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 <br /> <br /> <h2>Homework Statement </h2><br /> Find the volume of the solid of revolution:<br /> F(x)=2x+3 on [0,1]<br /> Revolved over the line x=3 and y=5<br /> <br /> <h2>Homework Equations</h2><br /> Shell Method: 2\pi\int^{b}_{a}x[f(x)-g(x)]dx<br /> obviously just sub y for dy <br /> Disk Method: \pi\int^{b}_{a}[F(x)^{2}-G(x)^{2}dx<h2>The Attempt at a Solution</h2><br /> line x=3: 2\pi\int(3-x)(2x+3)dx =115.19<br /> <br /> answer key is unfortunately in disk method which I don&#039;t like as much:<br /> \pi\int^{3}_{0}(9-4)dy + \pi\int^{5}_{3}(3-((y-3)/2))^{2}-4dy <br /> <br /> =78.91line y=5: 2\pi\int^{5}_{0}(5-y)(1-((y-3)/2)) =130.8996<br /> <br /> answer key/ disk method: \pi\int^{1}_{0}(25-(5-(2x+3))^{2}dx <br /> <br /> =77.206

 

[DaOneEnOnly]

The last part since I wasn't allowed so many <br /> <br /> <h2>The Attempt at a Solution</h2><br /> line x=3: 2/pi/int(3-x)(2x+3)dx =115.19<br /> <br /> answer key is unfortunately in disk method which I don&#039;t like as much:<br /> \pi\int^{3}_{0}(9-4)dy + \pi\int^{5}_{3}(3-((y-3)/2))^{2}-4dy <br /> <br /> =78.91<br /> <br /> line y=5: 2\pi\int^{5}_{0}(5-y)(1-((y-3)/2)) =130.8996<br /> <br /> answer key/ disk method: \pi\int^{1}_{0}(25-(5-(2x+3))^{2}dx <br /> <br /> =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