Answer:
For revolution about the y-axis:
- shell: dV = 2π·x·(f2(x) -f1(x))·dx
- disk: dV = π·(f2(y)^2 -f1(y)^2)·dy
For revolution about the x-axis, swap x and y.
Step-by-step explanation:
The methods for computing the volume of revolution of a plane figure are ...
The name of the method describes the shape of the differential of volume.
__
<u><em>Cylindrical shell</em></u>
For revolution about the y-axis, the differential of volume is a cylindrical shell of radius x, thickness dx, and height y2 -y1, where y2 = f2(x) and y1 = f1(x). (y2 > y1) Integration is over x.
dV = 2π·x·(f2(x) -f1(x))·dx
For revolution about the x-axis, the variables x and y are interchanged.
__
<em><u>Disk</u></em>
For revolution about the y-axis, the differential of volume is a donut or disk of inner radius f1(y) = x1 and outer radius f2(y) = x2 (x1 ≤ x2). The thickness is dy. Integration is over y.
dV = π·(f2(y)^2 -f1(y)^2)·dy
For revolution about the x-axis, the variables x and y are interchanged.
__
The preferred method is the one that gives simple integrable functions over a single range of integration. For some geometries, the integration range must be split, because the shape of the differential volume cannot be described by a simple rectangle whose length is the difference of functions of the integration variable.
When the differential of volume can be drawn either way, the choice of methods may be arbitrary. I usually choose the method that gives the simplest integrand.
_____
<em>Additional comment</em>
When the axis of revolution is something other than a coordinate axis, the radius of revolution becomes a different function than a simple x (or y). For example, if the axis of revolution is x=a, the radius of the cylindrical shell will be a-x (for x ≤ a) or x-a (for a ≤ x).
Here's an example of the use of the disk method.
brainly.com/question/17446389
