Find the length of the curve.
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Find the surface area generated when the curve is revolved about the x-axis.
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Suppose that we have a thin distribution of matter, a plate, laid out in the -plane in the shape of some region . If the mass density of the plate varies from point to point, then the determination of the center of mass of the plate requires double integration. If, however, the mass density of the plate is constant throughout , then the center of mass of the plate depends only on the shape of and falls on a point that we call the centroid of . Unless has a very complicated shape, we can calculate the centroid of by ordinary one-variable integration.
We will use two guiding principles to find the centroid of a region . The first is obvious. The second we take from physics.
Principle 1: Symmetry. If the region has an axis of symmetry, then the centroid lies somewhere along that axis. (It follows from Principle 1 that, if the region has a center, then that center is the centroid.)
Principle 2: Additivity. If the region, having area
consists of a finite number of pieces with areas
and centroids
, then
and
We are now ready to bring the techniques of calculus into play. Let's denote
the area of
by
.
The centroid
of
can be obtained from the following formulas:
To derive these formulas we choose a partition
of
This breaks up
into
subintervals. Choosing
as the midpoint of
,
we form the midpoint rectangles
.The
area of
is
and the centroid of
is its center
By Principle 1, the centroid
of the union of all these rectangles satisfies the following equations:
(Here
represents the area of the union of the
rectangles.) As
,
the union of rectangles tends to the shape of
and the equations we just derived tend to the formulas
Find the centroid of the quarter-disc
Solution.
The quarter-disc is symmetric about the line
We know therefore that
Here
Since
The centroid of the quarter-disc is the point
Find the centroid of the right triangle with vertices
Solution
There is no symmetry that we can use here. The hypotenuse lies on the line
Hence
and
Since
we have
and
Let the region
be between the graphs of two continuous functions
and
In this case, if
has area
and centroid
then
Proof.
Let be the area below the graph of and let be the area below the graph of Then in obvious notation
and
Therefore
and
Find the centroid of the region bounded by
Solution. Here there is no symmetry we can appeal to. We must carry out the
calculations.
Therefore
All the formulas that we have derived for volumes of solids of revolution are simple corollaries to an observation made by a brilliant, ancient Greek, Pappus of Alexandria (circa 300 A.D.).
A plane region is revolved about an axis that lies in its plane. If the region
does not cross the axis, then the volume of the resulting solid of revolution
is the area of the region multiplied by the circumference of the circle
described by the centroid of the region:
where
is the area of the region and
is the distance from the axis to the centroid of the region.
As special cases of Pappus's Theorem we have
1. The Washer Method Formula. If
and if the region bounded by
and
is revolved about the
-axis,
the resulting solid has volume
2. The Shell Method Formula. If
and if the region bounded by
and
is revolved about the
-axis,
the resulting solid has volume
Note that
and
Find the volume of the solid generated by revolving this region bounded by
about the line
Solution
Here
and
.
Hence
Find the volume of the doughnut (called torus in mathematics) generated by
revolving the circular disc
about (a) the
-axis,
(b) the
-axis.
Solution
The centroid of the disc is the center This lies units from the -axis and units from the -axis. The area of the disc is Therefore
(a)
(b)
Find the centroid of the half-disc
by appealing to Pappus's theorem.
Solution. Since the half-disc is symmetric about the -axis, we know that All we need to find is .
If we revolve the half-disc about the x-axis, we obtain a solid ball of volume
.
The area of the half-disc is
By Pappus's theorem
Simple division gives .
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