Calculating Volume Using the Disk Method for Revolving Regions

[jsun2015]

Homework Statement

Find the volume of the solid generated by revolving the region bounded by the parabola y=x^2 and the line y=1 about the line y=1

Homework Equations

V= integral of pi*r^2 from a to b with respect to variable "x"

The Attempt at a Solution

pi(integral of 1-(x^2-1)^2 from 0 to 1 dx)
but The answer is 15pi/16

 

[Dick]

Homework Statement

Find the volume of the solid generated by revolving the region bounded by the parabola y=x^2 and the line y=1 about the line y=1

Homework Equations

V= integral of pi*r^2 from a to b with respect to variable "x"

The Attempt at a Solution

pi(integral of 1-(x^2-1)^2 from 0 to 1 dx)
but The answer is 15pi/16

Why do you think the 'r^2' part in your volume equation is 1-(x^2-1)^2 and why do you think the limits to the integration are 0 to 1?

 

[jsun2015]

r^2
1 represents outer radius (the larger radius)
x^2-1 represents the inner radius (smaller radius)

limits to integration
the radius of the sphere is on the region x= 0 to 1

 

[jsun2015]

Why do you think the 'r^2' part in your volume equation is 1-(x^2-1)^2 and why do you think the limits to the integration are 0 to 1?

r^2
1 represents outer radius (the larger radius)
x^2-1 represents the inner radius (smaller radius)

limits to integration
the radius of the sphere is on the region x= 0 to 1

 

[Dick]

r^2
1 represents outer radius (the larger radius)
x^2-1 represents the inner radius (smaller radius)

limits to integration
the radius of the sphere is on the region x= 0 to 1

That's confusing me. The formula you gave was actually for the method of disks, which is what I would use here. If you are rotating the region between y=x^2 and y=1 around y=1, then the inner radius is 0, isn't it? And I don't see why you are putting one of the limits to 0. y=x^2 and y=1 cross at x=1 and x=(-1), yes?

 

[jsun2015]

Yes. Yes. I came up with my original answer because at the outer region 1= radius so we have one and at the inner region 0 = radius as (1^2-1)^2=0

I have only studied the washer method.

 

[jsun2015]

That's confusing me. The formula you gave was actually for the method of disks, which is what I would use here. If you are rotating the region between y=x^2 and y=1 around y=1, then the inner radius is 0, isn't it? And I don't see why you are putting one of the limits to 0. y=x^2 and y=1 cross at x=1 and x=(-1), yes?

Yes. Yes. I came up with my original answer because at the outer region 1= radius so we have one and at the inner region 0 = radius as (1^2-1)^2=0

I have only studied the washer method.

 

[Dick]

Yes. Yes. I came up with my original answer because at the outer region 1= radius so we have one and at the inner region 0 = radius as (1^2-1)^2=0

I have only studied the washer method.

The disk method is the same as the washer method with an inner radius of 0. Isn't the outer radius ALWAYS 1-x^2? And the inner radius 0?

 

[jsun2015]

I correct myself, I have studied disc method. Yes. yes.

 

[jsun2015]

The disk method is the same as the washer method with an inner radius of 0. Isn't the outer radius ALWAYS 1-x^2? And the inner radius 0?

I correct myself, I have studied disc method. Yes. yes.