Integration - Cylindrical Shells and Washer method

I

Idealistic

Junior Member
Joined
Sep 7, 2007
Messages
97
I have the function y = x[sup:3gbsot7d]3[/sup:3gbsot7d] where x is [0, 1], I have y = 1 where x is [0, 1], and the curve is rotated about y = 1. They want to know the area under y = x[sup:3gbsot7d]3[/sup:3gbsot7d] using the washer method and cylindrical shells.

the formula for cylindrical shells is, integral of 2pi(y)xdx,

but because the shells have to be parallel to the axis of rotaion, I have to express x in terms of y.

The function becomes x = y[sup:3gbsot7d]1/3[/sup:3gbsot7d].

I have:

integral of 2pi(1)ydy - 2pi(y[sup:3gbsot7d]1/3[/sup:3gbsot7d])ydy (from y is [0, 1])- this represent the total area subtract the area given from x = y[sup:3gbsot7d]1/3[/sup:3gbsot7d] where y is [0, 1] which gives the area under y = x[sup:3gbsot7d]1/3[/sup:3gbsot7d].

piy[sup:3gbsot7d]2[/sup:3gbsot7d] - 6/7(piy[sup:3gbsot7d]7/3[/sup:3gbsot7d]) from y is [0, 1]

pi - 6pi/7 = pi/7

But when I do the washer method, my washers are perpendicular to the axis of rotation (y = 1) so my equation is:

integral of pi(1)dx - pi(1 - x[sup:3gbsot7d]3[/sup:3gbsot7d])[sup:3gbsot7d]2[/sup:3gbsot7d]dx (from x is [0, 1]) - comes from (integral of piy[sup:3gbsot7d]2[/sup:3gbsot7d]dx) but my radius is going to be 1 minus x[sup:3gbsot7d]3[/sup:3gbsot7d] which is the vertical distance from y = 1 to y = x[sup:3gbsot7d]3[/sup:3gbsot7d].

pix - (pix - (1/2)pix[sup:3gbsot7d]4[/sup:3gbsot7d] + (1/7)pix[sup:3gbsot7d]7[/sup:3gbsot7d]) from x is [0, 1]

pi/2 - pi/7 = 5pi/14

The problem is i'm not getting the same answer for both.
 
You're revolving about y=1, not one of the axes.

You have \(\displaystyle y=x^{3}\) revolved about y=1, with x limits 0 to 1?.

shells:

\(\displaystyle 2\pi\int_{0}^{1}(1-y)y^{\frac{1}{3}}dy\)

washers:

\(\displaystyle {\pi}\int_{0}^{1}(x^{3}-1)^{2}dx\)
 
but for the washers, the axis of rotation is y = 1, wouldnt the radius equal 1 - x[sup:2ti9iarc]3[/sup:2ti9iarc]? the area between y = 1 and y = x[sup:2ti9iarc]3[/sup:2ti9iarc] and then subtracted from integral of pi(1)[sup:2ti9iarc]2[/sup:2ti9iarc] to give the area under y = x[sup:2ti9iarc]3[/sup:2ti9iarc]?

and with the shell method, why is it (1 - y)?
 
Here's a animated graph so you can see. See the line y=1?.
 

Attachments

  • solid1.gif
    solid1.gif
    167.8 KB · Views: 1,052
i see, but the area I want Is the bowl shape that would cover that bullet shape, the area between y = 0 and y = x[sup:1sgl0iw7]3[/sup:1sgl0iw7], its just revolved around y = 1. The result shape would look like a bowl, or more so a half pipe. Which is why I though radius for each washer was going to be 1 - x[sup:1sgl0iw7]3[/sup:1sgl0iw7], and subtracting that washer area (the bullet shape you showed me) from the whole "cann" to get the area under x[sup:1sgl0iw7]3[/sup:1sgl0iw7] which looks like a half pipe.
 
That is how I interpreted your problem. Perhaps someone else has another. Revolving y=x^3 about y=1 between x=0 and x=1. That is its graph
 


I suspect that you've confused the heck out of Cody; I know that you've confused the heck out of me.

You keep posting that you're looking for areas, but, if you're using shells and washers, then you are not looking for any areas.

You're looking for a volume.

Also, it's not the curve that's being rotated; it's a region that's being rotated. We call this resulting object a "solid of revolution".

This particular solid of revolution looks neither like a bowl nor half a pipe. And you are definitely not looking for an area that "covers the bullet shape". That would be a surface area, which has nothing to do with shells and washers.

This object looks like a solid cylinder with a bullet-shaped indentation in the direction of its axis.

I get a volume of (5/14) * Pi cubic units using both methods.

Regarding washers: do you know?

(1 - x^3)^2 = (x^3 - 1)^2

Since both expressions are the same, we can use either one for the diameter of the hole. Your work on the washer method looks good to me, except for integrating a constant! (I don't think that's necessary.)

Clearly, a solid cylinder with radius 1 and height 1 has a volume of Pi.

Therefore, we only need to subtract the volume of the indentation from the volume Pi. I used the following.

\(\displaystyle V \; = \; \pi - \pi \cdot \int_{0}^{1}(1 - x^{3})^{2} \; dx\)

Regarding shells: do you know?

Each shell can be "unrolled" to form a rectangular solid.

The length is 2 * Pi * (1 - y).

The width is dy.

The height is 1 - y^(1/3).

It looks like your mistake in the shell method is that you wrote the length as 2 * Pi * y, instead.

I used the following.

\(\displaystyle V \; = \; 2 \cdot \pi \cdot \int_{0}^{1}(1-y^{\frac{1}{3}}) \cdot (1 - y) \; dy\)

Next time you post, please use the Preview button and proofread what you've typed. Many of your descriptions in this discussion are really off the mark.

MY EDIT: Corrected misstatement regarding relation between area and volume
 
galactus said:
You are better than I at interpreting …
Click to expand...


I don't think so!

I was just willing to spend 40 minutes fighting my way through the jungle of confusion.

 
Well, you done a fine job deciphering it. :D

I used Maple to create that animated diagram. It worked nice.
 


When I animate with Maple V, the speed is much slower.

Is the frame-rate adjustable? My discounted version of Maple has neither a printed manual nor help files.

I was wondering if the higher speed is a function of posting it here. (I don't know anything about animated .GIF files. I've only worked with Maple animation within Maple worksheets; I've never saved one.)

 
Yes, you can increase or decrease the speed. I left mine a little too fast, but it's OK.

I have Maple 11. I create the graph, Export it as a gif and Save. Then, upload it to the site. It is actually very easy.

Maple 11 has a Volumes of Revolution tutorial where it creates the graph and you can animate it if you wish.

Gotta go. Be back tonight....probably.
 
Hey thanks for helping me out. I'm sorry for butchering the question like that but it was honestly the best I could do at describing the volume I wanted. Next time i'll be sure to use appropriate wording. As far as the math goes, my washer method was okay but I see my mistake in the shell method. I forgot to account for the height as 1 - y. Thanks for pointing out that (1 - x[sup:29hoths1]3[/sup:29hoths1]) and (x[sup:29hoths1]3[/sup:29hoths1] - 1) are the same thing lol I think I was tired when I missed that one.
 
Idealistic said:
… Thanks for pointing out that (1 - x[sup:22atlr1d]3[/sup:22atlr1d])[sup:22atlr1d]2[/sup:22atlr1d] and (x[sup:22atlr1d]3[/sup:22atlr1d] - 1)[sup:22atlr1d]2[/sup:22atlr1d] are the same thing …
Click to expand...


You are welcome.

(Do your best to get at least 8 hours of sleep every 24 hours, if you want to help enable your brain to solidify each day's new connections.)

 
You must log in or register to reply here.
Top