Expert: Paul Klarreich - 4/29/2007

Question
QUESTION: Let R be the region in Quadrant 1 bounded by the graphs of y=lnx and y=1. Find the exact volume of the solid generated by rotating R about the y-axis.

So, I had set up dV= piy^2dx-piy^2dx. This then would be   pi(lnx)^2 dx - pi(1)^2dx.  But, have not yet learned how to integrate lnx, so I don't think this is right. Can you please help?  Thanks.
ANSWER: Questioner:   Colleen
Category:  Calculus
 
Subject:  Volume by cylindrical shells.
Question:  Let R be the region in Quadrant 1 bounded by the graphs of y=lnx and y=1. Find the exact volume of the solid generated by rotating R about the y-axis.

So, I had set up dV= piy^2dx-piy^2dx.

>>
>> No, I think that is for rotating about the x-axis, not the y-axis.
>>

This then would be   pi(lnx)^2 dx - pi(1)^2dx.  But, have not yet learned how to integrate lnx, so I don't think this is right. Can you please help?  Thanks.
 
...........................................
Hi, Colleen,

Your region is not so well defined.  I will assume it is the 'trapezoidal' region bounded on:

The Left: by the y-axis.
The Top : by y = 1.
The right and the bottom: by  y = ln x
Also the bottom: by the x-axis from x=0 to x=1

|
1|------------------/|   
|          / |  
|          / < y = ln x, which is really curved,   
|          /     but I can't make that here.  
|          /    |
|          /     |
0|-------------------+--------------
|          1     e


So you have two portions:
1. The square that is 1-by-1 which rotates into an ordinary cylinder, with r = 1, h = 1, so has V = pi r^2 h = 1 pi, so forget that until later.

2. The 'sort of triangular' region bounded on

The Left: by x = 1.
The Top : by y = 1.
The right and the bottom: by  y = ln x

This is the part we will try to integrate.  You will take a vertical slice, then rotate that around the y-axis to make a cylindrical 'shell'.  Its properties are:

Height = 1 - f(x) = 1 - ln x
Radius = x
Thickness = dx.

The volume is the 'area', which is a rectangle, times the thickness.
The area is the circumference times the height:  2pi x(1 - ln x)

dV = 2pi x(1 - ln x) dx

And you have to integrate:

{e
| 2pi x(1 - ln x) dx
}1

{
| 2pi (x - x ln x) dx
}

The integral of the  x term is easy.  To integrate  x ln x:

(we'll put the rest of it in later)

{
| x ln x dx
}

HAVE YOU LEARNED INTEGRATION BY PARTS?  If you have not, you must tell me and we will have to do this problem in a completely different way.

Let  u = ln x,  du = 1/x dx
dv = x dx,  v = x^2/2

 x^2 ln x    {  x^2   1
= --------- - | ----- --- dx
     2       }   2    x

 x^2 ln x    {  x
= --------- - | --- dx
     2       }  2

 x^2 ln x     x^2
= --------- -  ---
     2         4

which you evaluate from  x = 1 to x = e.

I think you can handle the rest of it.


---------- FOLLOW-UP ----------

QUESTION: I do not know how to integrate by parts. I think it is in the next chapter. Thanks again for the help.

Answer
Questioner:   Colleen
Category:  Calculus
Private:  No
 
Subject:  Volume by cylindrical shells.
Question:  QUESTION: Let R be the region in Quadrant 1 bounded by the graphs of y=lnx and y=1.

Find the exact volume of the solid generated by rotating R about the y-axis.

So, I had set up dV= piy^2dx-piy^2dx. This then would be   pi(lnx)^2 dx - pi(1)^2dx.  But, have

not yet learned how to integrate lnx, so I don't think this is right. Can you please help?  

Thanks.
ANSWER: Questioner:   Colleen
Category:  Calculus

Subject:  Volume by cylindrical shells.
Question:  Let R be the region in Quadrant 1 bounded by the graphs of y=lnx and y=1. Find the

exact volume of the solid generated by rotating R about the y-axis.

So, I had set up dV= piy^2dx-piy^2dx.

>>
>> No, I think that is for rotating about the x-axis, not the y-axis.
>>

This then would be   pi(lnx)^2 dx - pi(1)^2dx.  But, have not yet learned how to integrate lnx, so

I don't think this is right. Can you please help?  Thanks.

...........................................
Hi, Colleen,

Your region is not so well defined.  I will assume it is the 'trapezoidal' region bounded on:

The Left: by the y-axis.
The Top : by y = 1.
The right and the bottom: by  y = ln x
Also the bottom: by the x-axis from x=0 to x=1

|
1|------------------/|   
|          / |  
|          / < y = ln x, which is really curved,   
|          /     but I can't make that here.  
|          /    |
|          /     |
0|-------------------+--------------
|          1     e


So you have two portions:
1. The square that is 1-by-1 which rotates into an ordinary cylinder, with r = 1, h = 1, so has V

= pi r^2 h = 1 pi, so forget that until later.

2. The 'sort of triangular' region bounded on

The Left: by x = 1.
The Top : by y = 1.
The right and the bottom: by  y = ln x

This is the part we will try to integrate.  You will take a vertical slice, then rotate that

around the y-axis to make a cylindrical 'shell'.  Its properties are:

Height = 1 - f(x) = 1 - ln x
Radius = x
Thickness = dx.

The volume is the 'area', which is a rectangle, times the thickness.
The area is the circumference times the height:  2pi x(1 - ln x)

dV = 2pi x(1 - ln x) dx

And you have to integrate:

{e
| 2pi x(1 - ln x) dx
}1

{
| 2pi (x - x ln x) dx
}

The integral of the  x term is easy.  To integrate  x ln x:

(we'll put the rest of it in later)

{
| x ln x dx
}

HAVE YOU LEARNED INTEGRATION BY PARTS?  If you have not, you must tell me and we will have to do

this problem in a completely different way.

Let  u = ln x,  du = 1/x dx
dv = x dx,  v = x^2/2

x^2 ln x    {  x^2   1
= --------- - | ----- --- dx
    2       }   2    x

x^2 ln x    {  x
= --------- - | --- dx
    2       }  2

x^2 ln x     x^2
= --------- -  ---
    2         4

which you evaluate from  x = 1 to x = e.

I think you can handle the rest of it.


---------- FOLLOW-UP ----------

QUESTION: I do not know how to integrate by parts. I think it is in the next chapter. Thanks again

for the help.

.........................................
OK, Colleen,

Here's what we will have to do, for:

2. The 'sort of triangular' region bounded on

The Left: by x = 1.
The Top : by y = 1.
The right and the bottom: by  x = e^y

This is the part we will try to integrate.  You will take a HORIZONTAL slice, then rotate that

around the y-axis to make an ANNULAR DISK.  Its properties are:

Outer radius:  e^y
Inner radius = 1
Thickness = dy.

The area of this disk is  pi(R^2 - r^2), where R and r are the radii.
So we integrate dV = pi ( (e^y)^2 - 1 ) dy  from  y=0 to y=1

{1
| pi ( (e^y)^2 - 1 ) dy =
}0

{
| pi ( e^2y - 1 ) dy =
}

and this should be a fairly easy integration.