Calculus II
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W04 - HW Solutions

01

(1) Integral formula for arclength:

(2) Work out integrand:

(3) Integrate:

02

(1) Integral formula for surface area, revolution about -axis:

(2) Work out integrand:

Then:

So:

(3) Perform -sub with and so :

(4) Integrate:

03

(1) Integral formula for arclength:

(2) Work out integrand:

Therefore, the integrand:

(3) Integrate:

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.

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:

  • 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

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 :

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 . 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.