A disc of radius r has Riemann area pi r squared; in particular the unit disc has area pi

Given: A real $r>0$ and the graph-area convention of Riemann area between two continuous graphs and the disc as a vertically simple region.

If $\phi$ is differentiable with integrable derivative and $f$ is continuous on an interval containing its image, then ${\int }_{\phi (c)}^{\phi (d)}f={\int }_c^d(f\circ \phi ){\phi }^{\mathrm{\prime }}$ (Substitution: if $\phi$ is differentiable on $[c,d]$ with ${\phi }^{\mathrm{\prime }}$ integrable and $f$ is continuous on an interval containing $\phi ([c,d])$, then ${\int }_{\phi (c)}^{\phi (d)}f={\int }_c^d(f\circ \phi ){\phi }^{\mathrm{\prime }}$).

For every real $x$, the quarter-turn shift formulas are $\sin (x+\pi \mathrm{/}2)=\cos x$ and $\cos (x+\pi \mathrm{/}2)=-\sin x$, and in particular $\sin (\pi \mathrm{/}2)=1$ and $\cos (\pi \mathrm{/}2)=0$. For all real $x,y$, the sine and cosine addition formulas hold; moreover, ${\sin }^{2}x+{\cos }^{2}x=1$ for every real $x$ (Quarter-turn values and shifts by pi/2 and pi, The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine).

$(\sin t{)}^{\mathrm{\prime }}=\cos t$ and $(\cos t{)}^{\mathrm{\prime }}=-\sin t$ (The derivatives of sine and cosine are cosine and minus sine).

The Riemann integral is linear (Integrable functions on $[a,b]$ form a set closed under sums and scalar multiples, and ${\int }_a^b(\lambda f+\mu g)=\lambda {\int }_a^{b}f+\mu {\int }_a^{b}g$).

If $G$ is differentiable at every point of $[a,b]$, $f=G^{\mathrm{\prime }}$ there, and $f$ is integrable, then ${\int }_a^{b}f=G(b)-G(a)$ (The second fundamental theorem: if $G$ is differentiable on $[a,b]$ with $G^{\mathrm{\prime }}=f$ and $f$ is integrable, then ${\int }_a^{b}f=G(b)-G(a)$).

The integral of the constant $1$ on $[a,b]$ is $b-a$ (If $m\le f\le M$ on $[a,b]$ then $m(b-a)\le L(f,P)\le \underset{‾}{{\int }_a^b}f\le \overline{{\int }_a^b}f\le U(f,P)\le M(b-a)$ for every partition $P$; in particular every constant function is integrable, with ${\int }_a^{b}c=c(b-a)$).