Lengths of symmetric curves Suppose a curve is described by y=f(x) on the interval [−b, b], where f′ is continuous on [−b, b]. Show that if f is odd or f is even, then the length of the curve y=f(x) from x=−b to x=b is twice the length of the curve from x=0 to x=b. Use a geometric argument and prove it using integration.
Table of contents
- 0. Functions7h 54m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms36m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
- 12. Techniques of Integration7h 41m
- 13. Intro to Differential Equations2h 55m
- 14. Sequences & Series5h 36m
- 15. Power Series2h 19m
- 16. Parametric Equations & Polar Coordinates7h 58m
9. Graphical Applications of Integrals
Introduction to Volume & Disk Method
Problem 6.R.66b
Textbook Question
Surface area and volume Let f(x) = 1/3 x³ and let R be the region bounded by the graph of f and the x-axis on the interval [0, 2].
b. Find the volume of the solid generated when R is revolved about the y-axis.
Verified step by step guidance1
Identify the region R bounded by the curve \(f(x) = \frac{1}{3}x^{3}\) and the x-axis on the interval \([0, 2]\). This means the region lies between \(x=0\) and \(x=2\), above the x-axis, and under the curve \(y = \frac{1}{3}x^{3}\).
Since the solid is generated by revolving the region R about the y-axis, use the method of cylindrical shells. The formula for the volume using cylindrical shells is:
\(V = \int_{a}^{b} 2\pi x \cdot f(x) \, dx\)
where \(x\) is the radius of the shell and \(f(x)\) is the height.
Substitute the given function \(f(x) = \frac{1}{3}x^{3}\) and the interval limits \(a=0\) and \(b=2\) into the volume integral:
\(V = \int_{0}^{2} 2\pi x \cdot \frac{1}{3}x^{3} \, dx\).
Simplify the integrand inside the integral:
\$2\pi x \cdot \frac{1}{3}x^{3} = \frac{2\pi}{3} x^{4}\(, so the integral becomes
\)V = \int_{0}^{2} \frac{2\pi}{3} x^{4} \, dx$.
Evaluate the integral by integrating \(x^{4}\) with respect to \(x\), then multiply by the constant \(\frac{2\pi}{3}\), and finally apply the limits from 0 to 2 to find the volume.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Volume of Solids of Revolution
This concept involves finding the volume of a 3D solid formed by rotating a 2D region around an axis. Common methods include the disk/washer method and the shell method. For rotation about the y-axis, the shell method is often more straightforward when the function is given in terms of x.
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Shell Method
The shell method calculates volume by summing cylindrical shells formed by revolving vertical slices of the region around the y-axis. Each shell's volume is 2π(radius)(height)(thickness), where radius is the distance from the y-axis, height is the function value, and thickness is a small change in x.
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Setting up Definite Integrals
To find the volume, you must express the integral with correct limits and integrand based on the region and axis of rotation. For the shell method, integrate with respect to x over [0, 2], using the radius and height functions derived from the problem's given function and interval.
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