Question

Find the area of the surface generated by revolving the given curve about the \(x\) -axis \(y=\sqrt{25-x^{2}},-2 \leq x \leq 3\)

Short answer

The surface area is \( 50\pi \).

Step-by-step solution

Understand the Problem

We are given the curve \( y = \sqrt{25-x^{2}} \), which is the upper semicircle of radius 5 centered at the origin, and we need to find the area of the surface created when this curve is revolved around the \(x\)-axis, between \(x = -2\) and \(x = 3\).

Formula for Surface Area of Revolution

The formula for the surface area when a curve \( y = f(x) \) is revolved around the \(x\)-axis is \[ S = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^{2}} \, dx \] where \( [a, b] \) is the interval over which the curve is defined.

Find the Derivative

The derivative of \( y = \sqrt{25-x^{2}} \) is found using the chain rule. \( \frac{dy}{dx} = \frac{-x}{\sqrt{25-x^{2}}} \).

Set Up the Integral

Substitute into the surface area formula: \( y = \sqrt{25-x^{2}} \) and \( \frac{dy}{dx} = \frac{-x}{\sqrt{25-x^{2}}} \). The integral becomes:\[ S = \int_{-2}^{3} 2\pi \sqrt{25-x^{2}} \cdot \sqrt{1 + \left(\frac{-x}{\sqrt{25-x^{2}}}\right)^{2}} \, dx \].

Simplify the Integral

Simplify the expression inside the integral:\( \sqrt{1 + \left(\frac{-x}{\sqrt{25-x^{2}}}\right)^{2}} = \sqrt{1 + \frac{x^2}{25-x^{2}}} = \frac{5}{\sqrt{25-x^{2}}} \).Then the integral becomes \( S = \int_{-2}^{3} 2\pi \cdot 5 \, dx \).

Evaluate the Integral

Evaluate the simplified integral:\[ S = 10\pi \int_{-2}^{3} \, dx = 10\pi [x]_{-2}^{3} = 10\pi (3 - (-2)) = 10\pi \times 5 = 50\pi \].

Key concepts

Integral Calculus

Integral calculus is a core branch of mathematics that primarily deals with integrals and their properties. It helps us calculate areas, volumes, central points, and many other essential concepts in both mathematics and physics. One of its most profound applications is in finding the surface area of revolution, which involves rotating a curve around an axis.
To achieve this, integral calculus uses the concept of definite integrals, allowing us to measure the precise area underneath a curve within a specific interval. For instance, in the exercise where we revolve the curve \( y = \sqrt{25-x^{2}} \) about the \(x\)-axis, integral calculus helps set up and evaluate the integral that gives us the surface area.
Essentially, integrals sum up an infinite number of infinitesimally small quantities, like thin slices of a shape, to find whole values such as total area or volume. It's through this process that we've calculated the surface area of a semicircle rotated around an axis in the problem.

Definite Integral

A definite integral is an integral with specific upper and lower limits, represented as \( \int_{a}^{b} f(x) \, dx \). It gives the net area under a curve \( f(x) \) from \( x = a \) to \( x = b \).
In the context of solving surface area of revolution problems, the definite integral helps in evaluating the total surface area between given boundaries. For the exercise, the surface area is found by evaluating the integral between \( x = -2 \) and \( x = 3 \).

• The integral \( \int_{-2}^{3} 2\pi \cdot 5 \, dx \) represents an accumulative process from \( -2 \) to \( 3 \).
• We interpret the limits \(a\) and \(b\) as the interval over which we're finding the surface area.
• After evaluating this definite integral, we obtain a precise numeric value, 50\(\pi\), signifying the total surface area.

Chain Rule

The chain rule is an essential principle of differentiation used to find the derivative of a composite function. When dealing with the equation \( y = \sqrt{25-x^{2}} \), which is a part of a surface area problem, we rely on the chain rule to calculate \( \frac{dy}{dx} \).
The chain rule states that the derivative of a composite function \( f(g(x)) \) can be found by taking the derivative of the outer function \( f \) and multiplying it by the derivative of the inner function \( g \). In our exercise, we start by differentiating the outer function, \( u^{1/2} \), and then the inner function, \( 25-x^2 \).

• The derivative of \( u^{1/2} \) becomes \( \frac{1}{2\sqrt{25-x^{2}}} \).
• The derivative of \( 25-x^2 \) is \(-2x\).
• Combining these results using the chain rule gives \( \frac{dy}{dx} = \frac{-x}{\sqrt{25-x^{2}}} \).

Understanding the chain rule allows us to properly set up and solve the integral needed for calculating the surface area.