Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 15

The radial interval is nonempty exactly when $1+\cos \varphi \ge 3/2$, or $\cos \varphi \ge 1/2$. Choose the representative angular range $-\pi /3\le \varphi \le \pi /3$. The region of integration is thereforeIt is the right-hand cap inside the cardioid and outside the circle of radius $3/2$, symmetric about the horizontal axis. Its boundary intersections have coordinates $(3/4,\pm 3\sqrt{3}/4)$, and the cap reaches $(2,0)$. The radial segment on the horizontal axis runs from $3/2$ to $2$.

The polar area element is $rdrd\varphi$. Integrating first in radius and using symmetry givesThis is the specialization $a=1$, $c=3/2$ of the polar cap between a cardioid and a circle area formula.

Substitute $x=r\cos \varphi$, $y=r\sin \varphi$ and include the Jacobian determinant factor $r$. The integrand times the area element becomesBoth terms containing one factor $\sin \varphi$ are odd in $\varphi$, while the radial bounds and angular interval are even. They therefore integrate to zero. The remaining double integral isThe domain stays away from $r=0$, so the denominator of the original integrand causes no singularity in this region.

Translate the variables by $u=x-x_0$, $v=y-y_0$, $w=z-z_0$. The numerator becomes $u+x_0$. The part odd in $u$ integrates to zero, while the constant part factors into three Gaussian integrals. ThusThe absolute integrability of a polynomial times a Gaussian justifies this separation by Fubini's theorem. Neither $y_0$ nor $z_0$ affects the result; when $x_0=0$, odd symmetry makes the integral zero.