Question

Find the area of the surface formed by revolving the curve about the given line. $$ \begin{array}{llll} \underline{\text { Polar Equation }} & & \text { Interval } & & \text { Axis of Revolution } \\ r=6 \cos \theta & & 0 \leq \theta \leq \frac{\pi}{2} & & \text { Polar axis } \\\ \end{array} $$

Short answer

The area of the surface formed is \( \frac{3\pi}{2} \) units

Step-by-step solution

Conversion from Polar to Cartesian

The relationship between polar and Cartesian coordinates is such that \(x = rcos(\theta)\) and \(y = rsin(\theta)\). Under polar coordinates given as \(r=6cos(\theta)\) where \(0 \leq \theta \leq \frac{\pi}{2}\), the Cartesian coordinates are \(x = rcos(\theta) = 6cos^{2}(\theta)\) and \(y = rsin(\theta) = 6cos(\theta)sin(\theta) = 3sin(2\theta)\).

Formulate Area equation

The integral that represents the area of the surface of revolution, using arc length, is given by \(A = 2\pi \int yp\,dq\) from \(q = 0\) to \(q = \frac{\pi}{2}\) which gives: \(A = 2\pi \int _{0}^{\pi/2} 3sin(2\theta) d\theta\). Using the derivatives formulas for sine and cosine we get.

Evaluate integral

The integral evaluates as \(A = 2\pi [ -\frac{3}{2} cos(2\theta) ]_{0}^{\pi/2}\) simplifying gives \(A = 2\pi [0 - (-\frac{3}{2})] = \frac {3\pi}{2}\)