91Ó°ÊÓ

Polar coordinates are used to describe the position of points in a plane, distinct from the traditional Cartesian coordinates. In this system, a point is defined by the distance from a reference point, known as the pole, and an angle from a reference direction, usually the positive x-axis. The position is represented as \( (r, \theta) \):

• \( r \) is the radial coordinate, indicating the distance from the pole.
• \( \theta \) is the angular coordinate, representing the counterclockwise angle from the reference direction.

In the exercise, the polar equation \( r = a \cos \theta \) describes a type of curve known as a limaçon. To find the surface area of the curve when revolved, we utilize polar coordinates as they effectively handle rotational symmetries.

Arc length is the distance along the curve between two points, which becomes crucial when dealing with curves in polar coordinates. Calculating the arc length \( ds \) for a polar curve requires considering both the change in \( r \) and the change in angle \( \theta \). The formula for the differential arc length in polar coordinates is:\[ ds = \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \: d\theta \]This formula accounts for the rate of change in radius and also the arc caused by change in angle. In our solution, we calculate \( r' = -a \sin(\theta) \), and incorporate this derivative into the arc length calculation. The square root of the sum of squares ensures only positive values are considered, making sure the length is always non-negative.

Integrals can sometimes seem intimidating, but at their core, they sum up small slices, such as area or length, to find a total. In this problem, a definite integral calculates the total area—specifically, the surface area formed by revolving the curve around a line. The definite integral is set up by defining limits, which are the start and end angles (in this case, from 0 to \( \frac{\pi}{2} \)). The integral stretches along the interval, and computes:\[ A = 2\pi \int_{0}^{\frac{\pi}{2}} (a \cos \theta) \sin \theta \sqrt{a^2} \: d\theta \]This integral setup is crucial because it compacts an infinite sum of small elements into a treatable algebraic expression. Evaluating this expression gives the precise value of the surface area as your final result.

The double angle formula helps simplify trigonometric integrals by expressing them in terms of a single angle instead of products of functions. One crucial identity is:\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]In this specific problem, we employ the formula to simplify our integral. From:\[ 2\pi \int_{0}^{\frac{\pi}{2}} (a \cos \theta) \sin \theta \: d\theta \] we substitute using the double angle identity:\[ 2\pi a^2 \int_{0}^{\frac{\pi}{2}} 0.5 \sin(2\theta) \: d\theta \]This substitution simplifies the computation, converting it to a basic trigonometric integral. It effectively reduces complex trigonometric products into easier, single-function integrals, allowing for straightforward evaluation. Once evaluated, it yields a clear result for the surface area calculation.