 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2015-resit/8/b/solution

A = 10 x^{4} - 2 x y^{3}

and

B = - 3 x^{2} y^{2}

. Their mixed derivatives agree:

A_{y} = - 6 x y^{2} = B_{x}

. Both components are smooth on all of

R^{2}

Choose a horizontal segment from

(0, 0)

(2, 0)

followed by a vertical segment to

(2, 1)

. On the first,

y = 0

and

d y = 0

; on the second,

x = 2

and

d x = 0

. Thus

I = \int_{0}^{2} 10 x^{4} d x + \int_{0}^{1} (- 12 y^{2}) d y = 64 - 4 = 60 .

The selected path and its orientation are shown below. No contribution comes from a differential whose coordinate is held constant on a segment.

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