Question

Find the surface area of the described solid of revolution. The sphere formed by revolving \(y=\sqrt{1-x^{2}}\) on [-1,1] about the \(x\) -axis.

Short answer

The surface area of the sphere is \(4\pi\).

Step-by-step solution

Understand the Problem

We need to find the surface area of a solid formed by rotating a semicircle with the equation \(y = \sqrt{1-x^2}\), over the interval \([-1, 1]\), around the \(x\)-axis. This is the upper half of a circle with radius 1.

Recall the Formula for Surface Area

The surface area \(A\) of a solid of revolution obtained by rotating the curve \(y=f(x)\) from \(x=a\) to \(x=b\) around the \(x\)-axis is given by the integral: \[ A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + [f'(x)]^2} \, dx \]

Calculate the Derivative

Find the derivative of \(y = \sqrt{1-x^2}\). Using the chain rule, we get \[ f'(x) = \frac{d}{dx} (1-x^2)^{1/2} = \frac{-x}{\sqrt{1-x^2}}. \]

Substitute into Surface Area Formula

Substitute \(y = \sqrt{1-x^2}\), and \(f'(x) = \frac{-x}{\sqrt{1-x^2}}\) into the surface area formula:\[ A = 2\pi \int_{-1}^{1} \sqrt{1-x^2} \sqrt{1 + \left(\frac{-x}{\sqrt{1-x^2}}\right)^2} \, dx \].

Simplify the Integral

Simplify the square root inside the integral:\( \sqrt{1 + \left( -\frac{x}{\sqrt{1-x^2}} \right)^2} = \sqrt{1 + \frac{x^2}{1-x^2}} = \sqrt{\frac{1}{1-x^2}}. \)Therefore, the expression becomes: \[ A = 2\pi \int_{-1}^{1} \sqrt{1-x^2} \cdot \frac{1}{\sqrt{1-x^2}} \, dx = 2\pi \int_{-1}^{1} 1 \, dx. \]

Evaluate the Integral

The simplified integral is easy to evaluate: \[ A = 2\pi \int_{-1}^{1} 1 \, dx = 2\pi [x]_{-1}^{1} = 2\pi (1 - (-1)) = 4\pi. \]

Key concepts

Solid of Revolution

A solid of revolution is created when we rotate a 2D shape around an axis, generating a 3D object. In this particular exercise, we rotated a semicircle, described by the function \(y = \sqrt{1-x^2}\), around the \(x\)-axis. This results in a sphere, a classic example of a solid of revolution. When picturing this, imagine drawing an arc of the semicircle and then spinning it around the axis. This rotation helps in visualizing how a flat shape turns into a full-bodied object.
Rotating around different axes or with different shapes can create a variety of solids. The method for finding the surface area of these 3D shapes requires understanding the geometry of the 2D shape and its axis of rotation.

Integral Calculus

Integral calculus plays a crucial role in calculating the surface area of solids of revolution. The integral used in this scenario assesses the cumulative effect when an infinitesimally small segment of the curve is rotated to form the shape.
The formula applied in this problem is:

• \( A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + [f'(x)]^2} \, dx \)

This integral captures both the curve \(f(x)\) and the change or slope \(f'(x)\) along the range of integration. Integrals are a fantastic tool as they can sum up infinitely small pieces to obtain a whole, in this case, the surface area of a sphere.

Chain Rule

The chain rule is a fundamental method in calculus used to differentiate composite functions. It is especially useful when dealing with functions nested within other functions.
For the function \(y = \sqrt{1-x^2}\), it can be thought of as the outer function \((u)^{1/2}\) and the inner function \(u = 1-x^2\). The derivative, using the chain rule, is calculated as:

• First, differentiate the outer function: \(\frac{1}{2}(u)^{-1/2}\).
• Then, multiply by the derivative of the inner function: \(-2x\).
• This results in: \(-x/\sqrt{1-x^2}\).

The chain rule enables us to tackle more complex problems involving multiple functions dependent on one another.

Sphere Surface Area

The exercise leads us to the surface area of a sphere, a well-known geometric formula. By rotating the semicircle, we see that an entire sphere is formed, and its surface area can be calculated as \(4\pi\) using the integration method described.
The standard formula for the surface area of a sphere is \(4\pi r^2\), where \(r\) is the radius. For this problem, the semicircle radius was 1, so directly applying the formula fits perfectly with the calculated result of \(4\pi\).
This example beautifully ties together calculus principles with geometrical intuition, demonstrating how rotation and integration help reveal fundamental properties of 3D shapes.