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

The Attempt at a Solution

line x=3: 2$$/pi$$$$/int$$(3-x)(2x+3)dx =115.19

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

=78.91


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

answer key/ disk method: $$/pi$$$$/int$$$$^{1}$$$$_{0}$$(25-(5-(2x+3))$$^{2}$$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!