Ris the region bounded by the functions f(x) = 2? +1 and the g(x) = – 2 and the lines x=1 and x=3. Represent the volume when R is rotated around the y-axis. Volume = dx

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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**Volume of a Solid of Revolution: Rotating Around the y-axis**

In this problem, we are given a region \( R \) that is bounded by the functions \( f(x) \) and \( g(x) \) and the vertical lines \( x = 1 \) and \( x = 3 \). The goal is to find the volume of the solid that is formed when this region is rotated about the y-axis.

The functions are:
\[ f(x) = x^2 + 1 \]
\[ g(x) = x - 2 \]

And the vertical boundaries are:
\[ x = 1 \]
\[ x = 3 \]

To represent the volume of the solid when \( R \) is rotated about the y-axis, we use the method of disks or washers. The volume \( V \) of the solid can be calculated using the integral formula:
\[ \text{Volume} = \int_{1}^{3} \left[ \text{(Outer radius)}^2 - \text{(Inner radius)}^2 \right] \ dx \]

In this context, the outer radius at any point \( x \) is given by \( f(x) = x^2 + 1 \) and the inner radius is given by \( g(x) = x - 2 \).

Thus, the volume integral setup is:
\[ \text{Volume} = \int_{1}^{3} \left[ (x^2 + 1)^2 - (x - 2)^2 \right] \ dx \]
Transcribed Image Text:**Volume of a Solid of Revolution: Rotating Around the y-axis** In this problem, we are given a region \( R \) that is bounded by the functions \( f(x) \) and \( g(x) \) and the vertical lines \( x = 1 \) and \( x = 3 \). The goal is to find the volume of the solid that is formed when this region is rotated about the y-axis. The functions are: \[ f(x) = x^2 + 1 \] \[ g(x) = x - 2 \] And the vertical boundaries are: \[ x = 1 \] \[ x = 3 \] To represent the volume of the solid when \( R \) is rotated about the y-axis, we use the method of disks or washers. The volume \( V \) of the solid can be calculated using the integral formula: \[ \text{Volume} = \int_{1}^{3} \left[ \text{(Outer radius)}^2 - \text{(Inner radius)}^2 \right] \ dx \] In this context, the outer radius at any point \( x \) is given by \( f(x) = x^2 + 1 \) and the inner radius is given by \( g(x) = x - 2 \). Thus, the volume integral setup is: \[ \text{Volume} = \int_{1}^{3} \left[ (x^2 + 1)^2 - (x - 2)^2 \right] \ dx \]
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