Find the volume of the solid formed by rotating the region

In summary, the student attempted to solve a homework problem using the "disk method", but realized that the region is not (pi)(2x - [x^2])^2 , but rather the area of the outer circle minus the area of the inner circle. This caused incorrect results, as the integral is not given by the expression he is using.
  • #1
zcabral
30
0

Homework Statement


Find the volume of the solid formed by rotating the region inside the first quadrant enclosed by
y=x^2
y=2x
about the x-axis.



Homework Equations



V= pi* integral [a,b] r^2

The Attempt at a Solution



So i used intergral 0 to 2 (2x-x^2)^2 dx
this is wat i got...
pi*(((2^5)/5)-(4(2^4)/4)+(4(2^3)/3))
its wrong so wat did i do wrong?
 
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  • #2
zcabral said:
V= pi* integral [a,b] r^2

You are using the "disk method", yes? But keep in mind that your disk has a hole in it: the area is not (pi)(2x - [x^2])^2 , but rather the area of the outer circle minus the area of the inner circle.
 
  • #3
but it doesn't technically have a hole that goes all the way through rite? wouldn't that still make it a disk? i know it has a like a big chunk out of it
 
  • #4
zcabral said:
but it doesn't technically have a hole that goes all the way through rite? wouldn't that still make it a disk? i know it has a like a big chunk out of it

Every slice of the volume will have a hole in it, so each slice is a ring (or annulus). The area is going to be the area of the disk defined by the outer radius minus the hole which has the inner radius. That difference is not given by the expression you are using. How should it be written?
 
  • #5
ok so this is what i did now...
pi [integral 0 to 2 (2x)^2-(2x-x^2)^2]
=
pi*-x^5/5 + 4x^4/4 |[0,2]
 
  • #6
zcabral said:
ok so this is what i did now...
pi [integral 0 to 2 (2x)^2-(2x-x^2)^2]
=
pi*-x^5/5 + 4x^4/4 |[0,2]

The "outer radius" of each ring is provided by the curve y = 2x , since that is farther from the x-axis than y = x^2 on the interval [0,2]. So the area of the full disc is (pi)[(2x)^2] . The "hole" is bounded by the curve y = x^2 , so that is the "inner radius" of the ring we want to include in our integration. So the area of the hole in each disc is (pi)[(x^2)^2].

The integral will then be (pi) [integral 0 to 2 of {(2x)^2} - {(x^2)^2} dx]

= (pi) [integral 0 to 2 of (4x^2 - x^4) dx].
 
  • #7
took a stab at it. i think yr on the right track, this was more so to see if i could still do it lawl.

http://i21.photobucket.com/albums/b277/riceboy89/problem.jpg

(im going to d/l latex)
 
Last edited by a moderator:

1. What is the formula for finding the volume of a solid formed by rotating a region?

The formula for finding the volume of a solid formed by rotating a region is V = π∫ab (f(x))2 dx, where f(x) is the function representing the cross-sectional area of the solid and a and b are the limits of integration.

2. How do you determine the limits of integration for finding the volume of a solid formed by rotating a region?

The limits of integration are determined by the boundaries of the region being rotated. If the region is bounded by the x-axis, the limits will be the x-coordinates of the endpoints. If the region is bounded by two functions, the limits will be the x-values where the two functions intersect.

3. What is the difference between finding the volume of a solid formed by rotating a region about the x-axis and the y-axis?

The main difference is the function used to represent the cross-sectional area of the solid. When rotating about the x-axis, the function will be in terms of x, while rotating about the y-axis, the function will be in terms of y.

4. Can you use the disk method to find the volume of a solid formed by rotating a region?

Yes, the disk method can be used to find the volume of a solid formed by rotating a region, as long as the cross-sectional area of the solid is a circle. In this case, the formula would be V = π∫ab (f(x))2 dx, where f(x) is the function representing the radius of the circle.

5. Are there any other methods besides the disk method for finding the volume of a solid formed by rotating a region?

Yes, there are other methods such as the shell method and the washer method. These methods are used when the cross-sectional area of the solid is not a circle. The shell method uses cylindrical shells to find the volume, while the washer method uses a combination of disks and washers.

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