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- Example: Volume between the functions y=x and y=x 3 from x=0 to 1. These are the functions: Rotated around the x-axis: The disks are now 'washers': And they have the area of an annulus: In our case R = x and r = x 3. In effect this is the same as the disk method, except we subtract one disk from another. And so our integration looks like.
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First of all, you want to be a bit more clear. The shape is revolved about the line x = 4. The boundaries of the shape being revolved are y = x3, y = 0, and x = 2.
If you draw a diagram, you should see that beween using disks or cylindrical shells, the shells are the best choice.
the height of the region being revolved at position x is x3. And that position is a distance 4 - x from the axis. Thus the volume of the shell is
Disk Graph 2 1 3 X 4
dV = 2πrh dx = 2π(4 - x)x3 dx = 2π(4x3 - x4).
![Disk graph 2 1 3 x 4 Disk graph 2 1 3 x 4](https://tutorial.math.lamar.edu/classes/calci/VolumeWithRings_Files/image001.gif)
The interval of integration is [0,2].
Thus the volume will be
2π∫02 (4x3 - x4) dx
= 2π(x4 - x5/5)|02 = 2π(16 - 32/5) = 96π/5
First of all, you want to be a bit more clear. The shape is revolved about the line x = 4. The boundaries of the shape being revolved are y = x3, y = 0, and x = 2.
Disk Graph 2 1 3 X 4
If you draw a diagram, you should see that beween using disks or cylindrical shells, the shells are the best choice.
the height of the region being revolved at position x is x3. And that position is a distance 4 - x from the axis. Thus the volume of the shell is
dV = 2πrh dx = 2π(4 - x)x3 dx = 2π(4x3 - x4).
The interval of integration is [0,2].
Thus the volume will be
2π∫02 (4x3 - x4) dx
Disk Graph 2 1 3 X 48
Disk Graph 2 1 3 X 4 5
= 2π(x4 - x5/5)|02 = 2π(16 - 32/5) = 96π/5