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Past exam of the mathematics course of the University of Oxford

2017

Prelims

Mathematics V

{c}

The <triangle> is $0\leq y\leq x\leq1$. In <polar coordinates> this becomes
$$
0\leq\theta\leq\frac\pi4,\qquad 0\leq r\leq\sec\theta,
$$
because the outer side is $x=r\cos\theta=1$. The <vector field> and the <Jacobian determinant> give
$$
\mathbf f=r\cos^2\theta\,\mathbf e_1+r\cos\theta\sin\theta\,\mathbf e_2,\qquad dx\,dy=r\,dr\,d\theta.
$$
Evaluate the <double integral> component by component:
$$
\begin{aligned}
\iint_R\mathbf f\,dx\,dy
&=\frac13\int_0^{\pi/4}\bigl(\sec\theta\,\mathbf e_1+\sec\theta\tan\theta\,\mathbf e_2\bigr)d\theta\\
&=\frac13\left[\log(\sec\theta+\tan\theta)\,\mathbf e_1+\sec\theta\,\mathbf e_2\right]_0^{\pi/4}.
\end{aligned}
$$
Consequently,
$$
\boxed{\iint_R\mathbf f\,dx\,dy=\frac{\log(1+\sqrt2)}3\,\mathbf e_1+\frac{\sqrt2-1}3\,\mathbf e_2.}
$$
The origin causes no ambiguity in this <double integral>: the <continuous extension> from part (a) is bounded there, and changing a value at a single point does not change the <integral>.

Solution

Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics V / 1 / b / Solution

Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics V / 1 / b / Solution

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