Washer and Shell Method (Newbie)

Hi, I’m pretty new using the Washer and Shell’s method.

So far I don’t really know much of the distinction between the two.

This is what I know that both of them requires:

Each one requires an outer radius and inner radius.

The only way I know how to find the outer radius and inner radius is by graphing them, and see where the points intersect, and make a cone out of them. Sometimes whenever I know the curve goes like the other way, I instantly know it’s an inner radius. What I mean is, like if I had x^{2} and √x, then x^{2} is the inner radius because it goes outward, so it doesn’t make like a round semi-circle. Am I right?

Though sometimes I get confused by the y=0 part, where it’s mainly just a line, so I’m not really sure how to make the shape. If anyone can give me some tips, or know other ways of telling the outer and inner radius, let me know.

After finding the outer and inner radius, I believe using the Washer and Shell’s method both get you the same answer, right?.

__Also, when it revolves around the x-axis, you would set both of the inner and outer radius equal to 'x' right? And if it was revolving around y, then it would be set equal to 'y'?__

Like for example: x+y^{2}=9, x=0. So I would set this up using the Shell method by doing the equation above, but I will make the first function equal to x like this: x=9-y^{2}, right?.

This is what I know about the Washer’s:

The equation is πR^{2}w. So I would use that, and subtract it from the inner radius, πr^{2}w.

So it would be like π**∫R**^{2}-r^{2}, then just plug in the radius, integrate it, and plug in the ‘b’ and ‘a’ and subtract the ‘a’ from ‘b’ right? That’s all it is to the Washer method?

This is what I know about the Shell’s:

It’s almost the same as the Washer’s except…

It’s like this: **2π∫x(R-r)**

Then plug in the inner and outer radius, and then integrate, and plug in the ‘b’ and ‘a’ and subtract the ‘a’ from ‘b’ right? I’m not really sure about the Shell method the most, since it has the 2π on the outside, and an x on the inside for every equation?

If anyone know an easy way to study the Washer, and Shell, or give me some tips, please let me know! J

Thanks!

Re: Washer and Shell Method (Newbie)

Quote:

Originally Posted by

**Chaim** Hi, I’m pretty new using the Washer and Shell’s method.

So far I don’t really know much of the distinction between the two.

This is what I know that both of them requires:

Each one requires an outer radius and inner radius.

No, this is not so. The Washer method requires two radii. A "washer" is the region between two circles. (A hardware "washer" is used to support a nut and bolt- the inner hold is for the bolt to go through and the flat area between the inner and outer edges helps support it.)

A "shell" is completely different. It consists of a cylinder that extends the length of the axis of rotation. It has only one radius.

Quote:

The only way I know how to find the outer radius and inner radius is by graphing them, and see where the points intersect, and make a cone out of them. Sometimes whenever I know the curve goes like the other way, I instantly know it’s an inner radius. What I mean is, like if I had x^{2} and √x, then x^{2} is the inner radius because it goes outward, so it doesn’t make like a round semi-circle. Am I right?

I have no idea what "cone" you mean. There is no cone involved in either method.

Quote:

Though sometimes I get confused by the y=0 part, where it’s mainly just a line, so I’m not really sure how to make the shape.

I don't know what you mean here. "y= 0" is, of course, the x-axis and certainly is a line. What "shape" are you talking about?\

Quote:

If anyone can give me some tips, or know other ways of telling the outer and inner radius, let me know.

After finding the outer and inner radius, I believe using the Washer and Shell’s method both get you the same answer, right?.

\

A given two-dimensional figure, revolved around an axis, **has** as specific number as volume so, whatever method you use to find the volume, you will get that number, of course.

Quote:

__Also, when it revolves around the x-axis, you would set both of the inner and outer radius equal to 'x' right? And if it was revolving around y, then it would be set equal to 'y'?__

__ __

Not [b]equal[/b[] to y but equal to a distance measure parallel to the y axis.

[quote]Like for example: x+y^{2}=9, x=0. So I would set this up using the Shell method by doing the equation above, but I will make the first function equal to x like this: x=9-y^{2}, right?.

To find the volume given by rotating $\displaystyle x= 9- y^2$, x> 0, around the x- axis, using the shell method, for each y, we would have a shell extending from (0, y) to $\displaystyle (9-y^2, y)$ and so of length $\displaystyle 9- y^2$. It radius, since it is rotated around the x-axis, is the distance from the x-axis to the line with given y so is y. That gives a cylindrical shell of circumference $\displaystyle 2\pi y$ and so surface area $\displaystyle 2\pi y(9- y^2)$. Taking a "thickness" of dy, the volume of that thin shell is $\displaystyle 2\pi y(9- y^2)dy$ and to find the entire volume, integrate that from y= 0 to y= 3 (the parabola crosses the x- axis at (0, 3) in the first quadrant).

This example would not be appropriate for the "washer" method since there no "inner" boundary- the figure goes all the way down to the axis. You could use the "disk" method.

Quote:

This is what I know about the Washer’s:

Quote:

The equation is πR^{2}w. So I would use that, and subtract it from the inner radius, πr^{2}w.

No, you subtract the area for the inner radius from the area for the outer radius. Further the area for a given radius, r, is NOT $\displaystyle \pi r^2w$. You don't say what "w" is but if it is a distance, what you give would have units of a volume, not an area. The area is $\displaystyle 2\pi rw$ where, now, w is the length of the shell.

If a 'washer' has outer radius R and inner radius r, then the area of the entire outer disk is $\displaystyle \pi R^2$ while the inner disk, the "hole" in the washer, would have area of $\displaystyle \pi r^2$. The area of the washer itself would be $\displaystyle \pi R^2- \pi r^2= \pi (R^2- r^2)$. If, for example, the axis were the x-axis, then the two raidii would be the y coordinates of two different boundaries, as functions of x, $\displaystyle y_2(x)$, say, for the outer radius and $\displaystyle y_1(x)$ for the inner radius. Then a washer would have area $\displaystyle \pi(y_2^2(x)- y_1^2(x))$. Taking the "thickness" of the washer to be dx, the volume of each washer is $\displaystyle \pi(y_2(x)- y_1(x))dx$ and to find the total volume you need to integrate that between whatever are the smallest and largest values of x.

Quote:

So it would be like π**∫R**^{2}-r^{2}, then just plug in the radius, integrate it, and plug in the ‘b’ and ‘a’ and subtract the ‘a’ from ‘b’ right? That’s all it is to the Washer method?

It would be better to sketch and think about each problem separately rather than trying to give a general formula but, yes, that is the basic idea

[quote]

This is what I know about the Shell’s:

It’s almost the same as the Washer’s except…

It’s like this: **2π∫x(R-r)**

Then plug in the inner and outer radius, and then integrate, and plug in the ‘b’ and ‘a’ and subtract the ‘a’ from ‘b’ right? I’m not really sure about the Shell method the most, since it has the 2π on the outside, and an x on the inside for every equation?[quote]

Yes, if you are rotating around the y-axis, then each "shell" rotates in a circle with radius x and so circumference $\displaystyle 2\pi x$. The area of a cylinder with circumference $\displaystyle 2\pi x$ and length (R- r) (those are NOT "radii" so the use of "R" and "r" is confusing- better would be the y(x) values of the boundaries) is $\displaystyle 2\pi x(R- r)$

Quote:

If anyone know an easy way to study the Washer, and Shell, or give me some tips, please let me know! J

Thanks!

Draw, at least, two dimensional figures of the region, then draw lines representing the radii of the disks or the edge of the shell and think about what they would look like rotated around the axis.