Finding Volume of Solid Revolved Around x=3, y=5

[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

 

[DaOneEnOnly]

The last part since I wasn't allowed so many [tex]

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

=78.91line 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

 

[DaOneEnOnly]

The last part since I wasn't allowed so many [tex]

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

 

[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

 

[HallsofIvy]

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

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?

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

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

=77.206