It is a good question about disk method Can you help me to solve

In summary, the disk method will find the radius of the disk if x0 is at the point of intersection of y=x and y=-x02, and x0 is the distance from the point of intersection to the disk center.
  • #1
erogol
14
0
Disk method question

Use the disk method to find the volume of the solid obtained by rotating the region bounded by y=x^2 and y=x about the line y=x.
 
Last edited:
Physics news on Phys.org
  • #2


From a point (x0, x02) on y= x2, find the line through that point perpendicular to y= x: y= -(x- x0)+ x02. Find where that line crosses y= x and calculate the distance from (x0, x02) to the point of intersection, as a function of x0. That will be the radius of the disk. Since the length of the segment y= x for y from x0 to x0+ [itex]\Delta x[/itex] is [itex]\sqrt{2}\Delta x[/itex] , in place of dx you will want to use [itex]\sqrt{2}dx[/itex] as the "thickness" of the disk when you integrate.

Alternatively, you can rotate the graph through -45 degrees to make the line y= x rotate to y= 0. Since, in this case, x= x' cos 45- y'sin 45, y= x' sin 45+ y' cos 45 or [itex]x= \frac{\sqrt{2}}{2}x'+ \frac{\sqrt{2}}{2}y'[/itex], [itex]y= \frac{\sqrt{2}}{2}x'- \frac{\sqrt{2}}{2}y'[/itex]. Replacing x and y in y= x with those gives [itex]y= \frac{\sqrt{2}}{2}x'- \frac{\sqrt{2}}{2}y'= \frac{\sqrt{2}}{2}x'+ \frac{\sqrt{2}}{2}y'[/itex] which reduces to [itex]\sqrt{2}y'= 0[/itex] or y'= 0. Replacing x and y in [itex]y= x^2[/itex] gives the equation of that curve in the x'y' coordinate system and then rotate around the x' axis.
 
  • #3


HallsofIvy said:
From a point (x0, x02) on y= x2, find the line through that point perpendicular to y= x: y= -(x- x0)+ x02. Find where that line crosses y= x and calculate the distance from (x0, x02) to the point of intersection, as a function of x0. That will be the radius of the disk. Since the length of the segment y= x for y from x0 to x0+ [itex]\Delta x[/itex] is [itex]\sqrt{2}\Delta x[/itex] , in place of dx you will want to use [itex]\sqrt{2}dx[/itex] as the "thickness" of the disk when you integrate.

Alternatively, you can rotate the graph through -45 degrees to make the line y= x rotate to y= 0. Since, in this case, x= x' cos 45- y'sin 45, y= x' sin 45+ y' cos 45 or [itex]x= \frac{\sqrt{2}}{2}x'+ \frac{\sqrt{2}}{2}y'[/itex], [itex]y= \frac{\sqrt{2}}{2}x'- \frac{\sqrt{2}}{2}y'[/itex]. Replacing x and y in y= x with those gives [itex]y= \frac{\sqrt{2}}{2}x'- \frac{\sqrt{2}}{2}y'= \frac{\sqrt{2}}{2}x'+ \frac{\sqrt{2}}{2}y'[/itex] which reduces to [itex]\sqrt{2}y'= 0[/itex] or y'= 0. Replacing x and y in [itex]y= x^2[/itex] gives the equation of that curve in the x'y' coordinate system and then rotate around the x' axis.


I cannot understand that. Can you write integral of first solution, please. Thank you.
 

1. What is the disk method?

The disk method is a mathematical technique used to calculate the volume of a solid shape formed by rotating a two-dimensional region around a specific axis.

2. How do you use the disk method to solve problems?

To use the disk method, you first need to determine the axis of rotation and the boundaries of the two-dimensional region. Then, you can use the formula V = π∫(R(x)^2 - r(x)^2)dx, where R(x) is the outer radius and r(x) is the inner radius.

3. What are the limitations of the disk method?

The disk method can only be used for solids with circular cross sections. It also assumes that the two-dimensional region is revolved around a single axis.

4. Can you provide an example of using the disk method?

Sure, let's say we want to find the volume of a cone with a radius of 2 and a height of 4 using the disk method. We can revolve a semi-circle with a radius of 2 around the y-axis. The outer radius will be 2 and the inner radius will be 0. Then, we can use the formula V = π∫(2^2 - 0^2)dx = 4π.

5. Are there any real-life applications of the disk method?

Yes, the disk method can be used in various fields such as engineering, physics, and architecture. For example, it can be used to calculate the volume of a water tank, the mass of a rotating object, or the amount of material needed to create a curved surface.

Similar threads

Replies
1
Views
1K
Replies
3
Views
2K
Replies
1
Views
2K
Replies
3
Views
1K
  • Calculus
Replies
11
Views
2K
Replies
5
Views
1K
Replies
2
Views
2K
Replies
3
Views
1K
Replies
2
Views
801
Replies
3
Views
219
Back
Top