How Do You Apply the Cylindrical Shells Method for Rotating a Bounded Region?

[nlsherrill]

1. From Stewart Calculus and Concepts 4th edition, page 454 #15

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis.

15. y=4x-x^2, y=; rotate about x=1

Homework Equations

3. I was able to find the volume of the function y=4x-x^2 by integrating from 0 to 4, but the line y=3 cuts the function off near the top, and I don't know how to take some of that volume "out". Are 2 integrals needed for this? I also tried the washer method of integrating the top function(y=4x-x^2 minus the bottom function(y=3) and I still didn't get it. I've been stuck on this one for about 2 hours now. Any hints?

 

[Raskolnikov]

Are 2 integrals needed for this? Any hints?

Nope, we don't need 2 integrals. Since this is rotation about a vertical axis (x=1), we have

V = \int_a^b 2\pi*r*h*dx
Where,
r = distance from axis
h = height of shell.

That's all we need. As for hints:
Since we're rotating about x=1, we're essentially "losing" 1 unit of x in the sense that r = x - 1.
Also, h is the difference of the two y(x) functions (top minus bottom).

Can you figure out the rest? I think you should end up with V = 4/3.

 

[nlsherrill]

Nope, we don't need 2 integrals. Since this is rotation about a vertical axis (x=1), we have

V = \int_a^b 2\pi*r*h*dx
Where,
r = distance from axis
h = height of shell.

That's all we need. As for hints:
Since we're rotating about x=1, we're essentially "losing" 1 unit of x in the sense that r = x - 1.
Also, h is the difference of the two y(x) functions (top minus bottom).

Can you figure out the rest? I think you should end up with V = 4/3.

The back of the book says the answer is 8*pi/3

I know how to do some of these but this one I must just be missing something fundamental.

I don't know how to do a definite integral, but its from 0 to 4 of2\pi\int(x-1)(4x-x^2-3)dx

 

[Raskolnikov]

The back of the book says the answer is 8*pi/3

I know how to do some of these but this one I must just be missing something fundamental.

ooops...yea, I forgot the factor of 2pi out in front. So 4/3 * 2pi = 8pi/3 is correct.

I don't know how to do a definite integral, but its from 0 to 4 of2\pi\int(x-1)(4x-x^2-3)dx

Close! Everything inside the integral is good, but your limits are a bit off, i.e. NOT from 0 to 4. Remember, the limits of your integral will be the 2 x-values that bound the region. Between which two x-values does the region lie?

 

[nlsherrill]

ooops...yea, I forgot the factor of 2pi out in front. So 4/3 * 2pi = 8pi/3 is correct.


Close! Everything inside the integral is good, but your limits are a bit off, i.e. NOT from 0 to 4. Remember, the limits of your integral will be the 2 x-values that bound the region. Between which two x-values does the region lie?

all I know is the functions intersect at x=1 and x=3. Are those the limits? If so then I don't understand why.

 

[nlsherrill]

Okay I got the answer finally.

The whole time I was trying to solve it I was thinking for some reason that they were looking for the bottom half of the curve, not the top part bisected by y=3. O well, lesson learned not to make blind assumptions! Thank you Raskolnikov