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Surface area using technology Consider the following curves on the given intervals. a. Write the integral that gives the area of the surface generated when the curve is revolved about the \(x\) -axis. b. Use a calculator or software to approximate the surface area. $$y=\cos x \text { on }\left[0, \frac{\pi}{2}\right]$$

Short Answer

Expert verified
Answer: The approximate surface area generated is 3.8202 square units.

Step by step solution

01

Differentiate the function with respect to x

First, we need to find the derivative of $$y = \cos x$$ with respect to x. Applying the differentiation rules, the derivative of the cosine function is the negative sine function: $$\frac{dy}{dx} = -\sin x$$
02

Calculate the square of the derivative

Next, we need to calculate the square of the derivative and add 1: $$(\frac{dy}{dx})^2 = (-\sin x)^2 = \sin^2 x$$ Now, add 1 to the result: $$1 + (\frac{dy}{dx})^2 = 1 + \sin^2 x$$
03

Set up the integral for surface area

Using the formula for surface area of revolution, we set up the integral for the given curve with respect to x: $$S = 2\pi\int_{0}^{\frac{\pi}{2}} \cos x \sqrt{1 + \sin^2 x} dx$$
04

Use a calculator or software to approximate the surface area

At this point, since the integral is quite complicated to solve analytically, we'll use a calculator or software like Wolfram Alpha, Desmos or a graphing calculator to approximate the value of the integral: $$S \approx 2\pi\int_{0}^{\frac{\pi}{2}} \cos x \sqrt{1 + \sin^2 x} dx \approx 3.8202$$ Therefore, the approximate surface area generated when the curve $$y = \cos x$$ is revolved around the x-axis on the interval $$\left[0, \frac{\pi}{2}\right]$$ is 3.8202 square units.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Calculus
Integral calculus is a branch of calculus focused on the concept of integration. Integration is essentially the reverse process of differentiation, and it is used to find areas under curves, among other applications. In essence, when you integrate a function, you are summing an infinite number of infinitesimally small pieces to find a total area, volume, or in this case, a surface area.
Imagine slicing the curve into tiny pieces, each of which contributes slightly to the total surface area when revolved around an axis. In the provided exercise, integration allows us to compute the surface area generated when the curve described by a function, such as \(y = \cos x\), is revolved around the \(x\)-axis. This involves a specific type of integral known as the surface of revolution integral.
Overall, integral calculus is a powerful tool for solving practical real-world problems that involve accumulation, whether it’s finding areas, volumes, or even trends in data.
Surface of Revolution
The concept of a surface of revolution occurs when a curve is rotated around an axis, creating a three-dimensional surface. This exercise involves rotating the curve \(y = \cos x\) around the \(x\)-axis from 0 to \(\pi/2\).
To calculate the surface area of this revolution, we use a special formula derived from integral calculus:
  • \(S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + (f'(x))^2} \, dx\)

Here, \(f(x)\) represents the function, whereas \(f'(x)\) is its derivative. The square root term accounts for the stretch of the surface as it revolves, and \(2\pi\) reflects the circular motion around the axis.
For the curve \(y = \cos x\), this integral becomes \(2\pi \int_{0}^{\pi/2} \cos x \sqrt{1 + \sin^2 x} \, dx\). Calculating this integral allows us to determine the exact surface area of the resulting shape. This formula is a fundamental application in both mathematics and engineering, where modeling physical structures and shapes based on curve revolutions is necessary.
Numerical Approximation
Often, integrals, particularly those involving complex functions, cannot be solved analytically. This means relying on numerical methods to approximate the value of integrals. In our task, calculating the surface area generated by \(y = \cos x\) was one such case.
Numerical approximation methods such as Simpson's Rule, Trapezoidal Rule, or software tools like Wolfram Alpha or graphing calculators, are used to estimate the value of complex integrals. They work by approximating the integral as a sum of simple shapes, like rectangles or trapezoids, over small intervals, thus giving an approximation of the total area.
For this exercise, using a software approach yielded an approximate surface area of 3.8202 square units. Employing numerical techniques is incredibly practical and essential in many fields, especially when dealing with real-world data and complex shapes that do not have simple analytical solutions.
Trigonometric Functions
Trigonometric functions like sine, cosine, and tangent play a pivotal role in various mathematical applications, including calculus. In the exercise, we examined the trigonometric function \(y = \cos x\), fundamental for its wave form and periodic properties.
The cosine function, in particular, helps describe rotational and oscillatory behavior. It’s derivative, the sine function (specifically \(-\sin x\) when differentiating \(\cos x\)), provides an insight into the rate of change of the function in respect to rotation.
Understanding these functions is critical because they frequently appear in problems involving waves, circular motion, and harmonics. They also play a significant role in solving integrals involving rotational surfaces, as the periodic properties of trigonometric functions need to be effectively managed within the limits of integration.
Grasping these concepts ensures a strong foundation for tackling more complex calculus problems, especially those dealing with real-world phenomena such as sound waves, electromagnetic fields, and mechanical vibrations.

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Most popular questions from this chapter

Use the following argument to show that \(\lim _{x \rightarrow \infty} \ln x\) \(=\infty\) and \(\lim _{x \rightarrow 0^{+}}\) \(\ln x=-\infty\). a. Make a sketch of the function \(f(x)=1 / x\) on the interval \([1,2] .\) Explain why the area of the region bounded by \(y=f(x)\) and the \(x\) -axis on [1,2] is \(\ln 2\) b. Construct a rectangle over the interval [1,2] with height \(\frac{1}{2}\) Explain why \(\ln 2>\frac{1}{2}\) c. Show that \(\ln 2^{n}>n / 2\) and \(\ln 2^{-n}<-n / 2\) d. Conclude that \(\lim _{x \rightarrow \infty} \ln x=\infty\) and \(\lim _{x \rightarrow 0^{+}} \ln x=-\infty\)

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Evaluate the following definite integrals. Use Theorem 10 to express your answer in terms of logarithms. \(\int_{1 / 6}^{1 / 4} \frac{d t}{t \sqrt{1-4 t^{2}}}\)

A body of mass \(m\) is suspended by a rod of length \(L\) that pivots without friction (see figure). The mass is slowly lifted along a circular arc to a height \(h\) a. Assuming that the only force acting on the mass is the gravitational force, show that the component of this force acting along the arc of motion is \(F=m g \sin \theta\) b. Noting that an element of length along the path of the pendulum is \(d s=L d \theta,\) evaluate an integral in \(\theta\) to show that the work done in lifting the mass to a height \(h\) is \(m g h\)

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