Volume of Solid of Revolution: 115.19 and 77.206

[DaOneEnOnly]

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

 

[mighty2000]

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!