How Do You Calculate the Volume of a Rotated Metal Ash Tray?

[lionely]

Homework Statement

The area under y = (x^2/9) + 1 from x = 0 to x = 3 , and the area enclosed by the y= 0 , y=2 , x=3 , and x=4, are rotated about the y-axis , and the solid generated represents a metal ash tray , the units being cm. Calculate the volume of a metal.

Homework Equations

The Attempt at a Solution

for the volume between x=3 and x=4 i wanted to rotate it about the y-axis but I got confused... so I just did about the x-axis. Also I only sketched the part of the curve I would need.
I can't see why I am not getting the answer, shouldn't I just find those two volumes and ADD them together to get the volume of the ash tray? This is so frustrating...

Is there something I am missing? All that I see done in the examples in this textbook is draw a rectangle perpendicular to the axis of rotation and then boom you have your formula,

either ∫πy² dx or ∫πx² dy

 

[SammyS]

Homework Statement

The area under y = (x^2/9) + 1 from x = 0 to x = 3 , and the area enclosed by the y= 0 , y=2 , x=3 , and x=4, are rotated about the y-axis , and the solid generated represents a metal ash tray , the units being cm. Calculate the volume of a metal.

Homework Equations

3. The Attempt at a Solution [/B]
for the volume between x=3 and x=4 i wanted to rotate it about the y-axis but I got confused... so I just did about the x-axis. Also I only sketched the part of the curve I would need.
I can't see why I am not getting the answer, shouldn't I just find those two volumes and ADD them together to get the volume of the ash tray? This is so frustrating...

Is there something I am missing? All that I see done in the examples in this textbook is draw a rectangle perpendicular to the axis of rotation and then boom you have your formula,

either ∫πy² dx or ∫πx² dy

[ IMG]http://i59.tinypic.com/2w5lvkj.jpg[/PLAIN]

It seems that you only know the "Disk" method. An extension of this is the "Washer" method, sometimes referred to as the "Donut" method.

A useful method for this problem as well as you recent problem, https://www.physicsforums.com/threads/volume-of-revolution-question.824553/#post-5177463, is the "Shell" method.

I suggest you study these methods and then come back to this problem.

Rotating about the x-axis rather than the y-axis gives a very different result.

 

[lionely]

Okay, I find that so weird, because the textbook I am using only has that method. So it should mean that method can be used. I will look up those methods though..
Also how would you even rotate an area under y=2 around the y axis. Since the integral would be something like ∫πx2 dy. There is no x to square , is there?

 

[SammyS]

Okay, I find that so weird, because the textbook I am using only has that method. So it should mean that method can be used. I will look up those methods though..
Also how would you even rotate an area under y=2 around the y axis. Since the integral would be something like ∫πx2 dy. There is no x to square , is there?

You would need to use subtraction. Find the volume of the large cylinder, (radius 4cm, symmetry axis is the y-axis). Then subtract the volume of the piece inside the parabolic portion.

In fact, for this problem, that's not too bad a technique.

 

[lionely]

OKAY so what I did now was instead of really using calculus, I rotated the area bounded by y=2 from x=4 to x=0. That's a cylinder with radius 4 and height 2
so the volume would be 32Pi. Then from my previous work the volume bounded by the parabola x=3 and x=0 about the y-axis was 9pi/2.

Subtracting the volume of the parabolic portion like you said from the larger cylinder I get 27(Pi/2). WHich is the answer, but I don't think i fully understand. Okay the way I was visualizing the shape made by the parabolic portion was like maybe a bowl of some sort? and then the volume that I would get from y=2 ,x=3 and x=4 would be the outside of the ash tray, so that is why I thought I would just add them...

Subtracting the volume formed form the parabolic portion wouldn't that give the larger cylinder a huge hole? Sorry I am just trying to visualize this ash tray haha...

 

[MidgetDwarf]

OKAY so what I did now was instead of really using calculus, I rotated the area bounded by y=2 from x=4 to x=0. That's a cylinder with radius 4 and height 2
so the volume would be 32Pi. Then from my previous work the volume bounded by the parabola x=3 and x=0 about the y-axis was 9pi/2.

Subtracting the volume of the parabolic portion like you said from the larger cylinder I get 27(Pi/2). WHich is the answer, but I don't think i fully understand. Okay the way I was visualizing the shape made by the parabolic portion was like maybe a bowl of some sort? and then the volume that I would get from y=2 ,x=3 and x=4 would be the outside of the ash tray, so that is why I thought I would just add them...

Subtracting the volume formed form the parabolic portion wouldn't that give the larger cylinder a huge hole? Sorry I am just trying to visualize this ash tray haha...

Like previous posters said, the washer or shell method would make this a trivial problem. Do you want the proof for the shell method?

 

[HallsofIvy]

Personally, I would do this as two separate parts, first the section under the parabola from x= 0 to x= 3.

The remaining portion can also be done in two parts. The entire rectangle, from x= 0 to x= 4 rotates into a disk with radius 4 and thickness 2 so volume [itex]\pi (4)^2(2)= 32\pi[/itex]. The part that we don't want, from x= 0 to x= 3 is a disk with radius 3 and thickness 2 so has volume [itex]\pi (3)^2(2)= 18\pi[/itex]. Subtract that inner portion, that we don't want from the entire disk to get the outer part.

 

[lionely]

I had got the answer from subtracting the section under the parabola from x =0 to x=3 from the cylinder formed by rotating the area under y=2. Thank you for all your responses

 

[SammyS]

I had got the answer from subtracting the section under the parabola from x =0 to x=3 from the cylinder formed by rotating the area under y=2. Thank you for all your responses

You should have subtracted the volume above the paraboloid from the cylinder volume.

It turns out that the volume above the paraboloid is the same as the volume above y=1 and below the paraboloid.