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Magazine Chapter 5: Problem 2
Find the volume generated by rotating about the \(x\) -axis the regions bounded by the graphs of each set of equations. $$y=x, x=0, x=2$$
Short Answer
Expert verified
The volume is \( \frac{8\pi}{3} \).
Step by step solution
01
- Identify the Bounded Region
The region bounded by the given equations is enclosed by the lines:\( y = x \), \( x = 0 \), and \( x = 2 \). This region forms a right triangle with vertices at \( (0,0) \), \( (2,0) \), and \( (2,2) \).
02
- Setup the Volume Integral
To find the volume of the solid generated by rotating the bounded region about the \( x\)-axis, use the disk method. The formula for the volume of a solid of revolution using the disk method is: \[ V = \int_{a}^{b} \pi [f(x)]^2 \, dx \]. Here, \( f(x) = x \) and the bounds are from \( x = 0 \) to \( x = 2 \).
03
- Substitute and Simplify the Integral
Substitute \( f(x) = x \) into the volume formula: \[ V = \int_{0}^{2} \pi (x)^2 \, dx \]. Simplify inside the integral: \[ V = \pi \int_{0}^{2} x^2 \, dx \].
04
- Evaluate the Integral
Evaluate the integral \( \int_{0}^{2} x^2 \, dx \): \[ \int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} \]. Substitute the bounds into the antiderivative: \[ \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \].
05
- Calculate the Final Volume
Multiply the result of the integral by \( \pi \): \[ V = \pi \cdot \frac{8}{3} = \frac{8\pi}{3} \]. Therefore, the volume of the solid generated by rotating the region bounded by the given equations about the \( x\)-axis is \( \frac{8\pi}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Disk Method
The disk method is a technique for finding the volume of a solid of revolution. Imagine slicing the solid perpendicular to the axis of rotation, creating thin disks. The volume of each disk is calculated and then summed up. This method uses the formula: V = pi int_ a ^ b f(x) ^ 2 dx . The width of each disk is an infinitesimally small thickness, represented by dx .
In our example, the region is defined by y = x, x = 0, and x = 2. When revolving this around the x-axis, the radius of each disk is simply f(x) = x.
- f(x): radius of the disk based on the function
- a and b: bounds of integration
- dx: thickness of each disk
In our example, the region is defined by y = x, x = 0, and x = 2. When revolving this around the x-axis, the radius of each disk is simply f(x) = x.
Definite Integral
A definite integral is used to calculate the accumulation of quantities, such as area under a curve or the volume of a solid. It is represented as int_ a ^ b f(x) dx . The limits a and b are the bounds within which the quantity is accumulated.
For our problem, we set up the integral to find the volume using the disk method: V = pi int_ 0 ^ 2 (x) ^ 2 dx . Here, we integrate x^2 from 0 to 2.
- a: lower limit of integration
- b: upper limit of integration
- f(x): function to be integrated
For our problem, we set up the integral to find the volume using the disk method: V = pi int_ 0 ^ 2 (x) ^ 2 dx . Here, we integrate x^2 from 0 to 2.
Solid Geometry
Solid geometry deals with three-dimensional figures. When dealing with volumes of solids of revolution, visualizing the resulting 3D shape is crucial. Rotation of a region around an axis creates a solid.
In our case, rotating a right triangle formed by y = x, x = 0, and x = 2 around the x-axis gives us a cone with the same base and height. Understanding the shape helps when setting up volume calculations.
In our case, rotating a right triangle formed by y = x, x = 0, and x = 2 around the x-axis gives us a cone with the same base and height. Understanding the shape helps when setting up volume calculations.
Volume Integrals
Volume integrals involve integrating a cross-sectional area along a certain path. For solids of revolution, the integral sums up the volume of infinitesimally thin disks to get the total volume.
We calculated the integral pi int_ 0 ^ 2 x^2 dx to find the total volume of the cone in our problem.
- Each disk's volume is: pi (radius)^2 thickness
- Sum all disk volumes using an integral
- Integral bounds are from the start to end of the solid along the axis of rotation
We calculated the integral pi int_ 0 ^ 2 x^2 dx to find the total volume of the cone in our problem.
Antiderivatives
An antiderivative is the reverse process of differentiation. It's used to evaluate definite integrals. The antiderivative of a function F(x) is a function whose derivative is F(x).
For integrating x^2, the antiderivative is (x^3)/3 . We used this to compute int_ 0 ^ 2 x^2 dx , resulting in (2^3)/3 - (0^3)/3 = 8/3. Multiplying by pi gives final volume 8pi/3.
- Find the antiderivative using basic integration rules
- Use it to evaluate the definite integral by applying the limits
For integrating x^2, the antiderivative is (x^3)/3 . We used this to compute int_ 0 ^ 2 x^2 dx , resulting in (2^3)/3 - (0^3)/3 = 8/3. Multiplying by pi gives final volume 8pi/3.
