
#1
Jan1604, 11:30 PM

P: 19

Included is my attempt at the following question. I get an answer of 10pi, whereas the right answer is (10pi)/3 from my text book. Here is the question:
Rotate the triangle described by (1,0),(0,1),(1,0) around the axis x=2 and calculate the volume of the solid. I basically changed the problem to the following and continued as it is the same Rotate the triangle described by (1,0),(2,1),(3,0) around the yaxis and calculate the volume of the solid Thanks for the help 



#2
Jan1704, 12:10 AM

P: 661

It is so small to understand the pic u quoted so i'm giving u my solution




#3
Jan1704, 01:31 AM

P: 661





#4
Jan1704, 08:18 AM

P: 19

Volume of Revolution
I don't quite understand it and I dont understand why mines wrong :(
I'll keep trying to figure it out. 



#5
Jan1704, 08:26 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886

You may not have had it yet but this looks like an exercise in using Pappus' theorem: the volume of a solid of revolution is the area of a cross section times the circumference of the circle generated by the centroid of that cross section.
In this case, the cross section is a triangle with base of length 2 and height 1: area= (1/2)(2)(1)= 1. The centroid (for a triangle only) is the "average" of the vertices: ((1+0+1)//3,(0+1+0)/3)= (0, 1/3). The distance from (0, 1/3) to the line x= 2 is 2 1/3= 5/3. The centroid "travels in" (generates) a circle of radius 5/3 and so circumference (10/3)pi. The volume of the figure is (1)(10/3)pi= (10/3)pi. 



#6
Jan1704, 11:53 PM

P: 661





#7
Jan1804, 06:07 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886

Pappus' theorem is true of any "solid of revolution". You should be able to find it in any calculus textbook (that includes multiple integrals.)




#8
Jan1804, 08:32 AM

P: 19

I haven't learned that theorem so I don't really think I should use it.
I'm trying to use horizontal rectangles for my area so I formulated the following integral which is how I was taught how to do questions like this. [tex] \int \pi r_o^2  \pi r_i^2 dr [/tex] where [tex]r_o=(y+3)[/tex] [tex]r_i=(y+1)[/tex] and the limits of integration are from y=0 to y=1 so we have this [tex] \int_0^1 \pi (y+3)^2  \pi (y+1)^2 dy [/tex] it makes perfect sense to me but then once you work it out you get 10 pi. which is wrong So can someone point out what I did wrong and how to fix it so I dont do it again. Thanks very much. 



#9
Jan1804, 08:39 AM

P: 661




#10
Jan1804, 08:43 AM

P: 19

does mine not make sense for some reason
i am calculating the area of the circle when the farthest line is rotated and then subtracting the area of the circle when the closest line is rotated 



#11
Jan1804, 08:49 AM

P: 661

Yes it do makes sense dont you have gone to the previous post.
Thats what u have to do and its general too Your way do make sense 



#12
Jan1804, 08:52 AM

P: 661

When you rotate a point about a line u get a circle Not When you rotate a line It would be somewhat like a truncated cone 



#13
Jan1804, 08:53 AM

P: 19

but i get the wrong answer so can you point out whats wrong with my formulation?
keep in mind i moved the triangle to (1,0),(2,1),(3,0) because its the same problem correct? 



#14
Jan1804, 08:56 AM

P: 19

I'm rotating horizontal rectangles around the yaxis therefore each rectangle will make a circle 



#17
Jan1804, 09:33 AM

P: 19

I still want to know whats wrong with this answer because it does not yield 10pi/3
[tex]\int_0^1 \pi (y+3)^2  \pi (y+1)^2 dy[/tex] 


Register to reply 
Related Discussions  
volume of revolution  Calculus & Beyond Homework  1  
volume of revolution  Calculus & Beyond Homework  1  
Volume of Revolution  Calculus & Beyond Homework  2  
Volume of revolution  Calculus & Beyond Homework  5  
volume of revolution  Calculus & Beyond Homework  7 