Integral using disk/washer method

In summary, the conversation discusses finding the volume of a solid generated by a given function and bounded by two lines, which is then revolved around a given line. The correct method to use is determined to be the washer method, with the outer radius of the disk being the distance between the two given lines and the inner radius being the distance between the curve and the given line. The conversation also mentions struggling with understanding and practicing this method, and requests additional resources for learning and practicing it.
  • #1
icesalmon
270
13

Homework Statement



Find the volume of the solid generated by y = [itex]\sqrt{x}[/itex] bounded by y = 0, and x = 3
and revolved about the line x = 3

Homework Equations



volume generated by a disk/washer about the y-axis.


The Attempt at a Solution



I have pi(integral from 0 -> [itex]\sqrt{3}[/itex] of (y2)2dy)

I'm thinking maybe my bounds change? I'm also pretty sure I don't use the Washer Method because I don't have an inner radius, I have a 3 dimensional half circle lying on top of the x-axis.
 
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  • #2
Did you pay attention to the "revolved about the line x = 3" part? You aren't revolving around x=0. There is an inner radius.
 
  • #3
I got the answer while playing around with the function, I did pay attention to my axis of revolution, but I don't see how it cuts out anything to be taken away; I don't see why my integrand is ( 3 - y^2)^2 my radius is 3 but if I draw a slice perpendicular to my axis of revolution, x = 3, I don't see it cutting anything out after you revolve it around the line.
 
  • #4
icesalmon said:
I got the answer while playing around with the function, I did pay attention to my axis of revolution, but I don't see how it cuts out anything to be taken away; I don't see why my integrand is ( 3 - y^2)^2 my radius is 3 but if I draw a slice perpendicular to my axis of revolution, x = 3, I don't see it cutting anything out after you revolve it around the line.

The outer radius is 3, right? It's the distance between x=0 and x=3. The inner radius is the distance between the curve and x=3. It looks like a washer to me, not a disk.
 
  • #5
Okay, I have an additional qualm with another problem I feel like asking. I have to integrate y = 2(x^2) bounded by y = 0 and x =2 and I get my volume = 0, how is this at all possible? That my volume is zero, doesn't that mean that one of my representitive elements are zero? By representitive elements, I'm attempting to say height, width, or length.

nevermind, I can see that the volume carved out by my function and the left over inner radius volume are equal. Or am I wrong?
 
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  • #6
Dick said:
The outer radius is 3, right? It's the distance between x=0 and x=3. The inner radius is the distance between the curve and x=3. It looks like a washer to me, not a disk.
No, it's a disk. You seem to be thinking about rotating the area between x= 0 and the curve around x= 3. The problem talks about the region bounded by x= y2, y= 0, and x= 3, rotated around x= 3. For a given y, that is a disk having as radius, the line from x=y2 to x= 3, a length of 3- = 3 y2. The area of such a disk would be [itex]\pi (3- y^2)^2[/itex]. The volume of such a disk, of thickness dy, would be [itex]\pi (3- y^2)^2 dy[/itex]. y goes from 0 up to [itex]\sqrt{3}[/itex].
 
  • #7
Okay, well I'm floundering with this and I want to move on to try my hand at shell method and other applications of the integral, work, hydrostatics, centers of mass/moments of inertia. And while they all utilize the same style of integration, I can't help but feel like they might help me see whatever it is I'm not seeing with the washer method. Should I just stay where I am and continue with Washers, I'm not really sure but I'm chewing up all my time on this. I've reviewed the section, reading through it and I've looked at a lot of examples on the internet and youtube which isn't helping all too much. Any additional information out there might be helpful, if anyone knows.
 
  • #8
icesalmon said:
Okay, well I'm floundering with this and I want to move on to try my hand at shell method and other applications of the integral, work, hydrostatics, centers of mass/moments of inertia. And while they all utilize the same style of integration, I can't help but feel like they might help me see whatever it is I'm not seeing with the washer method. Should I just stay where I am and continue with Washers, I'm not really sure but I'm chewing up all my time on this. I've reviewed the section, reading through it and I've looked at a lot of examples on the internet and youtube which isn't helping all too much. Any additional information out there might be helpful, if anyone knows.

Halls is right. I was wrong. I was reading 'y=0' as 'y axis'. That's wrong. It is a disk. Pay attention to Halls post, not mine.
 
  • #9
I'm clear on this example, really I do understand it. I moved on from that problem and I'm just getting maybe 15% of the problems I've answered actually correct, it's just something I'm not seeing or phrasing correctly. And instead of just having someone solve all my problems, I want to utilize external sources to learn more because my book isn't helping and everything else I've tried isn't working. I just need something additional, do you know where I can read about the washer method, all I want to practice is setting up my inner and outter radii. My book and all these internet sources, the ones I've tried so far, do all the same examples. They aren't helping me. Do you know where I can try more of these or get step-by-step walk throughs on more difficult setups?

I appreciate your efforts with me here, but I'm too stubborn to give up on this.
 
  • #10
icesalmon said:
I'm clear on this example, really I do understand it. I moved on from that problem and I'm just getting maybe 15% of the problems I've answered actually correct, it's just something I'm not seeing or phrasing correctly. And instead of just having someone solve all my problems, I want to utilize external sources to learn more because my book isn't helping and everything else I've tried isn't working. I just need something additional, do you know where I can read about the washer method, all I want to practice is setting up my inner and outter radii. My book and all these internet sources, the ones I've tried so far, do all the same examples. They aren't helping me. Do you know where I can try more of these or get step-by-step walk throughs on more difficult setups?

I appreciate your efforts with me here, but I'm too stubborn to give up on this.

I would say post an example of a problem you are having trouble with here. Probably not everyone will give as bad an advice as I did.
 
  • #11
alright sure, y= 3/(1+x) bounded by y = 0 x = 0 and x = 3, this really shapes my region quite nicely. Integrate about the line y = 4, I have my outer radius as R(x) = 4, the total distance from my axis of revolution to the x-axis, the lower bound of my function y = 3/(1+x). I see my inner radius as the area in between y =4 and y = 3/(1+x) should I integrate to find the area between y = 4 and y = 3/(1+x) to find my inner radius and then subtract that away from the Outer Radius?
 
  • #12
icesalmon said:
alright sure, y= 3/(1+x) bounded by y = 0 x = 0 and x = 3, this really shapes my region quite nicely. Integrate about the line y = 4, I have my outer radius as R(x) = 4, the total distance from my axis of revolution to the x-axis, the lower bound of my function y = 3/(1+x). I see my inner radius as the area in between y =4 and y = 3/(1+x) should I integrate to find the area between y = 4 and y = 3/(1+x) to find my inner radius and then subtract that away from the Outer Radius?

Mmm. No, that doesn't sound right. The outer radius is 4, sure, The inner radius isn't the area between anything. The inner radius is 4-3/(1+x), isn't it? It's the distance between y=4 and y=3/(1+x). I would just integrate pi*(outer^2-inner^2) dx.
 
  • #13
Wouldn't that just mean the area under y = 4 minus the area under y = 3/(1+x). Don't I want the area under 3/(1+x)? Which would be the area under y = 4 minus the area in between y = 4 and y = 3/(1+x)

Disregard my previous statement prior to my edit, if you got to it. I was mistaken.
 
  • #14
Phew, okay I've gotten 4pi ( 3 - ln(4) ) I don't think that's right... what are you getting, using your outer radius as 4 and your inner radius as 4 - 3/(1+x) ?
 
  • #15
The answer IS pi(48ln(2) - (27/4). But (4(1+x) - 3)/(1+x) is NOT 1. The integrand isn't 15 either. From what you've shown I'm still not sure where you are going wrong. Can you show the details? If I integrate pi*(4^2-(4-3/(1+x))^2) I definitely get the correct answer. What are you doing?
 
  • #16
I forgot to square my radii, I'm in the process of doing that now. What were your limits of integration?

For my integrand, after squaring and simplifying and before I do a u-substitution, I have 24(x) + 15 / ( x^2 + 2x +1 )
 
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  • #17
icesalmon said:
I forgot to square my radii, I'm in the process of doing that now. What were your limits of integration?

x goes from 0 to 3, I would hope you would be using those limits.
 
  • #18
well if I do a u-substitution, I normally change my limits and don't change them back.
So after I simplify I get for my integrand, 24(x) + 15 / ( x^2 + 2x + 1 ) I let u = x^2 + 2x + 1 so that du = ( 2x + 2 ) dx and now dx = du / ( 2x + 2 ) as x: 0 -> 3 u: 1 -> 16 and I am left with something I can't work with.
 
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  • #19
icesalmon said:
well if I do a u-substitution, I normally change my limits and don't change them back.
So after I simplify I get for my integrand, 24(x) + 15 / ( x^2 + 2x + 1 ) I let u = x^2 + 2x + 1 so that du = ( 2x + 2 ) dx and now dx = du / ( 2x + 2 ) as x: 0 -> 3 u: 1 -> 16 and I am left with something I can't work with.

It would have been easier to work with if you had done less simplification. Try working from 24/(x+1)-9/(x+1)^2. Split the integral into two parts.
 
  • #20
alright, why can't I get the same answer using what I had? I don't understand the difference between simplifying and splitting up my integrals because I tried what you had here and it feels even more complicated than what I had before.
 
  • #21
icesalmon said:
alright, why can't I get the same answer using what I had? I don't understand the difference between simplifying and splitting up my integrals because I tried what you had here and it feels even more complicated than what I had before.

To get an answer from (24x + 15) / ( x^2 + 2x + 1 ) write it as (24x+24-9)/(x+1)^2= (24x+24)/(x+1)^2-9/(x+1)^2. You need to split it into two integrals that you do in different ways. The first one is a substitution and log, the second is a substitution and power law.
 
  • #22
http://calcchat.tdlc.com/free_solutions/main.html
Text: Larson 9e
Chapter: 7
Section: 2
Exercise: 11 B and C
I'm trying to compare the shaded regions here

For B it says the Outer Radius is 3 and the Inner Radius is y^2 so they subtract the area under x = y^2 from the distance of 3, fair enough.

For C it says the Outer Radius is (3-y^2), the distance of 3 back to the start of the gray region ( the area under x = y^2 ). This is crazy, how in the world can they be finding the volume of the gray region in both exercises if they are subtracting it away from the distance of 3 in Exercise B, they would be finding the volume of the white area for Exercise B.
 
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Related to Integral using disk/washer method

1. What is the disk/washer method for calculating integrals?

The disk/washer method is a technique used to find the volume of a solid of revolution by integrating cross-sectional areas. It involves slicing the solid into thin discs or washers, finding the area of each disc or washer, and then integrating those areas to find the total volume.

2. When should I use the disk method versus the washer method?

The disk method is used when the cross-sections of the solid are circular, while the washer method is used when the cross-sections are annular (donut-shaped). This means that if the solid has a hole in the middle, the washer method should be used. Otherwise, the disk method can be used.

3. How do I set up the integral for the disk/washer method?

The integral for the disk/washer method is typically set up as ∫(base to top) π(radius)^2 dx or ∫(base to top) π(outer radius)^2 - π(inner radius)^2 dx, depending on whether the disk or washer method is being used. The base and top refer to the limits of integration, and the radius can be found by solving for the distance from the axis of revolution to the function being rotated.

4. Can the disk/washer method be used for non-circular cross-sections?

No, the disk/washer method can only be used for circular or annular cross-sections. If the cross-sections are not circular, other methods such as the shell method or the method of cylindrical shells can be used instead.

5. Are there any common mistakes to watch out for when using the disk/washer method?

One common mistake is forgetting to square the radius when setting up the integral. Another mistake is using the wrong limits of integration, especially when the function being rotated intersects the axis of revolution. It is important to carefully visualize the cross-sections and the limits of integration to avoid these errors.

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