 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-iv/3/a/ii/solution

r_{θ} r_{ϕ} = (- sin θ / 2, cos θ cos ϕ, cos θ sin ϕ), = (0, - sin θ sin ϕ, sin θ cos ϕ) .

r_{θ} \land r_{ϕ} = (sin θ cos θ, \frac{s i n ^{2} θ c o s ϕ}{2}, \frac{s i n ^{2} θ s i n ϕ}{2}) .

d A = \frac{s i n θ}{2} 1 + cos^{2} θ d θ d ϕ .

The two poles and duplicated seam have zero surface area, so integrating over the stated parameter domain counts the surface area once. With the change of variables

u = cos θ

and the supplied definite integral,

area (S) = \frac{2 π}{2} \int_{0}^{π} sin θ 1 + cos^{2} θ d θ = 2 π \int_{- 1}^{1} 1 + u^{2} d u = 2 π + 2 π ln (1 + 2) .

Here

- ln (2 - 1) = ln (2 + 1)

because

(2 - 1) (2 + 1) = 1

 New to topics? Read the docs here!