Problem 7 - Thomas Calculus

Question

In Exercises \(1 - 8 :\) a. Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis. b. Graph the curve to see what it looks like. If you can, graph the surface too. c. Use your utility's integral evaluator to find the surface's area numerically. $$x = \int _ { 0 } ^ { y } \tan t d t , \quad 0 \leq y \leq \pi / 3 ; \quad y -axis$$

Step-by-Step Solution

Answer

Set up the integral and calculate using a numerical integration tool.

Step 1: Understand the Problem

The curve given in the exercise is parameterized in the form of an integral: \(x = \int_{0}^{y} \tan t \, dt\). This represents a curve in the xy-plane. We need to find the surface area when this curve is revolved around the y-axis from \(y = 0\) to \(y = \pi/3\).

Step 2: Derivative of the Curve

To construct the surface integral, determine the derivative \(\frac{dx}{dy}\). Since \(x = \int_{0}^{y} \tan t \, dt\), we have \(\frac{dx}{dy} = \tan y\) by the Fundamental Theorem of Calculus.

Step 3: Set up the Surface Area Integral

The formula for surface area when revolving around the y-axis is \( S = \int_{a}^{b} 2\pi x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy \). Plug in our expressions: \( x = \int_{0}^{y} \tan t \, dt \) and \( \frac{dx}{dy} = \tan y \). Thus, \( S = \int_{0}^{\pi/3} 2\pi \left(\int_{0}^{y} \tan t \, dt\right) \sqrt{1 + \tan^2 y} \, dy \).

Step 4: Simplification

Note that \(\sqrt{1 + \tan^2 y} = \sec y\). Hence, the integral becomes \( S = \int_{0}^{\pi/3} 2\pi \left(\int_{0}^{y} \tan t \, dt\right) \sec y \, dy \).

Step 5: Graph the Curve

Graph the curve \(x = \int_{0}^{y} \tan t \, dt\) from \(y = 0\) to \(y = \pi/3\). Compute \(x\) for several values of \(y\) to plot points and see the shape. The curve will start at the origin \((0,0)\) and move upwards.

Step 6: Calculate Numerically with a Calculator

Using a numerical integration tool or a calculator, evaluate the integral: \[ S = \int_{0}^{\pi/3} 2\pi \left(\int_{0}^{y} \tan t \, dt\right) \sec y \, dy \]. This will give us the exact numerical value of the surface area.

Key Concepts

integration

Integration in calculus is the process of finding the integral of a function. In our exercise, we deal with integrals to define a curve and calculate its properties. When we say integral, we are referring to the accumulation of values such as area or volume. Integration can be thought of as summing up an infinite number of infinitesimally small quantities.

The integral for our problem is defined as \(x = \int _ { 0 } ^ { y } \tan t \, dt\), which accumulates the tangent values from \(0\) to \(y\). This is an example of a definite integral, where we have specific upper and lower limits. This integral essentially "builds" our curve by summing the area under the curve of \(\tan t\).

When setting up integrals for surface areas or volumes, it's crucial to break down the problem to define what exactly you're summing over and the limits of integration.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a pivotal concept that connects differentiation and integration, two of the main operations in calculus. It tells us that the derivative of an integral function is the original function itself.

In our exercise, we used this theorem to simplify the problem. Since \(x = \int_{0}^{y} \tan t \, dt\), by applying the Fundamental Theorem of Calculus, the derivative \(\frac{dx}{dy}\) is simply \(\tan y\). This is because the Fundamental Theorem tells us that differentiating the integral with respect to \(y\) returns to the function inside the integral, which in this case is \(\tan t\), evaluated at \(t = y\).

The ability to derive one expression from another is vital in setting up and solving many types of calculus problems, particularly those involving areas and volumes.

revolving curves

When we talk about revolving curves, we are referring to the method of rotating a curve about a specific axis to create a three-dimensional surface or solid. The exercise involves revolving the curve, given by the integral, around the y-axis to generate a surface.

The formula for the surface area generated by revolving a curve \(x=f(y)\) around the y-axis is: \[ S = \int_{a}^{b} 2\pi x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy \] This formula accounts for the increase in radius and the arc length of each tiny rotated strip.

Revolving curves allows us to calculate areas or volumes of complex shapes by using rotational symmetries. This process transforms integrals of functions into descriptions of surfaces, giving a practical tool for engineering, physics, and broader scientific applications.

numerical integration

Numerical integration is a method of calculating the value of an integral when an analytical solution is either impossible or requires complicated calculations. In many real-world applications, integrals are evaluated numerically using approximation methods.

For our exercise, once we set up the integral for the surface area \( S \), we use numerical tools like calculators or software to evaluate it. This is particularly useful when the integral involves complex functions or when the integral can't be solved analytically.

Common numerical methods include the Trapezoidal Rule and Simpson's Rule, each using different approaches to approximate the area under the curve. By breaking the region into manageable segments and calculating the sum of areas of these segments, we achieve a precise approximate result.

• Trapezoidal Rule: Approximates the region using trapezoids.
• Simpson's Rule: Uses parabolic segments for better accuracy.

Numerical integration ensures that we can work with integrals in practical situations, providing solutions where traditional methods fall short.