Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 12

For polar coordinates, $x=r\cos \theta$ and $y=r\sin \theta$. The Jacobian determinant is $r$, so the area element isThis accounts for the increasing length of an angular strip as its distance from the origin increases.

Let $I$ be the positive Gaussian integral. Its improper integral converges because $e^{-x^2}\le e^{-|x|}$ for $|x|\ge 1$. Multiplying two copies and using the polar coordinates area element givesNonnegativity justifies the product double integral and the change of integration order. Since $I>0$,

The cardioid has a nonnegative radial boundary, so each ray from the origin contributes $0\le r\le a(1+\cos \theta )$. Integrating the polar coordinates area element yieldsThe three angular integrals are $2\pi$, zero and $\pi$, respectively, using the double-angle formula for the last. Thus

The volume element is the absolute Jacobian determinant times the coordinate differentials. In the Cartesian coordinate system, cylindrical coordinates $(r,\theta ,z)$ and spherical coordinates $(\rho ,\vartheta ,\phi )$, respectively,Here $\rho$ is distance from the origin, $\vartheta$ is the polar angle from the positive $z$ axis, and $\phi$ is the azimuthal angle. The spherical coordinates convention fixes which angle enters the sine factor.

At height $z>0$, the elliptic paraboloid has a circular cross-section $0\le r\le \sqrt{z}$, of area $\pi z$. The cylindrical coordinates volume element therefore reduces the volume integral toFor the specified $f$, the derivative is $e^{-z}(z^{-1/2}+\frac{1}{2}{z}^{-3/2})$. Thus $z{f}^{\prime }(z)$ is integrable at zero and decays exponentially at infinity. Apply integration by parts first on a truncated interval and then take the endpoint limits:Both boundary terms vanish because $zf(z)=-\sqrt{z}{e}^{-z}$. With $z=u^2$, integration by substitution gives the Gaussian integral:The integrand is singular at the apex but the resulting improper integral is convergent.

In spherical coordinates, the cone condition forces $z\ge 0$ and becomes $0\le \vartheta \le \alpha$. The unit ball gives $0\le \rho \le 1$. HenceFor an independent calculation in cylindrical coordinates, the lower boundary is $z=r\cot \alpha$ and the upper boundary is $z=\sqrt{1-r^2}$. They meet at $r=\sin \alpha$, since $r^2{\cot }^2\alpha =1-r^2$. ThusThe equality uses ${\sin }^3\alpha \cot \alpha ={\sin }^2\alpha \cos \alpha$. The positive cosine on the stated range removes any square-root sign ambiguity.