surface area of the ellipse x^2/a^2 + y^2/b^2 = 1 rotated around the x-axis

Question

What is the integral

Answer 1

This equation of the ellipse is symmetrical so that we need only consider one quadrant of the ellipse. The surface area of revolution will then be twice that of rotating the quadrant, which forms the right half of the ellipsoid.

The surface area of this half-ellipsoid is the summation of all the rotated arcs in the quadrant making up the ellipse. We need to know the length of an infinitesimal arc. Using Pythagoras and the infinitesimals dx and dy, ds²=dx²+dy², so (ds/dy)²=(dx/dy)²+1. If we rotate the arc at point P(x,y) around the x-axis, the y-coordinate is the radius of a vertical disc of width ds. The surface area is 2πyds (circumference 2πy times width) and the integral is:

A=∫2πyds, where ds=dy√((dx/dy)²+1). If we assume that the semi-major axis is a, then a>b. This may be important when considering square roots that appear later.

x²/a²+y²/b²=1, b²x²+a²y²=a²b², b²x²=a²(b²-y²), x=(a/b)√(b²-y²);

dx/dy=-(a/b)y/√(b²-y²), (dx/dy)²=(a/b)²y²/(b²-y²),

ds=dy√((a/b)²y²/(b²-y²)+1)=√{(b²(b²-y²)+a²y²)/(b²(b²-y²))}=(1/b)√{(b²(b²-y²)+a²y²)/(b²-y²)}

The integral becomes: A=(2π/b)∫y√{(b²(b²-y²)+a²y²)/(b²-y²)}dy. The limits are 0≤y≤b.

Let y=bsinθ, dy=bcosθdθ. When y=0, θ=0, and when y=b, θ=π/2; b²-y² becomes b²cos²θ.

√{(b²(b²-y²)+a²y²)/(b²-y²)} becomes √(b⁴cos²θ+a²b²sin²θ)/(bcosθ)=

√(b²cos²θ+a²sin²θ)/cosθ.

This changes the integral and limits:

(2π/b)∫(bsinθ)√(b²cos²θ+a²sin²θ)/cosθ(bcosθdθ)=

(2πb)∫sinθ√(b²cos²θ+a²(1-cos²θ)dθ for 0≤θ≤π/2,

(2πb)∫sinθ√(a²-(a²-b²)cos²θ)dθ. [Note that if a=b the ellipsoid becomes a sphere, and A=2πa², the surface area of a hemisphere. Also remember that a>b.]

We need to make another substitution to help solve this integral:

Let p²=a²-(a²-b²)cos²θ, so cosθ=√{(a²-p²)/(a²-b²)} for a>b.

When θ=0, p=b; when θ=π/2, p=a. From these we have new limits: b≤p≤a.

pdp=(a²-b²)cosθsinθdθ, so sinθdθ=pdp/[(a²-b²)cosθ], and substituting for cosθ:

sinθdθ=pdp/√{(a²-b²)(a²-p²)}.

The integral becomes A=2πb∫psinθdθ=

2πb∫p²/√{(a²-b²)(a²-p²)}dp=

(2πb/√(a²-b²))∫[p²/√(a²-p²)]dp.

Let B=∫[p²/√(a²-p²)]dp=-∫[-p²/√(a²-p²)]dp=-∫[(a²-p²-a²)/√(a²-p²)]dp,

B=-∫√(a²-p²)dp+a²∫dp/√(a²-p²).

Let p=asinφ, dp=acosφdφ, B=-∫(acosφ)²dφ+a²sin^-1(p/a);

Note that cos²φ=(1+cos(2φ))/2:

sinφ=p/a, sin(2φ)=2sinφcosφ=2(p/a)√(1-p²/a²)=2(p/a²)√(a²-p²).

B=-(a²/2)∫(1+cos(2φ))dφ+a²sin^-1(p/a)=

-(a²/2)(sin^-1(p/a)+½sin(2φ))+a²sin^-1(p/a)=

½a²sin^-1(p/a)-p√(a²-p²). 

A=(2πb/√(a²-b²))[½a²sin^-1(p/a)-p√(a²-p²)]_b^a.

A=(2πb/√(a²-b²))(¼πa²-½a²sin^-1(b/a)+b√(a²-b²)).

The total surface area=2A=(πb/√(a²-b²))(πa²-2a²sin^-1(b/a)+4b√(a²-b²)).

2A=πb{a²[π-2sin^-1(b/a)]/√(a²-b²)+4b}. Note that that 2A=4πa² when a=b (circle to sphere).