Work
Videos, BlackPenRedPen:
- Work performed: pumping water from trough
- Work performed: pumping water from rectangular tank
- Work performed: pumping water from conical tank
- Work performed: pumping water from spherical tank
01 Theory
Work is a measure of energy expended to achieve some effect. According to physics:
To compute the work performed against gravity while lifting some matter, decompose the matter into horizontal layers at height
The work performed on each single layer is summed by an integral to determine the total work performed to lift all the layers:
Work performed
02 Illustration
Example - Pumping water from spherical tank
Pumping water from spherical tank
Calculate the work done pumping water out of a spherical tank of radius
. Solution Divide the sphere of water into horizontal layers.
Coordinate
is at the center of the sphere.
Work done pumping out water is constant across any single layer
Find formula for weight of a single layer.
Area of the layer at
is because its radius is . Volume of the layer at
is then . Weight of the layer is then
. Plug in:
Find formula for vertical distance a given layer is lifted.
Layer at
must be lifted by to the top of the tank.
Work per layer is the product.
Product of weight times height lifted:
Total work done is the integral.
Integrate over the layers:
Supplement: what if the spigot sits
above the tank? Increase the height function from
to .
Supplement: what if the tank starts at just
of water depth? Integrate the water layers only: change bounds to
Link to original.
Example - Water pumped from a frustum
Water pumped from a frustum
Find the work required to pump water out of the frustum in the figure. Assume the weight of water is
.
Solution
Find weight of a horizontal slice.
Coordinate
at top, increasing downwards. Use
for radius of cross-section circle. Linear decrease in
from to : Area is
:
:
Find work to pump out a horizontal layer.
Layer at
is raised a distance of . Work to raise layer at
:
Integrate over all layers.
Integrate from top to bottom of frustum:
Final answer is
Link to original.
Example - Raising a building
Raising a building
Find the work done to raise a cement columnar building of height
and square base per side. Cement has a density of .
Solution
Divide the building into horizontal layers.
Work done raising up the layers is constant for each layer.
Find formula for weight of each layer.
Volume
Mass
Weight of layer
Find formula for work performed lifting one layer into place.
Work
Find total work as integral over the layers.
Total work
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Example - Raising a chain
Raising a chain
An
chain is suspended from the top of a building. Suppose the chain has weight density . What is the total work required to reel in the chain? Solution
Divide the chain into horizontal layers.
Each layer has vertical thickness
. Each layer has weight
in .
Find formula for distance each layer is raised.
Each layer is raised from
to , a distance of .
Compute total work.
Work to raise each layer is weight times distance raised:
Add the work over all layers:
Link to original
Example - Raising a leaky bucket
Raising a leaky bucket
Suppose a bucket is hoisted by a cable up an
tower. The bucket is lifted at a constant rate of and is leaking water weight at a constant rate of . The initial weight of water is . What is the total work performed against gravity in lifting the water? (Ignore the bucket itself and the cable.) Solution
Convert to static format.
Compute rate of water weight loss per unit of vertical height:
Choose coordinate
at base, at top. Compute water weight at each height
:
Work formula.
Total work is integral of force times infinitesimal distance:
Integrate weight times
. Plug in weight as force:
Compute integral:
Link to original
In the last example, the same bucket passes through each height, it is not sliced into layers.
- This integral adds work done to lift matter through each
as if in parallel, and is thus representing distance lifted. - Previous examples have integral adding work done to lift matter through some
or ; the matter was sliced into layers with featuring in the weight of a portion.
Improper integrals
Videos, Math Dr. Bob:
- Improper integrals: Infinite limits
- Improper integrals: Vertical asymptote
- Improper integrals:
- Improper integrals:
03 Theory
Improper integrals are those for which either a bound or the integrand itself become infinite somewhere on the interval of integration.
Examples:
- (a) the upper bound is
- (b) the integrand goes to
as the integrand is at the point
The limit interpretation of
The limit interpretation of
The limit interpretation of
These limits are evaluated using familiar methods.
An improper integral is said to be convergent or divergent according to whether it may be assigned a finite value through the appropriate limit interpretation.
For example,
04 Illustration
Example - Improper integral - infinite bound
Improper integral - infinite bound
Show that the improper integral
converges. What is its value? Solution
Replace infinity with a new symbol
. Compute the integral:
Take limit as
. Find limit:
Apply definition of improper integral.
By definition:
Conclude that
Link to originalconverges and equals .
Improper integral - infinite integrand
Improper integral - infinite integrand
Show that the improper integral
converges. What is its value? Solution
Replace the
where diverges with a new symbol . Compute the integral:
Take limit as
. Find limit:
Apply definition of improper integral.
By definition:
Conclude that
Link to originalconverges to .
Example - Improper integral - infinity inside the interval
Example - Improper integral - infinity inside the interval
Does the integral
converge or diverge? Solution
It is tempting to compute the integral incorrectly, like this:
But this is wrong. There is an infinite integrand at
. We must instead break it into parts. Identify infinite integrand at
. Integral becomes:
Interpret improper integrals.
Limit interpretations:
Compute integrals.
Using
:
Take limits.
We have:
Neither limit is finite. For
Link to originalto exist we’d need both of these limits to be finite.
05 Theory
Two tools allow us to determine convergence of a large variety of integrals. They are the comparison test and the
Comparison test - integrals
The comparison test says:
- When an improper integral converges, every smaller integral converges.
- When an improper integral diverges, every bigger integral diverges.
Here, smaller and bigger are comparisons of the integrand (accounting properly for signs), and the bounds are assumed to be the same.
For example, since
-integral cases Assume
and . We have:
Proving the
-integral cases It is easy to prove the convergence / divergence of each
-integral case using the limit interpretation and the power rule for integrals. (Or for , using .)
Additional improper integral types
The improper integral
also has a limit interpretation: The double improper integral
has this limit interpretation: Where
is any finite number. This double integral does not exist if either limit does not exist for any value of .
Double improper is not simultaneous!
Watch out! This may happen:
This simultaneous limit might exist only due to internal cancellation, in a case where the separate individual limits do not exist!
06 Illustration
Example - Comparison to
-integrals Comparison to p-integrals
Determine whether the integral converges:
- (a)
- (b)
Solution
(a)
Integrand tends toward
for large . Consider large
values:
Try comparison to
. Comparison attempt:
Validate. Notice
and when .
Apply comparison test.
We know:
We know:
We conclude:
(b)
Integrand tends toward
for large . Consider large
values:
Try comparison to
. Comparison attempt:
Validate. Notice
and when .
Apply comparison test.
We know:
We know:
We conclude:
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