Question

A parametrization of a circle of radius 1 centered at \((1,0)\) in the \(x y\) -plane is given by \(x=1+\cos t, y=\sin t\), for \(0 \leq t \leq 2 \pi .\) Find the surface area when this curve is revolved about the \(y\) -axis.

Short answer

The surface area is \(4\pi^2\).

Step-by-step solution

Understand the problem context

We're asked to find the surface area generated by revolving a circle parametrized by \(x=1+\cos t, y=\sin t\) around the \(y\)-axis. This involves using the surface area formula for a revolution around the \(y\)-axis.

Apply Surface Area of Revolution Formula

The surface area \(A\) of a curve \(y = f(x)\) rotated about the \(y\)-axis from \(t = a\) to \(t = b\) is given by \[ A = \int_{a}^{b} 2\pi x \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]. Here, \(x = 1 + \cos t\) and \( y = \sin t \). We must compute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

Compute derivatives

Compute \( \frac{dx}{dt} = -\sin t \) and \( \frac{dy}{dt} = \cos t \). We will use these derivatives for the integration.

Set up the integral for surface area

Substitute \( x = 1 + \cos t \), \( \frac{dx}{dt} = -\sin t \), and \(\frac{dy}{dt} = \cos t\) into the formula: \[ A = \int_{0}^{2\pi} 2\pi (1+\cos t) \sqrt{(-\sin t)^2 + (\cos t)^2} \, dt \]. Recognize that \((-\sin t)^2 + (\cos t)^2 = 1\), simplifying the integral.

Simplify the integral expression

Substitute back into the integral, resulting in \[ A = \int_{0}^{2\pi} 2\pi (1+\cos t) \, dt \]. Split the integral into two parts: \[ A = 2\pi \int_{0}^{2\pi} (1) \, dt + 2\pi \int_{0}^{2\pi} (\cos t) \, dt \].

Evaluate each integral.

Calculate the two separate integrals: \( \int_{0}^{2\pi} 1 \, dt = 2\pi \) and \( \int_{0}^{2\pi} \cos t \, dt = 0 \). So, the total surface area \( A = 2\pi \times 2\pi + 2\pi \times 0 = 4\pi^2 \).

Conclusion

The surface area of the circle, when revolved around the \(y\)-axis, is \(4\pi^2\). This matches the expected shapes created by revolution using known geometric shapes like a torus.

Key concepts

Surface Area of Revolution

When a curve is rotated around an axis, it creates a 3D surface. The surface area of this 3D object is known as the surface area of revolution. To find it, we rely on calculus, using a concept called integration. For the situation where a curve defined parametrically is rotated around a vertical axis like the y-axis, there's a specific formula we use. The formula is slightly different than when rotating around the x-axis since the curve's position and rotation direction affect the derived surface.To calculate the surface area, we use:- The integral involves a multiplicative factor of \(2\pi\), which accounts for the circular nature of the rotation.- It incorporates both the x and y coordinates of the curve, alongside their derivatives.This formula essentially sums up tiny surface strips generated by the rotation. By integrating these surface strips over a given interval, we accumulate the full surface area.

Parametric Equations

Parametric equations are a powerful tool in calculus, allowing us to express curves that aren't easily described using just x and y information. By introducing a third parameter, usually denoted as \( t \), we can describe more complex paths or shapes.- The parameter \( t \) navigates through points, often representing an angle or time.- For our circle, the parametrization \( x = 1 + \cos t \) and \( y = \sin t \) describe a circle with radius 1 and center at \((1,0)\).The advantage of parametric equations is that they provide flexibility in representing both closed shapes like circles and open paths. They are especially useful in cases where the shape is easier to describe or manipulate with an additional parameter. In a sense, parameterizing can be seen as a way to draw a shape by specifying a path along which a point travels.

Integration Techniques

Integration is central to finding the surface area of revolution. It involves summing infinitesimally small parts to find a total. Techniques of integration help us solve these sums effectively.- Basic understanding of how derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) work is key. It allows us to effectively replace complex expressions when needed.- By understanding trigonometric identities, we often simplify integrals. For instance, in our example, the Pythagorean identity \( (-\sin t)^2 + (\cos t)^2 = 1 \) plays a vital role in reducing complexity.The integral in our problem splits into two simpler integrals, thanks to its linear form. By treating the integral in parts, like the straightforward integral of \( \int_{0}^{2\pi} 1 \, dt \) that gives a constant result, we can handle even complex integrals with relative ease.Mastering these techniques not only helps solve calculus problems but also deepens understanding of fundamental mathematical ideas, revealing the interconnectedness of calculus concepts.