Question

In Exercises 55-62, find the area of the surface generated by revolving the curve about the indicated axis. $$x=y^{1 / 2}-(1 / 3)^{3 / 2}, \quad 1 \leq y \leq 3 ; \quad y$$

Short answer

The surface area \(SA\) is given by \(SA = 2\pi I\), where \(I\) is the value of the integral obtained from Step 5.

Step-by-step solution

Writing Down the Formula

The formula to calculate the surface area of revolution along the y-axis is \[SA = 2\pi\int_a^b x \sqrt{1 + (dy/dx)^2}dy\]. For this problem, \(a = 1\) and \(b = 3\). The function \(x(y)\) is given by \(y^{1 / 2}-(1 / 3)^{3 / 2}\).

Calculating dy/dx

First find the derivative of \(x(y) = y^{1 / 2}-(1 / 3)^{3 / 2}\) , which is \((dy/dx) = (1/2)y^{-1 / 2}\).

Setup the Integral

Substitute the function \(x(y)\) and \((dx/dy)\) into the surface area formula gives: \[SA = 2\pi\int_{1}^{3} \left(y^{1 / 2}-(1 / 3)^{3 / 2}\right) \sqrt{1 + \left(1/2y^{-1 / 2}\right)^2}dy\].

Computing the Integral

Evaluate the integral with the boundaries of \(1\) and \(3\). This should be done either analytically (if possible) or via numerical methods (like definite integral calculation on a calculator or programming language).

Evaluate the Integral

The integral may not be solvable in standard function elementary way. Numeric methods or software may be needed to compute the value of the integral. Let's call the solution for the integral \(I\).

Compute the Surface Area

Finally, multiply \(I\) by \(2\pi\) to get the surface area \(SA\).

Key concepts

Integrals in Calculus

The concept of integrals is a cornerstone in calculus, representing the accumulation of quantities and the area under curves. In particular, definite integrals are employed in a vast array of mathematical problems, including the calculation of areas and volumes. In the context of the given exercise, the integral is used to calculate the surface area of a shape created by revolving a curve around an axis.

Integral calculus is all about finding the total or 'integral' part when you only have small 'differential' pieces. Think of it like piecing together a puzzle - each small piece represents a differential element, and the complete picture is revealed when all these infinitesimal parts are combined using the process of integration. The evaluation of the integral often requires techniques such as substitution, integration by parts, or numerical methods when an integral is too complex to solve by hand.

Volume and Surface Area in Calculus

Calculating volume and surface area using calculus is a practical application of integrals. For volumes, we might use the disk, washer, or shell method, depending on the given problem’s symmetry and the axis of rotation. Surface area calculations involve a more intricate formula, which includes the function describing the revolving curve and the derivatives of this function.

The exercise's requirement to find a surface area of revolution is a classic example. This specific application uses a formula integrating over a range to consider the entire shape created as the curve spins around an axis. The integral's result, when multiplied by a constant - in this exercise, by the circumference of a circle, or in mathematical terms, by 2π - yields the desired surface area.

Derivatives and Differential Calculus

Derivatives represent the rate at which one value changes with respect to another. In differential calculus, we study how functions change and use derivatives to understand this behavior. The derivative gives us a slope, which can represent speed, acceleration, or rates of change in various science and engineering contexts.

In the step-by-step solution to the surface area exercise, calculating the derivative of the function with respect to y, designated as dy/dx, is essential. This derivative quantifies the rate at which x changes as y changes, and when squared and added to 1, it provides a measure of how much the curve bends as it revolves around the y-axis. This bending measure is crucial as it influences the total surface area – think of it like wrapping a sheet around a bent tube; the tighter the tube bends, the more material you'll need to cover it without any gaps.