic integration to find the exact value of \(\bar { y }\)
The lamina is freely suspended from \(A\) and hangs in equilibrium with \(O A\) at angle \(\theta ^ { \circ }\) to the downward vertical.
- Find the value of \(\theta\)