 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-2/12/b/ii/solution

First perform the inner integral, which gives the vertical width of the region:

I = \int_{- a}^{a} (1 - x^{2} - x^{2}) d x = 2 \int_{0}^{a} (1 - x^{2} - x^{2}) d x .

Using

\int 1 - x^{2} d x = \frac{1}{2} [x 1 - x^{2} + arcsin x]

gives

I = a 1 - a^{2} + arcsin a - \frac{2}{3} a^{3} .

At the intersection,

1 - a^{2} = a^{2}

, so

I = arcsin a + \frac{1}{3} a^{3}, A = 1, B = \frac{1}{3} .

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