4. Let Σ be the surface parametrised by φ(a, b) = (cos a cos b,sin a sin b, cos b + log(tan(b/2)). where 0 < a < 2π and 0 < b < π.

Question

Show that the vector n = (cos a cos b,sin a cos b, ; sin b) is a unit normal to the surface Σ at the point φ(a, b). (b) Compute the area of the surface Σ. [Possible hint: The function log(tan(b/2)) has a particularly nice derivative. Simplify it as much as possible

Comments

Please explain the significance of “; sin(b)” in the definition of the normal n. I don’t recognise the use of semicolon in the parametrised expression. Also, does “log” mean “ln”?

Answer 1

Derivative of cos(b)+log(tan(b/2)) wrt b assuming log is ln:

-sin(b)+½cot(b/2)sec²(b/2)=

-sin(b)+½(cos(b/2)/sin(b/2))/cos²(b/2)=

-sin(b)+1/(2sin(b/2)cos(b/2))=

-sin(b)+1/sin(b)=(1-sin²(b))/sin(b)=cos²(b)csc(b) or cos(b)cot(b).

n=∂φ/∂a × ∂φ∂b. (Vector cross product)

∂φ/∂a=(-sin(a)cos(b),cos(a)sin(b),0)

∂φ∂b=(-cos(a)sin(b),sin(a)cos(b),cos²(b)csc(b)),

n=(cos(a)cos²(b),sin(a)cos²(b)cot(b),-sin²(a)cos²(b)+cos²(a)sin²(b)).

|n|=√(cos²(a)cos⁴(b)+sin²(a)cos⁴(b)cot²(b)+sin⁴(a)cos⁴(b)-2sin²(a)cos²(b)cos²(a)sin²(b)+cos⁴(a)sin⁴(b)).

Let A=sin²(a), B=sin²(b), then 1-A=cos²(a), 1-B=cos²(b), cot²(b)=(1-B)/B=1/B-1.

|n|=√[(1-A)(1-B)²+A(1-B)²(1/B-1)+A²(1-B)²-2AB(1-A)(1-B)+(1-A)²B²].

|n|=√[1-2B+B²-A+2AB-AB²+A/B-2A+3AB-A-AB²

+A²-2A²B+A²B²-2AB+2A²B+2AB²-2A²B²+B²-2AB²+A²B²].

|n|=√[1-2B+2B²-4A-2AB²+A/B+3AB+A²].

|n|=√[1-2B+B²+A²+2AB+B²+AB-4A-2AB²+A/B].

|n|=√[(1-B)²+(A+B)²+A(B-4-2B²+1/B)].

|n|=√[cos⁴(b)+(sin²(a)+sin²(b))²+sin²(a)(sin²(b)-4-2sin⁴(b)+csc²(b))].

Unit normal vector=n/|n|.

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