Question

Surface area Find the area of the surface generated when the region bounded by the graph of \(y=e^{x}+\frac{1}{4} e^{-x}\) and the \(x\) -axis on the interval \([0, \ln 2]\) is revolved about the \(x\) -axis.

Short answer

Question: Find the approximate surface area generated when the region bounded by the graph of \(y=e^{x}+\frac{1}{4} e^{-x}\) and the \(x\)-axis on the interval \([0, \ln 2]\) is revolved about the \(x\)-axis. Answer: To approximate the surface area, we use the formula: $$A = 2\pi \int_0^{\ln2} (e^x+\frac{1}{4}e^{-x})\sqrt{1+(e^x-\frac{1}{4}e^{-x})^2} \, dx$$ After applying numerical integration techniques, such as the trapezoidal rule or Simpson's rule, we get an approximate value for the surface area A.

Step-by-step solution

Find the derivative the function \(y=e^{x}+\frac{1}{4} e^{-x}\)

To find the derivative of the function \(y=e^{x}+\frac{1}{4} e^{-x}\) with respect to \(x\), we start by finding the derivatives of each part of the function. Using the chain rule, we differentiate \(e^x\) and \(\frac{1}{4}e^{-x}\) with respect to \(x\). $$(e^x)' = e^x$$ $$(\frac{1}{4} e^{-x})' = -\frac{1}{4} e^{-x}$$ So, the derivative of the given function is: $$y' = e^x - \frac{1}{4} e^{-x}$.

Find \(({y'})^2\)

Now, we need to find the square of the derivative, \(({y'})^2\). $$({y'})^2 = (e^x - \frac{1}{4} e^{-x})^2$$

Find \(1+({y'})^2\)

Next, we need to find \(1+({y'})^2\). $$1+({y'})^2 = 1 + (e^x - \frac{1}{4} e^{-x})^2$$

Find the integral of the surface area formula

Now, we will plug \(y\) and \(1+({y'})^2\) into the surface area formula and integrate it over the given interval \([0,\ln2]\). $$A = 2\pi \int_0^{\ln2} (e^x+\frac{1}{4}e^{-x})\sqrt{1+(e^x-\frac{1}{4}e^{-x})^2} \, dx$$ This integral can be quite tedious to compute, so we will use numerical integration techniques, such as the trapezoidal rule or Simpson's rule, to approximate the value of the integral, and therefore, the surface area.

Approximate the integral using numerical integration techniques

By approximating the integral using numerical integration techniques, the value of the surface area A can be found. If we use, for example, Simpson's rule with sufficient subintervals, we can achieve good accuracy for the surface area. This will lead to an approximation of the surface area generated when the region bounded by the given function is revolved around the x-axis. From here, one could perform such numerical approximation techniques to compute the surface area of the given function.

Key concepts

Surface Area of Revolution

Calculating the surface area of a solid of revolution is a common task in calculus. When we revolve a curve around an axis, it creates a three-dimensional object with a curvilinear surface. To find this surface area, we integrate along the curve using the surface area formula:\[ A = 2\text{\pi} \(x\) \(y\) \(ds\) \], where \( x \) is the value on the axis of rotation and \( y \) is the function representing the curve. The term \( ds \) is an infinitesimal arc length element of the curve, given by:\[ ds=\text{\sqrt{1+(y')^2}} dx \], where \( y' \) is the derivative of the function with respect to \( x \).

In the context of our exercise, this formula allows us to calculate the surface area generated by revolving the region bounded by the graph of \( y = e^x + \frac{1}{4} e^{-x} \) and the \( x \)-axis from \( x = 0 \) to \( x = \text{\ln}(2) \).

Derivative of Exponential Functions

In calculus, the derivative of an exponential function follows a unique property; the derivative of \( e^x \) with respect to \( x \) is simply \( e^x \) itself. To differentiate a function like \( y = e^{x} + \frac{1}{4} e^{-x} \), we use the basic rules of differentiation. For the first term, the derivative of \( e^x \) is \( e^x \), and for the second term, utilizing the chain rule, the derivative of \( \frac{1}{4}e^{-x} \) is \( -\frac{1}{4}e^{-x} \).

Thus, the derivative or rate at which \( y \) changes with respect to \( x \) is \( y' = e^x - \frac{1}{4}e^{-x} \). Recognizing these patterns is crucial for solving problems related to exponential functions, including those involving the surface area of revolution.

Numerical Integration Techniques

Numerical integration is a pillar of computational mathematics and is the key to solve integrals that are difficult or impossible to evaluate analytically. Techniques for numerical integration are algorithms for approximating the definite integral of a function. When it's challenging to find the exact value of an integral because of the complexity of the function, numerical methods such as the trapezoidal rule or Simpson's rule come in handy.

These methods work by breaking the area under the curve into shapes (like trapezoids or parabolic slices) for which the area can be calculated and then summing these areas to approximate the total area under the curve. With fine enough subdivisions, numerical techniques can provide highly accurate approximations of definite integrals.

Simpson's Rule

One of the most powerful numerical integration techniques is Simpson's rule. This method provides a higher degree of accuracy compared to other techniques such as the trapezoidal rule. Simpson's rule is based on the idea of approximating the area under a curve by fitting parabolas to subintervals of the function.

The rule states that the integral of a function can be approximated with:\[ A \(≈\) \frac{h}{3}[y_0 + 4(y_1 + y_3 + y_5 + ... ) + 2(y_2 + y_4 + ... ) + y_n] \], where \( h \) is the width of the subintervals and \( y_0, y_1, ..., y_n \) are the function values at the corresponding subinterval endpoints. In practical terms, for our exercise, it means dividing the interval from 0 to \( \text{\ln}(2) \) into even segments and using the corresponding \( y \) values to estimate the area under the curve.

Chain Rule

The chain rule is an essential differentiation technique in calculus used whenever we are dealing with the composition of functions. It expresses the derivative of a composite function in terms of the derivatives of the individual functions.

For example, if a function \( y \) can be expressed as \( y = f(g(x)) \), then the derivative of \( y \) with respect to \( x \) is \( y' = f'(g(x)) \(g'(x\)) \). This approach allows us to unravel complex functions into simpler parts that can be differentiated individually. Applying the chain rule, such as when differentiating \( \frac{1}{4} e^{-x} \) in our exercise, provides the derivative needed to proceed with calculating the surface area of the solid of revolution.