Problem 152
Question
The surface of a large cup is formed by revolving the graph of the function \(y=0.25 x^{1.6}\) from \(x=0\) to \(x=5\) about the \(y\) -axis (measured in centimeters). [T] Use technology to graph the surface.
Step-by-Step Solution
Verified Answer
Use technology to graph the surface area described by the integral.
1Step 1: Identify the Surface of Revolution Formula
To find the surface area of a solid formed by revolving a curve about the y-axis, we use the formula for the surface area of a solid of revolution: \[S = 2\pi \int_{a}^{b} x \, \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx\]where \(x\) is expressed in terms of \(y\).
2Step 2: Differentiate the Curve Equation
Given the function \(y = 0.25x^{1.6}\), we need to find \(\frac{dy}{dx}\). Differentiate the function: \(\frac{dy}{dx} = 0.25 \times 1.6 \times x^{0.6} = 0.4x^{0.6}\).
3Step 3: Substitute into the Surface Integral
Substitute \(\frac{dy}{dx} = 0.4x^{0.6}\) into the formula for surface area:\[S = 2\pi \int_{0}^{5} x \, \sqrt{1 + (0.4x^{0.6})^2} \, dx\]which simplifies to:\[S = 2\pi \int_{0}^{5} x \, \sqrt{1 + 0.16x^{1.2}} \, dx\]
4Step 4: Use Technology to Evaluate the Integral
Input the integral expression \(S = 2\pi \int_{0}^{5} x \, \sqrt{1 + 0.16x^{1.2}} \, dx\) into a graphing calculator or computer software to evaluate it numerically.
Key Concepts
DifferentiationDefinite IntegralNumerical Integration
Differentiation
Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. In simple terms, differentiation helps us understand how a function changes at any given point. This is crucial for various applications, including finding the surface area of a revolution as in the given problem.
When differentiating a function like \(y = 0.25x^{1.6}\), we essentially determine the rate at which \(y\) changes with respect to \(x\). In this case, we apply the power rule of differentiation, which states that the derivative of \(x^n\) is \(nx^{n-1}\).
Thus, the derivative \(\frac{dy}{dx}\) becomes \(0.4x^{0.6}\). This derivative tells us how steep the curve is at any point \(x\).
Differentiation is not only crucial in finding how functions behave but also essential for solving complex problems like calculating surfaces of revolution.
When differentiating a function like \(y = 0.25x^{1.6}\), we essentially determine the rate at which \(y\) changes with respect to \(x\). In this case, we apply the power rule of differentiation, which states that the derivative of \(x^n\) is \(nx^{n-1}\).
Thus, the derivative \(\frac{dy}{dx}\) becomes \(0.4x^{0.6}\). This derivative tells us how steep the curve is at any point \(x\).
Differentiation is not only crucial in finding how functions behave but also essential for solving complex problems like calculating surfaces of revolution.
Definite Integral
The definite integral is a way of calculating the accumulation of quantities, which can be areas under curves, volumes, or—in our case—the surface area of a revolution. The integral plays a key role in determining the precise dimensions of a surface formed when a curve is revolved around an axis.
In the exercise, we use the definite integral to find the surface area \(S\) from \(x = 0\) to \(x = 5\). The integral \(\int_{0}^{5} x \, \sqrt{1 + 0.16x^{1.2}} \, dx\) represents accumulating the infinitesimally small amounts of surface area created as the curve rotates around the y-axis.
To set up this integral, we use the surface of revolution formula. This involves substituting the derivative into the integral, which provides the expression \(\sqrt{1 + \left( \frac{dy}{dx} \right)^2}\) that accounts for the slant height of the surface.
Through this process, the definite integral allows us to sum up all these infinitesimal parts to find the total surface area.
In the exercise, we use the definite integral to find the surface area \(S\) from \(x = 0\) to \(x = 5\). The integral \(\int_{0}^{5} x \, \sqrt{1 + 0.16x^{1.2}} \, dx\) represents accumulating the infinitesimally small amounts of surface area created as the curve rotates around the y-axis.
To set up this integral, we use the surface of revolution formula. This involves substituting the derivative into the integral, which provides the expression \(\sqrt{1 + \left( \frac{dy}{dx} \right)^2}\) that accounts for the slant height of the surface.
Through this process, the definite integral allows us to sum up all these infinitesimal parts to find the total surface area.
Numerical Integration
Numerical integration is used when a definite integral cannot be solved by standard analytical methods due to its complexity. It involves approximate methods for evaluating integrals, which are essential for solving problems like the surface of revolution.
In the given exercise, once the integral \(S = 2\pi \int_{0}^{5} x \, \sqrt{1 + 0.16x^{1.2}} \, dx\) is established, we use numerical methods to evaluate it. Tools like graphing calculators or specialized computer software facilitate this by using techniques such as the Trapezoidal Rule or Simpson's Rule.
These methods break down the integral into simpler calculations that approximate the curve and generate an accurate numerical result for \(S\). It's often more efficient and sometimes necessary to use these numerical methods when dealing with complex functions that don't have straightforward antiderivatives.
Thus, numerical integration bridges the gap when traditional calculus methods come up short, providing practical solutions for real-world applications.
In the given exercise, once the integral \(S = 2\pi \int_{0}^{5} x \, \sqrt{1 + 0.16x^{1.2}} \, dx\) is established, we use numerical methods to evaluate it. Tools like graphing calculators or specialized computer software facilitate this by using techniques such as the Trapezoidal Rule or Simpson's Rule.
These methods break down the integral into simpler calculations that approximate the curve and generate an accurate numerical result for \(S\). It's often more efficient and sometimes necessary to use these numerical methods when dealing with complex functions that don't have straightforward antiderivatives.
Thus, numerical integration bridges the gap when traditional calculus methods come up short, providing practical solutions for real-world applications.
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