Problem 2
Question
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. $$y=x^{2}, \quad 0 \leq x \leq 2 ; \quad x \text { -axis }$$
Step-by-Step Solution
Verified Answer
Approximately 34.204 square units.
1Step 1: Parametrize the Surface
The given curve is described by the function \( y = x^2 \). We need to find the surface area when this curve is revolved about the \( x \)-axis. Therefore, the curve will form a solid of revolution.
2Step 2: Surface Area Formula for Revolution
For a curve \( y = f(x) \) revolved around the \( x \)-axis, the surface area \( A \) is determined by the formula: \[ A = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] Here, \( y = x^2 \), so \( f(x) = x^2 \) and we need to compute \( \frac{dy}{dx} \).
3Step 3: Calculate the Derivative
Find the derivative of \( y = x^2 \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx}(x^2) = 2x \]
4Step 4: Substitute Values in the Formula
Substitute \( y = x^2 \) and \( \frac{dy}{dx} = 2x \) into the surface area formula: \[ A = 2\pi \int_{0}^{2} x^2 \sqrt{1 + (2x)^2} \, dx \] This simplifies to: \[ A = 2\pi \int_{0}^{2} x^2 \sqrt{1 + 4x^2} \, dx \]
5Step 5: Graph the Curve and the Surface
Graph the function \( y = x^2 \) which is a parabola opening upwards. When this parabola is revolved around the \( x \)-axis, it forms a 3D parabolic surface, similar to a bowl shape.
6Step 6: Numerically Evaluate the Integral
Use a calculator or computer software capable of numerical integration to find \[ A = 2\pi \int_{0}^{2} x^2 \sqrt{1 + 4x^2} \, dx \] The calculated area results in an approximate value of 34.204 square units.
Key Concepts
Integral CalculusParametric CurvesNumerical IntegrationDifferentiation
Integral Calculus
Integral calculus is a branch of mathematics focusing on finding the total size or value, such as area under a curve.
It is crucial when analyzing the area of a surface generated by revolving a curve around an axis.
This involves setting up appropriate integrals to capture the accumulated quantity in question.In this exercise, we deal with finding the surface area generated by revolving a parabolic curve.
The surface area can be found using a specific integral formula:
It is crucial when analyzing the area of a surface generated by revolving a curve around an axis.
This involves setting up appropriate integrals to capture the accumulated quantity in question.In this exercise, we deal with finding the surface area generated by revolving a parabolic curve.
The surface area can be found using a specific integral formula:
- Given by \[ A = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] This equation allows us to sum up all tiny bits of surface around the curve to get the total area.
Parametric Curves
Parametric curves are a way to express mathematical curves by defining both x and y coordinates in terms of another parameter, typically \( t \).
This can allow for more flexible representations of curves, especially useful in geometry and calculus.
For this exercise's purpose, although the function \( y = x^2 \) is expressed in its more basic form, parameterization can help understand complex curves when dealing with revolutions and integrating along curves.By defining a curve parametrically, the idea is to treat both x and y as functions of \( t \), allowing for more direct applications of calculus in surface evolutions.
This can simplify computations and provide new insights into the transformations of curves and their geometric properties during calculus calculations. While \( y = x^2 \) isn’t a parametric form, understanding parameterization expands our toolbox for addressing curves in rotational calculations.
This can allow for more flexible representations of curves, especially useful in geometry and calculus.
For this exercise's purpose, although the function \( y = x^2 \) is expressed in its more basic form, parameterization can help understand complex curves when dealing with revolutions and integrating along curves.By defining a curve parametrically, the idea is to treat both x and y as functions of \( t \), allowing for more direct applications of calculus in surface evolutions.
This can simplify computations and provide new insights into the transformations of curves and their geometric properties during calculus calculations. While \( y = x^2 \) isn’t a parametric form, understanding parameterization expands our toolbox for addressing curves in rotational calculations.
Numerical Integration
Numerical integration is a technique to approximate the value of integrals typically analytically infeasible.
We approximate an area, volume, or another integral mathematically using numerical methods.Techniques often involve calculating the integral's value based on sampling points along the function and using methods like the trapezoidal rule, Simpson's rule, or more advanced algorithms found in computer software.
This problem involves calculating the definite integral:
We approximate an area, volume, or another integral mathematically using numerical methods.Techniques often involve calculating the integral's value based on sampling points along the function and using methods like the trapezoidal rule, Simpson's rule, or more advanced algorithms found in computer software.
This problem involves calculating the definite integral:
- \[ 2\pi \int_{0}^{2} x^2 \sqrt{1 + 4x^2} \, dx \] Using numerical integration tools, the result is approximately 34.204 square units.
Differentiation
Differentiation is a key concept in calculus that deals with finding rates of change or slopes of curves.
In this surface area problem, differentiation helps us determine how the function \( y = x^2 \) changes for different values of x.The derivative, \( \frac{dy}{dx} = 2x \), tells us the slope of the curve at any point on \( y = x^2 \).
This derivative is essential in the integral for surface area revolution:
Mastery of differentiation techniques plays an integral role in solving real-world calculus problems efficiently.
In this surface area problem, differentiation helps us determine how the function \( y = x^2 \) changes for different values of x.The derivative, \( \frac{dy}{dx} = 2x \), tells us the slope of the curve at any point on \( y = x^2 \).
This derivative is essential in the integral for surface area revolution:
- It appears in the calculation formula:\[ A = 2\pi \int_{0}^{2} x^2 \sqrt{1 + (2x)^2} \, dx \]
Mastery of differentiation techniques plays an integral role in solving real-world calculus problems efficiently.
Other exercises in this chapter
Problem 1
Find the volumes of the solids. The solid lies between planes perpendicular to the \(x\) -axis at \(x=0\) and \(x=4 .\) The cross-sections perpendicular to the
View solution Problem 2
A spring has a natural length of \(10 \mathrm{cm} . \mathrm{An}\) 800-N force stretches the spring to \(14 \mathrm{cm}\). a. Find the force constant. b. How muc
View solution Problem 2
Find the center of mass of a thin plate of constant density \(\delta\) covering the given region. The region bounded by the parabola \(y=25-x^{2}\) and the \(x\
View solution Problem 2
Find the lengths of the curves. If you have a grapher, you may want to graph these curves to see what they look like. $$y=x^{3 / 2} \quad \text { from } \quad x
View solution