Using disk method with respect to y-axis

In summary, the conversation is about finding the volume of a region between y=sqrt(x) and x=4 rotated around the y-axis. The correct method to use is the method of washers, where the inner and outer radii are determined by the boundaries of the region. The mistake made at first was trying to use the disk method, which is only applicable when there is no hole in the region being rotated.
  • #1
Geekchick
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Homework Statement


y=[tex]\sqrt{x}[/tex], x=4, y=0


Homework Equations


[tex]\pi[/tex][tex]\int^{c}_{d}{R(y)}^{2}[/tex]dy



The Attempt at a Solution


I solve for x then pluged it into the formula

[tex]\int[/tex][tex]^{2}_{0}[/tex]{(y[tex]^{2}[/tex])[tex]^{2}[/tex]}

The answer I got was [tex]\frac{32\pi}{5}[/tex]

but my textbook says the answer should be [tex]\frac{128\pi}{5}[/tex]
 
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  • #2
It looks like what they want is the area between y=sqrt(x) and the vertical line x=4 rotated around the y-axis. You'll want to use washers. Or you could subtract your answer from the volume of the cylinder with radius 4 and height 2.
 
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  • #3
Yeah sorry, I thought I put that I was supposed to find the the volume as it rotates around the y-axis. which is what i tried to do by solving y=sqrt(x) for x=y^2 then I used that boundries 0-2 and integrated. How do you suppose I use washers? I thought the problem had to have two functions of x to do that.

Please help this problem has been bothering me since yesterday!
Thanks!
 
  • #4
If you cut the volume at a fixed value of y then you get a washer shaped region (a circle with a circular hole in the middle). The inner radius is where it hits y=sqrt(x) and, as you said, that's y^2. The outer radius is where it hits the line x=4. So the outer radius is 4. To use the method of washers, you integrate pi*(outer radius)^2-pi*(inner radius)^2.
 
  • #5
I got it! thank you so much! I can always depend on the PF members!
 
  • #6
Oh, I was wondering can the disk method ever be used to find the volume around the y-axis? If so how can I tell? thanks!
 
  • #7
Draw a sketch of the region. Do you see what you were doing wrong the first time? You were rotating the region between x=0 (the y-axis) and y=sqrt(x). Disks work fine for that. There's no hole in the middle. But what they wanted was the region between y=sqrt(x) and x=4. Now there's a hole in it and you should use washers. You were rotating the wrong region.
 
  • #8
Ok, so for the volume of the same function about the x-axis the graph essentialliy swings down and around the axis with no gap in between the values of integation 0-4. However, by trying to find the volume of the same function as it swings around the y-axis there is a gap between the values of integration 0-2 which is accounted for by the funtion x=y^2. Right?
 
  • #9
That sounds right. Because in your problem they wanted you to rotate the region OUTSIDE of the function, not inside.
 

Related to Using disk method with respect to y-axis

What is the disk method with respect to y-axis?

The disk method with respect to y-axis is a method used in calculus to find the volume of a solid of revolution. It involves slicing the solid into thin disks perpendicular to the y-axis and then integrating the area of these disks to find the total volume.

When is the disk method with respect to y-axis used?

The disk method with respect to y-axis is used when the solid of revolution has a cross-section that is perpendicular to the y-axis and can be represented by a function in terms of y.

What is the formula for the disk method with respect to y-axis?

The formula for the disk method with respect to y-axis is V = π∫ab (f(y))2 dy, where f(y) is the function that represents the cross-section of the solid, and a and b are the limits of integration along the y-axis.

How is the disk method with respect to y-axis different from the disk method with respect to x-axis?

The main difference between the disk method with respect to y-axis and the disk method with respect to x-axis is the orientation of the cross-sections. In the y-axis method, the cross-sections are perpendicular to the y-axis and have a height of f(y), while in the x-axis method, the cross-sections are perpendicular to the x-axis and have a width of f(x).

Can the disk method with respect to y-axis be used to find the volume of any solid of revolution?

No, the disk method with respect to y-axis can only be used to find the volume of solids of revolution with cross-sections that are perpendicular to the y-axis and can be represented by a function in terms of y. For solids with other orientations or cross-sections that cannot be represented by a function, other methods such as the shell method or the washer method may be used.

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