01
Surface area: revolved cubic
The curve
over is revolved around the -axis. Find the area of the resulting surface.
Solution
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
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.)
Solution
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:
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- 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.
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?
Solution
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
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.)
Solution
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.