 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/15/a/iii/solution

Substitute

x = r cos ϕ

y = r sin ϕ

and include the Jacobian determinant factor

r

. The integrand times the area element becomes

(cos ϕ + sin ϕ + r cos ϕ sin ϕ) d r d ϕ .

Both terms containing one factor

sin ϕ

are odd in

ϕ

, while the radial bounds and angular interval are even. They therefore integrate to zero. The remaining double integral is

I = \int_{- π /3}^{π /3} cos ϕ (1 + cos ϕ - \frac{3}{2}) d ϕ = \int_{- π /3}^{π /3} (cos^{2} ϕ - \frac{1}{2} cos ϕ) d ϕ = \frac{π}{3} - \frac{3}{4} .

The domain stays away from

r = 0

, so the denominator of the original integrand causes no singularity in this region.

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