Stepwise problems - Thu. 11:59pm
Arc length
01
01
Arc length - reversed
and roles Find the arc length of the curve that satisfies the equation
over . Link to originalSolution
01
(1) Integral formula for arclength:
(2) Work out integrand:
(3) Integrate:
Link to original
Surface areas of revolutions - thin bands
02
01
Surface area: revolved cubic
The curve
over is revolved around the -axis. Find the area of the resulting surface.
Link to originalSolution
02
(1) Integral formula for surface area, revolution about
-axis:
(2) Work out integrand:
Then:
So:
(3) Perform
-sub with and so :
(4) Integrate:
Link to original
Regular problems - Sun. 11:59pm
Arc length
03
02
Arc length - tricky algebra
Find the arc length of the curve
for . (Hint: expand under the root, then simplify, then factor; now it’s a square and the root disappears.)
Link to originalSolution
03
(1) Integral formula for arclength:
(2) Work out integrand:
Therefore, the integrand:
(3) Integrate:
Link to original
04
03
Arc length - tricky integration
Find the arc length of the curve
for . (Hint: the integral can be done using either: (i)
-sub then trig sub, or (ii) ‘rationalization’ then partial fractions.) Link to originalSolution
04
(1) Integral formula for arclength:
(2) Perform
-sub with and so and also: Now transform the integral to
:
(3) Integrate: partial fraction decomposition:
Number degree not lower → long division first:
Write general PFD formula:
Solve for
and . Cross multiply:
(4) Evaluate integral:
Note A: Instead of this
-sub and partial fractions, one can set and obtain . Then trig sub with leads to (eventually) the same final answer. Note B: This answer is sufficient. It is not necessary to simplify as in the last step.
Link to original
Surface areas of revolutions - thin bands
05
02
Surface area: cone
A cone may be described as the surface of revolution of a ray emanating from the origin, revolved around the
-axis. Let
for some . Find the surface area of the cone given by revolving the graph of around the -axis over . Can you also calculate this area using geometry? And verify the two methods give the same formula? (Hint: ‘unroll’ the cone into a sector.)
Link to originalSolution
05
(1) Integral formula for surface area, revolution around
-axis:
(2) Work out integrand:
(4) Evaluate integral:
(5) Verify with geometry:
Note that unrolling the cone forms a sector with radius
and arc length . The total circumference is . So the area of the sector is: Notes:
Link to original
- Sector radius is the lateral length of the cone: hypotenuse of right triangle with legs
(on -axis) and (on -axis). - Sector arc is
because is radius of the base.
06
03
Surface area: parabolic reflector
A parabolic reflector is given by rotating the curve
around the -axis for . What is the surface area of this reflector?
Link to originalSolution
06
Method 1: integrate in
(1) Integral formula for surface area:
(2) Integrate: perform
-sub with and so :
Method 2: integrate in
(1) Integral formula for surface area using
:
(2) Integrate: perform
-sub with and so : Link to original
07
04
Surface area: torus
A torus is created by revolving about the
-axis the circle with this equation: Find the surface area of this torus.
(Hint: compute for the top and bottom of the circle separately and add the results.)
Link to originalSolution
07
(1) Integral formula for surface area:
Compute
by solving for : Top semicircle:
Bottom semicircle: Bounds:
(2) Work out integrands:
(3) Simplify integral:
Top semicircle:
Bottom semicircle:
Therefore:
(4) Integrate: perform trig sub with
and so : Note: This answer can be written
Link to original. It is, therefore, the same as the surface area of a circular cylinder with radius and length . Bending the cylinder into a torus does not change its surface area.
