Volume of Solid of Revolution: 115.19 and 77.206

In summary, the volume of the solid of revolution can be found using the shell method by integrating from 0 to 1 of 2pi(3-x)(2x+3)dx, or using the disk method by integrating from 0 to 1 of pi[(2x+3)^2-5^2]dx. Alternatively, the volume can also be found by slicing with respect to y, using the line y=5 and integrating from 0 to 5 of pi[(5-(2x+3))^2]dy.
  • #1
DaOneEnOnly
8
0

Homework Statement


Find the volume of the solid of revolution:
F(x)=2x+3 on [0,1]
Revolved over the line x=3 and y=5

Homework Equations


Shell Method: 2[tex]\pi[/tex][tex]\int[/tex][tex]^{b}[/tex][tex]_{a}[/tex]x[f(x)-g(x)]dx
obviously just sub y for dy
Disk Method: [tex]/pi[/tex][tex]/int[/tex][tex]^{b}[/tex][tex]_{a}[/tex][F(x)[tex]^{2}[/tex]-G(x)[tex]^{2}[/tex]dx


The Attempt at a Solution


line x=3: 2[tex]/pi[/tex][tex]/int[/tex](3-x)(2x+3)dx =115.19

answer key is unfortunately in disk method which I don't like as much:
[tex]/pi[/tex][tex]/int[/tex][tex]^{3}[/tex][tex]_{0}[/tex](9-4)dy + [tex]/pi[/tex][tex]/int[/tex][tex]^{5}[/tex][tex]_{3}[/tex](3-((y-3)/2))[tex]^{2}[/tex]-4dy

=78.91


line y=5: 2[tex]/pi[/tex][tex/int[/tex][tex]^{5}[/tex][tex]_{0}[/tex](5-y)(1-((y-3)/2)) =130.8996

answer key/ disk method: [tex]/pi[/tex][tex]/int[/tex][tex]^{1}[/tex][tex]_{0}[/tex](25-(5-(2x+3))[tex]^{2}[/tex]dx

=77.206
 
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  • #2
7

Thank you for your post. I see that you have attempted to solve for the volume of the solid of revolution using both the shell and disk methods. While your calculations look correct, there are a few things that I would like to point out.

Firstly, when using the shell method, you do not need to integrate from 0 to 3, as the line x=3 only intersects with the function at x=1. Therefore, your integral should be from 0 to 1.

Secondly, when using the disk method, you need to subtract the smaller function from the larger function, as the formula you provided shows. In this case, the larger function is F(x)=2x+3 and the smaller function is y=5. So your integral should be from 0 to 1 of pi[(2x+3)^2-5^2]dx.

Lastly, when using the method of slicing by the line y=5, you need to integrate with respect to y, not x. So your integral should be from 0 to 5 of pi[(5-(2x+3))^2]dy.

I hope this helps clarify any confusion you may have had. Keep up the good work!
 

FAQ: Volume of Solid of Revolution: 115.19 and 77.206

1. What is the formula for finding the volume of a solid of revolution?

The formula for finding the volume of a solid of revolution is V = π∫(f(x))^2dx, where f(x) is the function that represents the cross-sectional area of the solid.

2. How do you find the cross-sectional area of a solid of revolution?

The cross-sectional area of a solid of revolution can be found by taking the area of the shape formed by rotating a cross-section of the solid about the axis of revolution.

3. What is the difference between a solid of revolution and a regular solid?

A solid of revolution is formed by rotating a curve or shape around an axis, while a regular solid is formed by stacking or combining 3-dimensional shapes.

4. Can the volume of a solid of revolution be negative?

No, the volume of a solid of revolution cannot be negative as it represents physical space and cannot have a negative value.

5. How do you use calculus to find the volume of a solid of revolution?

To find the volume of a solid of revolution, calculus is used to find the integral of the function representing the cross-sectional area. The integral represents the sum of infinitely small volumes, which when added together, give the total volume of the solid.

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