Question

f(x)=x^2+2 between 0 12 years ago

Answer 1

I am not sure how the y=3 comes into it and where I need to use it in the formula. I am using the disc method as this shape is like a cone when rotated around y=3.

Answer 2

3-y is the radius of the disc.

Answer 3

Ok so it is outer radius is (3-0)^2 - the inner radius (3-(x^2+2)^2
Correct?

Answer 4

Inner? Outer? These are not washers. These are disks. Or, which is it?

Are we rotating the portion adjacent to y = 3 or the portion adjacent to the x-axis?

Answer 5

We are rotating the function when the graph of x^2+2 is rotated around y=3 and my x values are 0 to 1

It doesn't tell me that it is the area enclosed between the graph and y=3 so I assume it is the portion below the graph x^2+2 and the x-axis. And then when this rotates upward it kind of becomes a cylinder with a cone in it? Does that make sense?

Answer 6

It's just not clear to me what is wanted.

Answer 7

The question is the above right at the top. I am just going from that.
When I plugged everything in maple it says
integral (8-x^4+2x*2)Pi dx from upper limit 1 to lower limit 0
the picture is a cylinder with a cone in it

I am just wondering how to come up with the above integral to solve.

Answer 8

I wish I had a drawing, or something. Anyway, here we have the top piece, the portion above y = x^2 + 2 and below y = 3 rotated about y = 3.

Cyllinders

\(2\pi\cdot\int\limits_{2}^{3}(3-y)\cdot \sqrt{y-2}\;dy = \dfrac{8\pi}{15}\)

Disks

\(\pi\cdot\int\limits_{0}^{1} (3-(x^{2} + 2))^{2}\;dx = \dfrac{8\pi}{15}\)

Answer 9

You just have to stare at it until it makes sense in your head. What makes each piece happen?

Answer 10

The outer piece is easily determined, before any other calculus, simply by subtracting the previous result from \(9\pi\). This gives \(\dfrac{127\pi}{15}\). The challenge, now, is to find the integrals.

Answer 11

I will have to come back to this one with a fresh mind. I think I've confused myself :(

Answer 12

Cyllinders

Two Pieces. First, the easy part. 0 < y < 2. This is just a right, circular cylinder.

\(9\pi - \pi = 8\pi\)

Second, for 2 < y < 3

\(2\pi\cdot\int\limits_{2}^{3}(3-y)\left(1-\sqrt{y-2}\right)\;dy = \dfrac{127\pi}{15}\)

Washers

\(\pi\cdot\int\limits_{0}^{1}3^{2} - (3-(x^{2}+2))^{2}\;dx = \dfrac{127\pi}{15}\)

I SUPER recommend doing EVERY problem both ways. This gives experience, corroboration, and confidence.

Come back when you're ready.

Answer 13

Whoops! For the outer part and cylinders, I meant to say that the integral increased by \(8\pi\) is the total result \(\dfrac{127\pi}{15}\)

Answer 14

@milkacha: I strongly recommend drawing both the function and the axis of rotation, y=3. Doing so will help you to identify the shape of the solid being generated, which in turn will help you decide which method to use to find the volume, and to identify the limits of integration. Comfortable using the Draw utility?

Answer 15

My interpretation is that the solid in question is ring-shaped; it has a maximum radius of 3, a max. diameter of 6, and a max. thickness of 1. Disks would be inappropriate to use here; washers WOULD be appropriate; but the easiest method would be cylindrical shells. Please include these in your drawing.

Answer 16

That's correct mathmale that sounds like my drawing.
I can probably do what tkhunny described and find the volume of the cylinder and then deduct the volume of the cone inside to check my answer.

Answer 17

Except that there's no cylinder in the usual sense here, and neither is there a cone.

Pretty please, sketch the graph of y=x^2 + 2. Darken the area of the graph between x=0 and x=1. Draw in the axis of rotation x=3. Draw the solid that would result if the darkened area were rotated about the axis x=3.

When we say "cylinder," we usually mean a RIGHT cylinder, not one with a sloped top.

Answer 18

Rotated around y=3

Answer 19

This is my diagram.

Answer 20

Next time, let's lead with that. :-)

Answer 21

@milkacha: Please summarize where you are now (in terms of solving this problem) and what you still need to learn or do to solve it. Your diagram is very good and makes a huge difference in terms of communicating this problem to your audience accurately. You do need to indicate clearly whether you're aiming to find the volume of the solid generated when (1) the region between the graph and the line y=3 is rotated around the line y=3 OR (2) the region between the graph and the x-axis is rotated around y=3. This is a critical piece of information that should be included in the statement of this math problem. Looking forward to hearing back from you.

Answer 22

Love the second drawing!! The cross section of the whole thing is a GREAT idea.

Answer 23

Ok it is a solid and it is the area between the curve and the axis we are rotating around so in the diagram it is not the shaded area, rather the area above. When rotated around y=3 it looks more like a cone shape with no hollow inside, so I should use the disc method.
So my integral becomes Pi (x^2-1)^2
Simple to solve and I get 8/15Pi
This was confirmed by Maple when I plug in all the variables.

Hopefully this is correct?

Answer 24

Rotating a region around an axis other than the x- or y-axis can be tricky. In this case, wit you rotating that small region around y=3, the radius of your disk would be 3-y, or 3-(x^2+2)=1-x^2. Check that out. If x=0, the radius is 1 (correct). If x=1, the radius is 0 (also correct).

The area of each disk is pi( r )^2, or pi*(1-x^2)^2.

How much sense does this make to you?