Volume of Revolution: Perpendicular Rectangles and the Axis of Rotation

[lionely]

Homework Statement

The area enclosed by y= 4/x^2 , y =1 and y =4 is rotated about the x-axis; find the volume generated.

I am really confused I keep getting (14pi/3). But the answer in the back of the book is not that at all.
An additional question, the only method I know is drawing a really thin rectangle of width dx or dy depending on which axis I am rotating it about. The question is , does the rectangle always have to be at right angles to the axis of rotation?

Homework Equations

The Attempt at a Solution

Here my attempt with my little sketch graph. The area I am supposed to be rotating is confusing because of that gap in the graph . I chose x = 2 and x=1 as the limits because that is what I got when I put in y=4 and y=1 into the function. I thought about using x=-2 and x=x-1 as well and adding them but that just gives me 28pi/3 and that is not the answer either. I am quite confused..

 

[Nathanael]

y² = 16/x⁴ not 16/x²

Calculate the volume from the middle region separately and add it to the volume from the other two regions.

 

[lionely]

I have 16/x^4 but I wrote it badly and the volume of the middle region is that between the lines y=4 and y=1?

 

[Nathanael]

I have 16/x^4 but I wrote it badly

Sorry.

the volume of the middle region is that between the lines y=4 and y=1?

The area between y=4, y=1, x=(...),x=(...) is the volume I'm talking about (well, it's a volume after you revolve it). It will be the shape of a cylinder (with a hole).

You want to find the volume of this cylinder separately (after finding the x-boundaries that I left out) and then add it on to the total volume, because when you integrate from x=1 to x=2 you are leaving this middle section unaccounted for.

 

[lionely]

For the volume between y=4 and y=1 , the bounds I used were x= 1 and x=-1 and I got (32Pi) and added that to the 28Pi/3 . but that's not the answer. Still confused.

 

[Nathanael]

When finding the volume of the cylinder, did you subtract the area of the hole?

 

[lionely]

I don't think so, when I rotated the area under y=4 from x=-1 to x=1 i got 32Pi ( This the area of the hole right? ) or is the area of the hole under the line y=1?

 

[Nathanael]

or is the area of the hole under the line y=1?

Yes.

You found the volume from y=0 to y=4, but that includes y=0 to y=1 ("the hole") so you must subtract it.

 

[lionely]

Oh I got the answer now, I find it really hard to visualize these things...
Thank you, also :

my additional question, the only method I know is drawing a really thin rectangle of width dx or dy depending on which axis I am rotating it about. The question is , does the rectangle always have to be at right angles to the axis of rotation?

thank you again! I will try harder with my remaining questions.

 

[Nathanael]

does the rectangle always have to be at right angles to the axis of rotation?

Yeah, the formula $V=\pi \int y^2 dx$ is implicitly dividing it up into many small rectangles which are perpendicular to the axis of rotation (and then adding them up).

You can make other formulas which apply to non-perpendicular rectangles, but the result is nice and simple if they are perpendicular.

For example if you want to take rectangles parallel to the y-axis and rotate them about the y-axis then I think this should work:

Spoiler

$2\pi\int\limits_a^b xydx$

(I give no explanation so that you can try to figure the formula out in the future, if you want.)

If you want to take rectangles that go off at some angle, then you are just making life difficult