 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2015/6/solution

The boundaries intersect where

x^{2} = 2 x + 8

x = - 2

and

x = 4

. Between these values, the straight line lies above the parabola. Therefore

I = \int_{- 2}^{4} \int_{x^{2}}^{2 x + 8} x d y d x = \int_{- 2}^{4} (- x^{3} + 2 x^{2} + 8 x) d x .

I = [- \frac{x ^{4}}{4} + \frac{2 x ^{3}}{3} + 4 x^{2}]_{- 2}^{4} = 36 .

A useful independent check uses the region's symmetry about

x = 1

: its vertical width is

9 - (x - 1)^{2}

. Its area is

\int_{- 3}^{3} (9 - u^{2}) d u = 36

, and its centroid has

x

coordinate one, so its first moment

\iint x d A

is again

36

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