Problem 43 - Essential Calculus

Question

In Exercises 43-46, find the area of the surface formed by revolving the curve about the given line. $$ \begin{array}{lll} \underline{\text { Polar Equation }} & \underline{\text { Interval }} & \underline{\text { Axis of Revolution }} \\ r=6 \cos \theta & 0 \leq \theta \leq \frac{\pi}{2} & \text { Polar axis } \end{array} $$

Step-by-Step Solution

Answer

The exact value of the surface area cannot be determined directly from the integral. It could be approximated using numerical methods.

Step 1: Compute the Derivative of the Function

The first step involves finding the derivative of \( r(\theta) \) with respect to θ. From calculus, we know that the derivative of \( cos\theta \) is \( -sin\theta \). Therefore, \( r'(\theta)=-6sin\theta \).

Step 2: Insert r and r' into the Surface Area Formula

The next step involves substituting \( r(\theta) \) and \( r'(\theta) \) into the formula for the surface area of a solid of revolution in polar coordinates. With \( r(\theta)=6cos\theta \) and \( r'(\theta)=-6sin\theta \), the formula becomes: \( 2\pi\int_0^{\frac{\pi}{2}} r(\theta)\sqrt{1+[r'(\theta)]^2} d\theta = 2\pi\int_0^{\frac{\pi}{2}} 6cos\theta\sqrt{1+(-6sin\theta)^2} d\theta \).

Step 3: Simplify and Evaluate the Integral

The next step is to simplify the integral and evaluate it. First, we simplify under the square root: \( 1+(-6sin\theta)^2 = 1+36sin^2\theta \). Substituting into the integral, we get: \( 2\pi\int_0^{\frac{\pi}{2}} 6cos\theta\sqrt{1+36sin^2\theta} d\theta \). Lastly, we evaluate this integral. Unfortunately, it is non-trivial and might require numerical methods to solve.

Key Concepts

Polar Equations

Understanding polar equations is fundamental when analyzing curves in polar coordinates. A polar equation typically expresses the relationship between the radius r and the angle θ in the polar coordinate system. The equation r = f(θ) describes how the radius changes as the angle varies, which in turn defines a curve on a polar grid. The curves often represent complex shapes that would be difficult to express in Cartesian coordinates.

For example, in the given exercise, the polar equation r = 6cosθ describes a circle with a radius that changes as a function of the angle θ. This specific form of the polar equation is a classic example, prevalent in coursework pertaining to polar coordinates. Here, the radius is at its maximum when θ = 0 and decreases to zero as θ approaches π/2, which elegantly shows the creation of a circular sector in the polar coordinate space.

Derivatives in Polar Coordinates

In polar coordinates, taking a derivative might seem a bit more intricate than in Cartesian coordinates, but it plays a crucial role in comprehending the dynamics of polar curves. When dealing with polar equations, the derivative dr/dθ describes how quickly the radius changes with respect to the angle θ. A steep derivative indicates a rapid change in radius for a small change in angle.

In the context of our exercise, the derivative of r with respect to θ, or r'(θ), informs us about the slope of the curve at any given point. This is pertinent when we calculate the surface area of the solid of revolution as it influences the nature of the surface. With r'(θ) = -6sinθ, we observe that the slope transitions from 0 to -6 as θ goes from 0 to π/2, affecting the shape's topology when revolved.

Integrals in Calculus

Integrals are the cornerstone of calculus, playing an indispensable role in diverse applications such as calculating areas, volumes, and, as in our case, the surface areas of solids of revolution. An integral aggregates an infinite sum of infinitesimally small data, typically the product of a function and a small change in the variable.

The integral in the provided solution encompasses the calculation of the surface area formed by revolving the polar curve around an axis. It is achieved by integrating the radius function multiplied by the square root of the sum of one and the square of its derivative, over the given interval of θ. The appearance of the integral, 2π∫ 6cosθ√(1+36sin²θ) dθ, essentially represents the continuous sum of circular frustums formed by the revolving curve, hence giving the surface area of the three-dimensional object. In cases where such integrals become complex, numerical methods or approximation techniques are often used to evaluate them.