Problem 19 - Thomas Calculus

Question

Find the areas of the surfaces generated by revolving the curves in Exercises \(13-23\) about the indicated axes. If you have a grapher, you may want to graph these curves to see what they look like. $$x=2 \sqrt{4-y}, \quad 0 \leq y \leq 15 / 4 ; \quad y -axis$$

Step-by-Step Solution

Answer

Surface area is calculated using the integral \( \int_{0}^{15/4} 4\pi \sqrt{8-y} \, dy \).

Step 1: Understand the Problem

We need to find the area of the surface generated by revolving the curve \( x=2 \sqrt{4-y} \) around the \( y \)-axis, for \( 0 \leq y \leq \frac{15}{4} \).

Step 2: Use the Surface Area Formula

The formula for the surface area of a solid of revolution about the \( y \)-axis is \( S = \int_{a}^{b} 2\pi x \sqrt{1+\left(\frac{dx}{dy}\right)^2} \, dy \). Here, \( x \) is the curve and \( \frac{dx}{dy} \) is the derivative of \( x \) with respect to \( y \).

Step 3: Find Derivative \( \frac{dx}{dy} \)

The given equation is \( x=2 \sqrt{4-y} \). To find \( \frac{dx}{dy} \), use the chain rule. \( \frac{dx}{dy} = \frac{d}{dy}[2(4-y)^{1/2}] = -\frac{2}{\sqrt{4-y}} \).

Step 4: Set Up the Integral

Substitute the expressions for \( x \) and \( \frac{dx}{dy} \) into the surface area formula: \[ S = \int_{0}^{\frac{15}{4}} 2\pi (2 \sqrt{4-y}) \sqrt{1+\left(-\frac{2}{\sqrt{4-y}}\right)^2} \, dy \].

Step 5: Simplify the Integral

Simplify the integrand. \( \sqrt{1+\left(-\frac{2}{\sqrt{4-y}}\right)^2} = \sqrt{1+\frac{4}{4-y}} = \sqrt{\frac{4-y+4}{4-y}} = \sqrt{\frac{8-y}{4-y}} \). So, the integral becomes \[ S = \int_{0}^{\frac{15}{4}} 4\pi \sqrt{8-y} \, dy \].

Step 6: Evaluate the Integral

Calculate the integral, \[ S = 4\pi \int_{0}^{\frac{15}{4}} (8-y)^{1/2} \, dy \]. To simplify the integration, use the substitution \( u = 8-y \), thus \( du = -dy \). Change the limits of integration accordingly: when \( y = 0, u = 8 \) and when \( y = \frac{15}{4}, u = 8-\frac{15}{4} = \frac{17}{4} \). The integral becomes \[ -4\pi \int_{8}^{17/4} u^{1/2} \, du \].

Step 7: Calculate the Integral

Compute the definite integral \( -4\pi \int_{8}^{17/4} u^{1/2} \, du \) using the antiderivative \( -4\pi \left[ \frac{2}{3}u^{3/2} \right] \). Evaluate this from \( u = 8 \) to \( u = \frac{17}{4} \) which simplifies to \[ S = -\frac{8\pi}{3} \left( \left( \frac{8}{4} \right)^{3/2} - (8)^{3/2} \right) \].

Step 8: Finish Calculations

Calculate the results: \( S = -\frac{8\pi}{3} ((\frac{\sqrt{17}}{2})^3 - 8\sqrt{8}) \) which further simplifies and can be calculated numerically.

Key Concepts

Integral Calculus

Integral calculus is a fundamental part of mathematics that deals with finding the integral of a function. It helps us calculate areas, volumes, and other related concepts. By integrating a function, we find the accumulation of quantities, which is essential in solving various problems in physics, engineering, and other fields.
Integral calculus explores two main processes: differentiation and integration.

• Differentiation breaks down functions into the smallest parts to identify the rate of change.
• Integration is the reverse process that combines these small parts into the total accumulated value.

To solve surface area problems involving revolved surfaces, as in the exercise, integral calculus is employed. We use a method called "integration" to sum an infinite number of infinitesimally small quantities.
This approach allows us to handle continuous data effectively, which is crucial for understanding complex surfaces generated by revolving curves.

Surface Area

The surface area of a solid is the total area that this surface occupies. When a curve revolves around an axis (like the y-axis in our example), it sweeps out a surface in three dimensions.
Understanding how to compute such a surface area is crucial in various applications such as engineering, physics, and architecture.

• In the context of a surface of revolution, we consider the surface formed by rotating a curve about a given line (the axis).
• For mathematical calculations, a special formula adapted for such surfaces is used.
  The formula integrates the length of these tiny circular ribbons, which sum up to make the total surface.

In our specific problem, the curve given by the equation \(x = 2\sqrt{4-y}\) is revolved about the y-axis. We need to calculate the surface area of the shape formed by this rotation from \( y=0 \) to \( y = \frac{15}{4} \).
The surface area formula we use is \( S = \int_{a}^{b} 2\pi x \sqrt{1+\left(\frac{dx}{dy}\right)^2} \, dy \), where \(x\) and \( \frac{dx}{dy} \) are obtained from the given curve.

Definite Integral

A definite integral computes the net area under a curve between two specified limits. It provides a numeric value representing this area, making it indispensable in numerous mathematical and practical applications.
In our exercise, we use definite integration to calculate the surface area of a revolved curve.

• We set up a definite integral using formulated boundaries (0 and \(\frac{15}{4}\) in our example) and then solve it to find the surface area.
• A definite integral enables us to compute this exact area rather than relying on approximations.

The process involves replacing variables with expressions that describe them relative to one another (substitution), as seen in the solution where \(u = 8 - y\) was substituted.
This greatly simplifies finding the integral's value.
After substitution and integration, we evaluate the antiderivative at both bounds and subtract the results to find the net area, allowing us to solve complex real-world problems efficiently.