 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-v-resit/question-1/a/i/solution

0 \leq r \leq sin^{2} θ

and

0 \leq θ \leq π

d A = r d r d θ

, so the area is

A = \iint_{R} 1 d A = \int_{0}^{π} \int_{0}^{s i n^{2} θ} r d r d θ = \frac{1}{2} \int_{0}^{π} sin^{4} θ d θ .

sin^{4} θ = \frac{3 - 4 c o s ( 2 θ ) + c o s ( 4 θ )}{8}

reduces the integral to its constant term: both cosine terms have zero integral over this interval. Therefore

Area (R) = \frac{3 π}{16} .

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