Question

In Exercises 43-46, find the area of the surface formed by revolving the curve about the given line. $$ \begin{array}{lll} \underline{\text { Polar Equation }} & \underline{\text { Interval }} & \underline{\text { Axis of Revolution }} \\ r=a \cos \theta & 0 \leq \theta \leq \frac{\pi}{2} & \theta=\frac{\pi}{2} \end{array} $$

Short answer

The area of the surface formed by revolving the curve about the given line is \( \pi a^2 \).

Step-by-step solution

Understanding the Polar Equation

Polar coordinates describe points in the plane using a distance from the origin (r) and an angle (\(\theta\)) from the positive x-axis. Here, the polar equation \( r = a cos \theta \) describes a circle of radius 'a' centered at the origin. Since \( \theta \) varies from 0 to \( \frac{\pi}{2} \), this represents a quarter of a circle on the first quadrant.

Interpreting the Axis of Revolution

The line \( \theta = \frac{\pi}{2} \) is the positive y-axis. So the curve is revolved around this y-axis, forming a cone-like shape with its tip at the origin of the polar coordinate plane and the wider part along the y-axis.

Setting up the Surface Area Integral

The surface area A of a surface of revolution in polar coordinates is given by the formula: \( A = 2\pi \int r sin \theta ds \), where ds is the differential arc length. The differential arc length in polar coordinates is given by \( ds = \sqrt{r^2+(dr/d\theta)^2} d\theta \). Substituting \( r=a cos\theta \) and \( dr/d\theta = -a sin\theta \) into the arc length formula gives \( ds = a d\theta \). Substituting these into the surface area formula gives: \( A = 2\pi \int_{0}^{\pi /2} a^2 cos \theta sin \theta d \theta \).

Evaluating the Integral

Integrate to find the area: \( A = 2\pi a^2 \int_{0}^{\pi/2} sin 2\theta d\theta \). Making the substitution \( x = cos 2\theta \), \( dx = -sin 2\theta d\theta \), the integral becomes \( A = 2\pi a^2 [ -1/2 cos 2\theta ]_{0}^{\pi/2} = \pi a^2 \).

Key concepts

Polar Coordinates

Polar coordinates are an alternative to the more widely known Cartesian coordinate system. In polar coordinates, each point on a plane is determined by a distance from a reference point, known as the origin, and an angle from a reference direction, often the positive x-axis.

In the provided exercise, we're dealing with the polar equation \r(a = a \text{cos} \theta), which is a fundamental form representing a circle when plotted on polar coordinates. The distance '`r`' from the origin varies with the angle '`\theta`', painting a curve that can depict various shapes such as circles and spirals depending on the equation. Since \theta varies from 0 to \( \frac{\pi}{2} \), the quarter circle in question lies in the first quadrant, which is bound by the positive x-axis and positive y-axis.

Integrals in Calculus

Integrals are a cornerstone of calculus, allowing us to find quantities like area, volume, and, in our particular case, the surface area of a solid of revolution. Integration in calculus is, in essence, the reverse operation of differentiation.

When a curve described in polar coordinates is revolved around an axis, an integral can calculate the resulting surface area. Typical integration, as seen with Cartesian coordinates, follows the formula \(A = \int f(x)dx\), while in polar coordinates, we adjust this concept to work with polar equations and the angles that define them.

The differential element of arc length (ds) in polar coordinates is crucial, as it incorporates the rate at which 'r' changes with 'θ', which is a bit more complex than just integrating the function itself. By plugging in the right equations, you can find out the surface area of the shape generated by rotation.

Surface Area of a Cone

The surface area of a cone is one of the simpler forms you can calculate in three-dimensional geometry. A cone can be visualized as a triangle that's rotated around one of its straight sides, known as the axis.

In our exercise, revoling a curve around an axis similar to the positive y-axis, we are, in effect, creating a cone. The formula for the lateral surface area of a cone (not including the base) is \(A = \pi r l\), where 'r' is the radius of the base and 'l' is the slant height of the cone. In the context of a surface of revolution, however, things are slightly more complicated as we use integrals to consider the evolving radius as the shape translates through the curve.

Thus, the use of integrals and the specific geometry of the figure in polar coordinates create a scenario where the surface area of parts of a cone or cone-like solids can be accurately determined, extending the principle of the simple cone surface area formula to more complex, evolving shapes.