Chapter 10: Problem 2
Find the length of the curve, locate the centroid, and determine the area of the surface generated by revolving the curve about the \(x\) -axis. \(f(x)=2 x, \quad x \in[0,1]\).
Short Answer
Expert verified
The length of the curve is \(\sqrt{5}\), the centroid is located at \(\left(\frac{\sqrt{5}}{3}, \frac{2\sqrt{5}}{3}\right)\), and the surface area of the solid generated by revolving the curve around the x-axis is \(4\pi\sqrt{5}\).
Step by step solution
01
Finding the Length of the Curve
To find the curve's length, we'll need to use the arclength formula given by:
\[L = \int_a^b \sqrt{1 + [f'(x)]^2}dx\]
where \(a\) and \(b\) are the bounds of the interval given as \([0, 1]\).
First, we need to compute the derivative \(f'(x)\). Since \(f(x) = 2x\), we have:
\[f'(x) = 2\]
Now, let's plug in \(f'(x)\) into the arclength formula and compute the length:
\[L = \int_0^1 \sqrt{1 + (2)^2} dx = \int_0^1 \sqrt{5} dx\]
which equals \(\sqrt{5}(x)|_0^1 = \boxed{\sqrt{5}}\).
02
Locating the Centroid
To find the centroid, we'll use the following centroid formulas:
\[ x_c = \frac{\int_a^b x \cdot \sqrt{1 + [f'(x)]^2} dx}{L}\]
and
\[y_c = \frac{1}{2}\frac{\int_a^b f(x) \cdot \sqrt{1 + [f'(x)]^2} dx}{L}\]
For \(x_c\), we have:
\[x_c = \frac{\int_0^1 x \cdot \sqrt{1 + (2)^2} dx}{\sqrt{5}}\]
Let's evaluate the integral:
\[x_c = \frac{\sqrt{5}}{3}\cdot(1)|_0^1 = \boxed{\frac{\sqrt{5}}{3}}\]
Now, for \(y_c\), we have:
\[y_c = \frac{1}{2}\frac{\int_0^1 (2x) \cdot \sqrt{1 + (2)^2} dx}{\sqrt{5}}\]
\[y_c = \frac{1}{2}\cdot\frac{2\sqrt{5}}{3}\cdot(1)|_0^1 = \boxed{\frac{2\sqrt{5}}{3}}\]
So the centroid of the curve is located at \((\frac{\sqrt{5}}{3}, \frac{2\sqrt{5}}{3})\).
03
Determining the Surface Area
To find the surface area generated by revolving the curve around the \(x\)-axis, we'll use the surface area formula given by:
\[A = 2\pi \int_a^b f(x) \cdot \sqrt{1 + [f'(x)]^2} dx\]
Using the previously calculated values and plugging them into the surface area formula, we have:
\[A = 2\pi \int_0^1 (2x) \cdot \sqrt{1 + (2)^2} dx\]
\[A = 2\pi \cdot (2\sqrt{5}) \frac{1}{2} dx\]
\[A = 2\pi \cdot 2\sqrt{5} \cdot (1)|_0^1 = 4\pi\sqrt{5}\]
Thus, the surface area of the solid generated by revolving the curve around the \(x\)-axis is equal to \(\boxed{4\pi\sqrt{5}}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Curve Length
The length of a curve in calculus, also known as the arc length, is a measurement of the distance along a continuous curve between two points. To calculate this, we often use an integral. The formula to determine the arc length of the function \(f(x)\) on the interval from \(x = a\) to \(x = b\) is: \[ L = \int_a^b \sqrt{1 + [f'(x)]^2}dx \] Here's a breakdown of what each part of the formula represents:
- \(\sqrt{1 + [f'(x)]^2}\): This term computes the infinitesimal arc length by considering both the vertical and horizontal change as \(x\) changes slightly along the interval.
- \(f'(x)\): This is the derivative of the function \(f(x)\), representing the slope or rate of change at each point within the interval.
- \(\int_a^b\): This is the definite integral symbol, which calculates the total arc length over the entire interval from \(a\) to \(b\).
Centroid Calculation Explained
The centroid of a curve can be visualized as its center of mass, providing a balance point based on both shape and density. It's particularly useful in engineering and physics for understanding how objects weigh distributedly. Clinically, the formulas for finding the centroid \((x_c, y_c)\) of a curve are: \[ x_c = \frac{\int_a^b x \cdot \sqrt{1 + [f'(x)]^2} dx}{L} \] \[ y_c = \frac{1}{2}\frac{\int_a^b f(x) \cdot \sqrt{1 + [f'(x)]^2} dx}{L} \] These formulas take into account both the curve's shape and its arc length \(L\). Here's what to consider:
- \(x \cdot \sqrt{1 + [f'(x)]^2}\) and \(f(x) \cdot \sqrt{1 + [f'(x)]^2}\): These expressions incorporate the curve's geometry in weighing the moment contributions in horizontal and vertical directions.
- \(L\): This is the previously calculated arc length, ensuring the centroid reflects the entire curve, not just a segmented portion.
Unraveling the Surface Area of Revolution
The surface area of revolution refers to the surface area of a three-dimensional object formed by rotating a curve around an axis. It's often used to solve practical problems in engineering and physics, such as fluid dynamics and material properties. The formula for finding a surface area formed by revolving a curve \(f(x)\) around the \(x\)-axis is: \[ A = 2\pi \int_a^b f(x) \cdot \sqrt{1 + [f'(x)]^2} dx \] Key elements to understand this formula include:
- \(2\pi\): This constant arises from the circular nature of revolution—the result of spinning a curve around the \(x\)-axis.
- \(f(x) \cdot \sqrt{1 + [f'(x)]^2}\): It combines both the height of the function \(f(x)\) and the rate of change \(f'(x)\), accounting for the distance each point moves during the spin.
- \(\int_a^b\): The integral computes the sum of the infinitesimal surface elements created by the curve's rotation over the interval \([a, b]\).