Disk Washer Method: Vol of Solid Generated by y=1/sqrt(x)

  • Thread starter Justabeginner
  • Start date
  • Tags
    Disk Method
In summary: I really appreciate all your help!In summary, we need to find the volume of the solid generated by revolving the region bounded by y=1/√x and y=0 for 1≤x≤2 about the line y=-1. To do this, we use the washer method, which involves subtracting the inner radius (1) from the outer radius (1/√x + 1) and squaring this difference. We then integrate this expression with respect to x, from 1 to 2, and multiply by π. This gives us a final volume of π(ln2 + 4√2 - 4).
  • #1
Justabeginner
309
1

Homework Statement


Find the volume of the solid generated by revolving the region bounded by y=1/sqrt(x) and y=0 for 1≤x≤2 about the line y=-1


Homework Equations


Disk Method? ∏ * ∫[f(x)]^2 dx


The Attempt at a Solution


A and B (lower and upper limits will be 1 and 2)?
∏ * ∫(1/sqrt(x)^2 dx
∏ * ∫1/x dx
∏ * ln x
Since limits are 2 and 1, my final answer is: ∏ * ln 2

Is this right? Thanks.
 
Physics news on Phys.org
  • #2
Justabeginner said:

Homework Statement


Find the volume of the solid generated by revolving the region bounded by y=1/sqrt(x) and y=0 for 1≤x≤2 about the line y=-1


Homework Equations


Disk Method? ∏ * ∫[f(x)]^2 dx
This formula is applicable only if the rotation is around the x-axis.

The formula would be better written as ∏ * ∫[radius]2dx
Justabeginner said:

The Attempt at a Solution


A and B (lower and upper limits will be 1 and 2)?
∏ * ∫(1/sqrt(x)^2 dx
∏ * ∫1/x dx
∏ * ln x
Since limits are 2 and 1, my final answer is: ∏ * ln 2

Is this right? Thanks.

No. Your integrand doesn't reflect the fact that the region is being rotated around the line y = -1. You can't use a disk in this case - you need to use a washer if you use vertical strips, or you could use shells if you go with horizontal strips.
 
  • #3
Would the correct way to set it up then be

∏ * ∫([1/sqrt(x)] + 1)^2) ?

And follow up with:
∏* ∫[1/x + 2/sqrt(x) + 1]
∏ * ([ln x] + 4sqrt(x) + x) ?

Thank you so much for all your help.
 
  • #4
Justabeginner said:
Would the correct way to set it up then be

∏ * ∫([1/sqrt(x)] + 1)^2) ?
No, this isn't set up correctly. You're using disks when you should be using washers.

For a washer, the volume is ##\pi [(R_{outer})^2 - (R_{inner})^2]Δx##.
Justabeginner said:
And follow up with:
∏* ∫[1/x + 2/sqrt(x) + 1]
∏ * ([ln x] + 4sqrt(x) + x) ?

Thank you so much for all your help.
 
  • #5
But I just don't understand, even after drawing this diagram, how there's an inner AND an outer radius? If you could explain that to me, I'd appreciate it.
 
  • #6
The region to be rotated is roughly trapezoidal in shape. Its bottom edge is the x-axis, the left and right edges are the lines x = 1 and x = 2, and the top edge is the curve y = 1/√x.

When the region is rotated around the line y = -1, you get a kind of ring-shaped solid that is hollow. The inner radius is 1, and the outer radius extends from the line y = -1 to the curve.
 
  • #7
Thank you! I am horrible at visualizing things in 3D and your description made it so much clearer.

But how would I determine the length of the outer radius? I know that the points on the curve are (1, 1) and (2, sqrt(2)/2)- how would that help me in this problem, or would it not be of any use?
 
  • #8
Points on the curve are of the form (x, 1/√x). The upper end of the outer radius is at this point, and the lower end is at (x, -1). How far apart are these points?
 
  • #9
They are (1/sqrt(x) + 1) units apart?
 
  • #10
Yes, so this is the outer radius. Don't forget, it needs to be squared.
 
  • #11
So what I get now is:
∏ *∫{[(1/sqrt(x) + 1)^2] - (1^2)} * Δx (Isn't delta x = 1 since x=1 and x=2 are bounding the region?)
∏ * ∫(1/x) + 2/sqrt(x) + 1 - (1)
∏ * ∫(1/x) + 2/sqrt(x)
∏ * (ln x + 4sqrt(x))
And since x=1 and x=2 are the left and right bounds, I'd plug that into solve for the final volume?

∏* [(ln 2 + 4*sqrt(2) - (ln 1 + 4 sqrt(1)]
∏ * [ln 2 + 4sqrt(2) - 4]

Is this correct?
 
  • #12
Justabeginner said:
So what I get now is:
∏ *∫{[(1/sqrt(x) + 1)^2] - (1^2)} * Δx (Isn't delta x = 1 since x=1 and x=2 are bounding the region?)
No, that's not what Δx means. It corresponds loosely with dx in the integral.

I don't have time to look at this, right now, but will later on if nobody else jumps in.
Justabeginner said:
∏ * ∫(1/x) + 2/sqrt(x) + 1 - (1)
∏ * ∫(1/x) + 2/sqrt(x)
∏ * (ln x + 4sqrt(x))
And since x=1 and x=2 are the left and right bounds, I'd plug that into solve for the final volume?

∏* [(ln 2 + 4*sqrt(2) - (ln 1 + 4 sqrt(1)]
∏ * [ln 2 + 4sqrt(2) - 4]

Is this correct?
 
  • #13
Okay thank you so much! (Yes I just realized, delta x can also be used for dx- I don't know why I even said that -_-)
 
  • #14
Justabeginner said:
So what I get now is:
∏ *∫{[(1/sqrt(x) + 1)^2] - (1^2)} * Δx (Isn't delta x = 1 since x=1 and x=2 are bounding the region?)
∏ * ∫(1/x) + 2/sqrt(x) + 1 - (1)
∏ * ∫(1/x) + 2/sqrt(x)
∏ * (ln x + 4sqrt(x))
And since x=1 and x=2 are the left and right bounds, I'd plug that into solve for the final volume?

∏* [(ln 2 + 4*sqrt(2) - (ln 1 + 4 sqrt(1)]
∏ * [ln 2 + 4sqrt(2) - 4]

Is this correct?
Yes, that's what I get, too.
 
  • #15
Thank you very much sir!
 

FAQ: Disk Washer Method: Vol of Solid Generated by y=1/sqrt(x)

What is the Disk Washer Method?

The Disk Washer Method is a mathematical technique that is used to calculate the volume of a solid generated by rotating a curve around a specific axis. It involves breaking down the solid into infinitely thin disks and finding the volume of each disk.

How is the Disk Washer Method applied?

In order to use the Disk Washer Method, the axis of rotation must be known and the curve must be a function of x or y. The equation for the volume of the solid is then found by integrating the function with respect to the axis of rotation.

What is the equation for the volume of a solid generated by y=1/sqrt(x)?

The equation for the volume of a solid generated by y=1/sqrt(x) is V=π∫(a,b) y^2 dx, where a and b are the x-values of the points of intersection between the curve and the axis of rotation.

What are the limitations of the Disk Washer Method?

The Disk Washer Method can only be used for solids generated by rotating a curve around a horizontal or vertical axis. It also cannot be used if the curve intersects the axis of rotation at more than two points.

How is the Disk Washer Method different from the Shell Method?

The Disk Washer Method and the Shell Method are two different techniques used to calculate the volume of a solid of revolution. While the Disk Washer Method involves using disks, the Shell Method involves using cylindrical shells to find the volume. The Shell Method is typically used when the curve is a function of y, while the Disk Washer Method is used when the curve is a function of x.

Back
Top