Calculating Volume of Revolved Solid Using Cylindrical Shells

[saxen]

This shouldn't be a problem but I just can't this right!

Homework Statement

Find the volume of the body that is generated when y= 1/x(sqrt(4-x^2)) ,1 <= x <2, is rotated around the y-axis

Homework Equations

Using cylindrical shells :

2 pi ∫x (g(x)-f(x)) dx

The Attempt at a Solution

2 pi ∫x (g(x)-f(x)) dx= 2 pi ∫1/sqrt(4-x^2) dx (Can I put f(x)=0?)

Using inverse substitution

x= 2sin t
dx= 2cos t dt

2 pi ∫dt

2 pi [t], Is it correct to use sin(2) and sin(1) as limits here?

Somewhere it all goes wrong, help is very much appreciated!

First post aswell, didn't find anywhere to introduce myself though. So hello all =)

 

[HallsofIvy]

This shouldn't be a problem but I just can't this right!

Homework Statement

Find the volume of the body that is generated when y= 1/x(sqrt(4-x^2)) ,1 <= x <2, is rotated around the y-axis

Homework Equations

Using cylindrical shells :

2 pi ∫x (g(x)-f(x)) dx

The Attempt at a Solution

2 pi ∫x (g(x)-f(x)) dx= 2 pi ∫1/sqrt(4-x^2) dx (Can I put f(x)=0?)

Using inverse substitution

x= 2sin t
dx= 2cos t dt

2 pi ∫dt

2 pi [t], Is it correct to use sin(2) and sin(1) as limits here?

Somewhere it all goes wrong, help is very much appreciated!

First post aswell, didn't find anywhere to introduce myself though. So hello all =)

The problem is that you don't have upper or lower bounds on the region. You appear, since you are taking the "height" of each shell to be y, to be assuming the region rotated around the y-axis to be bounded below by y= 0 and bounded above by y= that function.

And that's what's causing your confusion about "f(x)"- f should be the function giving the lower boundary. IF that lower boundary is y= f(x)= 0, then, yes, you can take f(x)= 0.

Since your original integral is "dx" and x is between 1 and 2, you should have
2\pi \int_1^2 \frac{dx}{\sqrt{4- x^2}}

With the substitution x= 2 sin(t), no, the limits are NOT "sin(1)" and "sin(2)". You are going "the wrong way". "1" and "2" are values of x. When x= 1= 2 sin(t), sin(t)= 1/2 so t= sin^{-1}(1/2)= \pi/6[/itex]. When x= 2= 2sin(t), sin(t)= 1 so t= \pi/2. Those are the limits of integration.

 

[saxen]

The problem is that you don't have upper or lower bounds on the region. You appear, since you are taking the "height" of each shell to be y, to be assuming the region rotated around the y-axis to be bounded below by y= 0 and bounded above by y= that function.

And that's what's causing your confusion about "f(x)"- f should be the function giving the lower boundary. IF that lower boundary is y= f(x)= 0, then, yes, you can take f(x)= 0.

Since your original integral is "dx" and x is between 1 and 2, you should have
2\pi \int_1^2 \frac{dx}{\sqrt{4- x^2}}

With the substitution x= 2 sin(t), no, the limits are NOT "sin(1)" and "sin(2)". You are going "the wrong way". "1" and "2" are values of x. When x= 1= 2 sin(t), sin(t)= 1/2 so t= sin^{-1}(1/2)= \pi/6[/itex]. When x= 2= 2sin(t), sin(t)= 1 so t= \pi/2. Those are the limits of integration.

I got it know, thank you very much! This was driving me crazy.