Question

Find the area of the surface generated by revolving the curve about each given axis. $$ \begin{array}{llll} x=t, y=2 t, & 0 \leq t \leq 4, & \text { (a) } x \text { -axis } & \text { (b) } y \text { -axis } \end{array} $$

Short answer

The surface area of the surface of revolution when revolved about the x-axis is \(16\pi\sqrt{5}\). When revolved about the y-axis, it is \(\frac{64\pi}{5}\sqrt{5}\).

Step-by-step solution

Calculating the Surface Area for Rotation about the x-axis

Firstly, the formula for surface area when rotated about the x-axis is given by \(2\pi\int_{a}^{b}y\sqrt{1+(f'(x))^2}dx\). Here, \(f'(x)\) is the derivative of \(f(x)\) or \(y\), that is \(2t\). Hence, \(f'(x)=2\). Now, calculate the integral: \(2\pi\int_{0}^{4}2t\sqrt{1+(2)^2}dt\).

Evaluating the Integral for the x-axis

Solving this integral leads to: \(2\pi[2t\sqrt{5}]_0^4 = 16\pi\sqrt{5}\).

Calculating the Surface Area for Rotation about the y-axis

Similarly, use the formula for the surface area when rotated about the y-axis: \(2\pi\int_{c}^{d}x\sqrt{1+(f^{-1}'(y))^2}dy\). Here, before proceeding, find \(f^{-1}'(y)\) by taking \(y=2t\) or \(t=y/2\) and differentiate to get \(f^{-1}'(y)=\frac{1}{2}\). The function \(x=f^{-1}(y)=t=y/2\). Now, calculate the integral from \(c=0\) to \(d=8\): \(2\pi\int_{0}^{8}\frac{y}{2}\sqrt{1+(\frac{1}{2})^2}dy\).

Evaluating the Integral for the y-axis

Solving this integral yields an answer of: \(2\pi[\frac{2}{5}y\sqrt{\frac{5}{4}}]_0^8 = \frac{64\pi}{5}\sqrt{5}\).

Key concepts

Calculus

Calculus is an essential branch of mathematics that deals with rates of change (differential calculus) and accumulation of quantities (integral calculus). It plays a critical role in solving problems involving complex shapes and motion. In the context of finding surface areas of revolution, calculus allows us to calculate the area of a surface generated by rotating a curve around an axis. This is done using the principles of integration, which enable us to sum infinitesimal contributions to the surface area made by rotating small segments of the curve.

Understanding calculus is fundamental when approaching such problems. It allows us to set up the correct integral to find the exact surface area, taking into account the varying radius of rotation and the curve's differential elements.

Integrals

Integrals are at the heart of calculus and are used to find the accumulation of quantities, such as areas, volumes, and, in our case, surface areas. Specifically, finding the surface area of revolution involves definite integrals. A definite integral represents the signed area under a curve. When calculating the surface area of a shape generated by revolution, we use integral calculus to add up all the infinitesimally small surface areas of the rotating curve.

The technique involves setting the limits of integration from the starting to the ending point of the curve, which, in our exercise, is from 0 to 4, and then from 0 to 8, for rotations around the x-axis and y-axis respectively. By evaluating these integrals, we obtain a numerical value that represents the surface area of the resulting geometric solid.

Revolution Surfaces

Revolution surfaces are three-dimensional shapes formed by rotating a two-dimensional curve about an axis. The rotation can be about the x-axis, the y-axis, or any other line (the axis of revolution). The surface of revolution is a smooth skin-like shape that covers the solid of revolution.

In our exercise, we are looking at a parametric curve described by the functions x=t and y=2t, generating different shapes when revolved around each of the two main axes. To compute the surface area, we treat each tiny segment of the curve as if it were a line generating a circular disc when rotated, and then we calculate the total area produced by these discs. This approach leads us to the use of integral calculus to sum the areas and find the solution. Understanding the geometry of revolution surfaces is key to setting up the correct integral formulas.

Derivatives

Derivatives represent the slope or rate of change of a function. In the study of surfaces of revolution, derivatives help us calculate the element of the surface area for a small part of the curve that is being rotated. We need to know how the function changes as it moves along the axis of rotation.

In the given exercise, derivatives come into play when finding the necessary part of the formula for surface area involving \( (f'(x))^2 \) or \( (f^{-1}'(y))^2 \). They play a crucial role in obtaining the value of the integrand (the function within the integral that describes the geometry of the surface). Calculating the derivative allows us to include the rate of change of the curve in respect to its length, which influences the varying radius of rotation and affects the overall surface area computation.