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 AP Calculus (AB)  Solids of Revolution  [ Disk Method | Examples of Solids | Washer Method ] |
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| AP Calculus (AB) Solids of Revolution [ Disk Method | Examples of Solids | Washer Method ] |
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This volume is also exactly what is produced in the following definite integral:
The key is that each y is the radius of a different disk. We can break any interval along the x-axis into subintervals were each subinterval has horizontal top. This is what was done when we solved area problems by finding the definite integral using Riemann sums. Using that technique, we should be able to find the volume generated by revolving any function about the x-axis (not just horizontal lines that make perfect cylinders). To do this we create many subintervals to find a "cross-sectional area" and then revolve this about the x-axis to produce "solids of revolution." By allowing the number of subintervals to approach infinity, we get a formula for volumes of a solid of revolution that uses integration:
Thus, the volume of a closed solid revolved about the x-axis is directly related to the area of the circles formed by the act of revolving the region. To find the volume, we take the the integral, from a to b, of p times the square of function that defines the radius of the region:
This process is illustrated below. Here, f (x) is defined as
The area below this curve and above the x-axis can be approximated using the rectangular approximation method, here using 8-subintervals and the left endpoints to calculate the height of the curve. If each of these rectangles is revolved around the x-axis, the second figure is produced. It looks like as strange stack of disks if various radii.
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Finding the volume of each cylinder in the "stack" of cylinders would be an easy (but dull) task. If the number of subintervals is increased, the process has not changed, only the amount of "number crunching" increases. The approximation of the volume will improve as the number of cylinders increases, which is (hopefully) obivious from the next pair of images:
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If we let the number of subintervals approach infinity, the area of the cross section is found by the definite integral and the volume of the region is found by the following integral:
This volume is visualized by the cross-sectional area of the function f on the interval [0, 2] and that region is revolved around the x-axis, as shown below:
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When the plane region revolved about an axis is the area between two curves, rather than just a single curve, the solid that is formed will look like a donut or a hollowed out pipe of varying thickness. The solid could even be closed at one end, like a pot formed by turning clay on a potter's wheel.
Consider the solid generated by revolving the following area:
When this region is revolved about the x-axis, the solid formed is "donut" shaped:
It bears repeating that the volume of a closed solid revolved about the x-axis is directly related to the area of the circles formed by the act of revolving the region. To find the volume, we take the the integral, from a to b, of p times the square of the function that defines the radius of the region:
In the case of this solid I recommend
imagining the solid from viewpoint of looking down the x-axis
to see the cross-sectional view. From this vantage point, the
method for determining the volume becomes more apparent. The
"circular disks" that comprise the solid are not solid
disks. Each has a hole in it. Thus, if we think of disks of certain
thickness with holes punched in them, we get a stack of "washers."
To find the area of each "washer," the simplest method
is to find the area of the outer / larger disk and subtract the
area of the inner /smaller disk. Algebraically, if the outer
radius is R and the inner is r, then the formula
for the area can be expressed as follows:
Thus, since the radii are changing as x changes, they can be written as functions of x:
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Therefore, with the area of the region between the curves defined by the intersection points at the left and right, the volume of the solid formed by revolving the region about the x-axis is found by evaluating the integral
where a = –1.176501... and b = 1.176501... . The intersection points can be found with the graphing calculator stored as A and B, and with the outer curve as Y1 and the inner as Y2, the volume is approximated with the following:
The AP Exam expects you be be able to use the graphing calculator to approximate integrals such as this. You also must show the "set-up" of the integral to get full credit.
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Matthew C. Whitney
This formula comes
from the method for calculating the volume of a right circular
cylinder, where the base is a circle. The volume of the cylinder
is the area of the base times the height, where the area of the
base is A = pr2 and the volume is V = pr2h.
It helps to picture a cylinder lying on its side, with its axis
of revolution on the x-axis, the base at the origin and
extend the cylinder to the right along the positive x-axis.
In the graph
at the left, each y-value on the curve serves as the radius
of infinitely many, incredibly thin circular disks that get stacked
up to produce the cylindrical solid. In this graph, the blue
line from the x-axis to the line y = 2 is any radius
of a cylinder with radius 2. If the blue line is revolved around
the x-axis, the result is a circular disk with radius
2. The area of each disk is:
.
