# Shell method

1. Mar 26, 2008

### DaOneEnOnly

1. The problem statement, all variables and given/known data
Find the volume of the solid of revolution:
F(x)=2x+3 on [0,1]
Revolved over the line x=3 and y=5

2. Relevant 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

2. Mar 26, 2008

### DaOneEnOnly

The last part since I wasn't allowed so many $$1. The problem statement, all variables and given/known data Find the volume of the solid of revolution: F(x)=2x+3 on [0,1] Revolved over the line x=3 and y=5 2. Relevant equations Shell Method: 2[tex]\pi$$$$\int$$$$^{b}$$$$_{a}$$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

3. 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$$\pi$$$$\int$$$$^{5}$$$$_{0}$$(5-y)(1-((y-3)/2)) =130.8996

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

=77.206

3. Mar 26, 2008

### DaOneEnOnly

The last part since I wasn't allowed so many $$3. The attempt at a solution line x=3: 2[tex]/pi$$$$/int$$(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$$\pi$$$$\int$$$$^{5}$$$$_{0}$$(5-y)(1-((y-3)/2)) =130.8996

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

=77.206

4. Mar 26, 2008

### DaOneEnOnly

the first number in the integral is the upper bound and the second is the lower.... I just can't get it to be formatted the right way for some reason.

EDIT: sry about the repeat... said database error so I thought it didn't go through.

EDIT: OMG there's 2 double posts.... srry

Last edited: Mar 26, 2008
5. Mar 27, 2008

### HallsofIvy

Staff Emeritus
Is this two separate problems? It doesn't appear to be from your work but what do you mean by "revolved over" two separate lines? In any case, "F(x)= 2x+3 on [0,1]" doesn't define a region. Do you mean the region bounded by y=2x+ 3, y= 0, x= 0, and x= 1?