Question

Find the area under one arch of the trochoid of exercise 34 in section 9.1 for the case \(d < r\)

Short answer

The area enclosed of the given curve is given as: \(A = \left( {2\pi {r^2}} \right) + \pi {d^2}\)

Step-by-step solution

Write the information from exercise 34

The equation of curve is:

\(\begin{array}{l}x = r\theta - d\sin \theta \\y = r - d\cos \theta \end{array}\)

Find\(f'(\theta )\)

\(\begin{array}{l}f(\theta ) = r\theta - d\sin \theta \\f'(\theta ) = r - d\cos \theta \end{array}\)

Find the area

The area enclosed by the curve\(x = f(\theta ),y = g(\theta )\)as \(\theta \)increases from\(\alpha \)to\(\beta \)is:

\(A = \int\limits_\alpha ^\beta {g(\theta )f'(\theta )} d\theta \)

Here\(f(\theta ) = r\theta - d\sin \theta ,g(\theta ) = r - d\cos \theta ,\alpha = 0,\beta = 2\pi \)

\(\begin{array}{l}A = \int\limits_\alpha ^\beta {g(\theta )f'(\theta )} d\theta \\A = \int\limits_0^{2\pi } {r - d\cos \theta (r - d\cos \theta )} d\theta \\A = \int\limits_0^{2\pi } {\left( {{r^2} - {d^2}{{\cos }^2}\theta - 2rd\cos \theta } \right)} d\theta \\A = \int\limits_0^{2\pi } {\left( {{r^2} - 2rd\cos \theta } \right)} d\theta + \int\limits_0^{2\pi } {{d^2}{{\cos }^2}\theta d\theta } \\A = \left( {{r^2}\theta - rd\sin \theta } \right)_0^{2\pi } + {d^2}\int\limits_0^{2\pi } {\frac{{\cos 2\theta + 1}}{2}d\theta } \\A = \left( {2\pi {r^2}} \right) + \frac{{{d^2}}}{2}\left( {\frac{{\sin 2\theta }}{2} + \theta } \right)_0^{2\pi }\\A = \left( {2\pi {r^2}} \right) + \frac{{{d^2}}}{2}\left( {2\pi } \right)\\A = \left( {2\pi {r^2}} \right) + \pi {d^2}\end{array}\)

Therefore the area enclosed of the given curve is given as: \(A = \left( {2\pi {r^2}} \right) + \pi {d^2}\)