Question

In Exercises 55-62, find the area of the surface generated by revolving the curve about the indicated axis. $$x=\sqrt{2 y-1}, \quad(5 / 8) \leq y \leq 1 ; \quad y$$

Short answer

The short answer will be the number obtained after the integral is evaluated, represented in terms of \(\pi\).

Step-by-step solution

Identify the variables and the limits of integration

First, identify the variables in the equation and the limits of integration. In this case, the equation is \(x=\sqrt{2 y-1}\) and the limits of integration are \(5/8 \leq y \leq 1\). These are the values for 'y' that will be plugged into the integral when calculating the area.

Compute the derivative

Next, the derivative of the function needs to be computed in order to use it in the surface area formula. The derivative of \(x=\sqrt{2 y-1}\) with respect to 'y' is \(dx/dy = 1/\sqrt{2y - 1}\).

Set up the integral for the surface area

The surface area \(A\) of the solid formed by revolving a curve \(y=f(x)\) around y-axis from \(a\) to \(b\) is given by \(A = 2\pi\int_a^b{x\sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy}\). Substituting \(x = \sqrt{2 y-1}\) and \(\frac{dx}{dy} = \frac{1}{\sqrt{2y - 1}}\) into the formula, the integral becomes \(A = 2\pi\int_{5/8}^1{\sqrt{2 y-1}\sqrt{1+\frac{1}{(2y - 1)}} dy}\).

Evaluate the integral

Finally, evaluate the integral. This could require techniques such as substitution or integrating by parts depending on the exact form of the integral. After calculation, the exact value of area might be represented in terms of \(\pi\).

Key concepts

Solid of Revolution

The concept of a solid of revolution is central to many problems in integral calculus, particularly when determining the volume or surface area of a 3D object created by rotating a 2D curve about an axis. To visualize a solid of revolution, imagine taking a curve on a graph and spinning it about an axis; the path traced by the curve generates a three-dimensional shape. In the provided exercise, the 2D curve defined by the equation \(x=\sqrt{2 y-1}\) is revolved around the y-axis to create such a solid.

Understanding the shape of the solid is crucial. For instance, revolving a line generates a cylinder, while more complex curves might produce shapes like cones, spheres, or uniquely formed objects. This kind of mental visualization not only aids in grasping the concept more fully but also in setting up the correct integral to calculate the surface area of the solid.

Integral Calculus

Integral calculus is a branch of mathematics focused on finding the total size, area, volume, or other measures of accumulation. A key tool within integral calculus is the integral itself, which can be interpreted as the area under a curve on a graph when dealing with functions of a single variable. Calculus is broadly divided into two complementary parts: differential calculus, concerned with rates of change (derivatives), and integral calculus, which is applied in totaling quantities.

When working with integrals, one typically deals with indefinite integrals, which represent a family of functions, and definite integrals, which evaluate to a specific numerical value. In our exercise, we're interested in the latter, as we need to find a concrete answer for the surface area. Integral calculus also involves various techniques for evaluating integrals, including substitution and integration by parts, which might be needed to solve more complex integrals like the one in the exercise.

Definite Integrals

Definite integrals are a cornerstone of integral calculus, providing the actual numerical value represented by the area under a curve between two specified points on the x-axis or y-axis. Notated with a lower limit 'a' and upper limit 'b', a definite integral \(\int_a^b{f(x)dx}\) is the net accumulation of the function \(f(x)\) over the interval [a, b].

The concept came into play in the given exercise when evaluating the surface area of our rotated curve between the limits of \(\frac{5}{8}\) and 1. These limits tell us the 'start' and 'stop' points for the accumulation process, which, in this case, are the points at which the curve intersects the y-axis and form the boundaries of our solid of revolution. Crucial to solving definite integrals is understanding the properties of integrals and commonly using the Fundamental Theorem of Calculus, which links integrals and derivatives in a way that allows us to evaluate the integral.

Calculus of Surface Areas

Calculating the surface area of a solid of revolution is another practical application of integral calculus. The formula to find this area is based on slicing the solid into infinitesimally thin disks or rings and then 'unrolling' these to form rectangles, the areas of which can be straightforwardly calculated. Summing up all these small area contributions, an integral is used to find the total surface area.

For a curve \(y=f(x)\) rotated about the y-axis, the surface area \(A\) can be calculated using the formula \((2\pi\int_a^b{x\sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy}\)), where \(a\) and \(b\) are the limits of integration, and \(\frac{dx}{dy}\) is the derivative of \(x\) with respect to \(y\). This derivative is vital because it represents the slope of the curve, which contributes to the 'steepness' of the surface and affects the area calculation. In our exercise, the derivative \(\frac{dx}{dy} = \frac{1}{\sqrt{2y - 1}}\) plays a crucial role in setting up the correct integral needed to compute the surface area.