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Solutions for Practice Test IV B for Math 1501, Calculus I
Problem 1:
A solid is generated by rotating a square of side length
a
about its diagonal.
Compute its volume.
Solution:
There are many ways to do this. Here is a diagram of the square. We have
chosen to orient the square so that
both
the
x
axis
and
the
y
axis run along diagonals.
(What could be more simple?)
We have shown a “slice” running through
x
in the diagram. If we rotate the ﬁgure
about the
x
axis, this segment generates a disk of volume
πr
2
(
x
)d
x
where
r
(
x
) is the radius of the disk at
x
. Clearly
r
(
x
)isthe
y
coordinate of the point at
the top of the segment generating the disk, and this is
y
=
a
√
2

x

since
y
=
a
√
2

x
is the equation of the line that bounds the square in the upper right
quadrant, and
y
=
a
√
2
+
x
is the equation of the line that bounds the square in the upper
left quadrant. Note that
x
ranges over

a
√
2
≤
x
≤
a
√
2
.
Hence the volume is
π
Z
a/
√
2

a/
√
2
±
a
√
2

x
²
2
d
x
=
π
√
2
6
a
3
.
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View Full Document Second Solution:
This time we use the shell method. If we rotate around the
y
axis,
which is also a diagonal, we generate a “piece of pipe” or a “shell”. The volume of the
shell is
2
πxh
(
x
)d
x,
where
h
(
x
) is the length of the “piece of pipe” or“shell”. This is given by
h
(
x
)=2
±
a
√
2

x
²
.
Now in this approach,
x
is the radius of rotation; that is, the distance of the shell
from the axis of rotation.
Distances are always positive, so both limits of integration will
be positive
. The smallest radius is 0, and the largest radius is
a/
√
2. Hence in doing the
integral,
x
ranges over
0
≤
x
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This note was uploaded on 02/07/2011 for the course MATH 1501 taught by Professor N/a during the Fall '08 term at Georgia Institute of Technology.
 Fall '08
 N/A
 Calculus

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