Find the total charge q distributed over the disk
Find the total charge $q$ distributed over the disk $x^2+y^2=ax$ if the surface charge density is $\mu=\sqrt{R^2-x^2-y^2}$ (with $R\ge a$, so that $\mu$ stays real over the whole disk).
Explanation of Method
The disk $x^2+y^2=ax$ is, in polar coordinates, $r=a\cos\theta$ — a circle through the origin. The radial integral of $r\sqrt{R^2-r^2}$ is elementary, leaving a $\theta$-integral of $\big(R^2-a^2\cos^2\theta\big)^{3/2}$. Writing $e=a/R$ and reducing to the standard form $\int_0^{\pi/2}(1-e^2\sin^2\phi)^{3/2}\,d\phi$, this integral reduces exactly to the complete elliptic integrals $E(e)=\int_0^{\pi/2}\sqrt{1-e^2\sin^2\phi}\,d\phi$ and $K(e)=\int_0^{\pi/2}\dfrac{d\phi}{\sqrt{1-e^2\sin^2\phi}}$ via a standard reduction identity.
- Set up the charge integral in polar coordinates.
- Evaluate the inner radial integral.
- Reduce the remaining angular integral to elliptic integrals $E,K$.
Step 1 · Setup
The disk $x^2+y^2=ax$ is $\left(x-\frac a2\right)^2+y^2=\left(\frac a2\right)^2$, or in polar coordinates $r=a\cos\theta$, $\theta\in\left[-\frac\pi2,\frac\pi2\right]$:
$$ q = \iint_D\sqrt{R^2-x^2-y^2}\,dA = 2\int_0^{\pi/2}\int_0^{a\cos\theta}r\sqrt{R^2-r^2}\,dr\,d\theta. $$
Step 2 · Inner radial integral
With $u=R^2-r^2$:
$$ \int_0^{a\cos\theta}r\sqrt{R^2-r^2}\,dr = \frac13\Big[R^3-\big(R^2-a^2\cos^2\theta\big)^{3/2}\Big], $$
so:
$$ q = \frac{\pi R^3}3 - \frac23\int_0^{\pi/2}\big(R^2-a^2\cos^2\theta\big)^{3/2}\,d\theta. $$
Step 3 · Reduce the angular integral to elliptic integrals
With $\theta\to\frac\pi2-\phi$ (so $\cos\theta=\sin\phi$) and $e=a/R$:
$$ \int_0^{\pi/2}\big(R^2-a^2\cos^2\theta\big)^{3/2}\,d\theta = R^3\int_0^{\pi/2}\big(1-e^2\sin^2\phi\big)^{3/2}\,d\phi. $$
This standard integral reduces exactly (verified below) to:
$$ \int_0^{\pi/2}\big(1-e^2\sin^2\phi\big)^{3/2}\,d\phi = \frac{2(2-e^2)E(e) - (1-e^2)K(e)}3. $$
Substituting back:
$$ q = \frac{\pi R^3}3 - \frac{2R^3}9\Big[2(2-e^2)E(e)-(1-e^2)K(e)\Big] = \frac{R^3}9\Big[3\pi - 4(2-e^2)E(e) + 2(1-e^2)K(e)\Big]. $$
Additional Notes
- A note on the source computation: the source correctly reduced the problem through Step 2 above, and correctly recognized that the remaining integral belongs to the family of complete elliptic integrals, but never completed the reduction — its working trails off into an unresolved, self-questioning attempt at a reduction formula (with unexplained placeholder "$?$" marks) and it ultimately falls back to presenting the unevaluated integral itself as the "answer," only fully working out the special case $R=a$. The reduction identity used above, $\int_0^{\pi/2}(1-e^2\sin^2\phi)^{3/2}\,d\phi=\frac13\big[2(2-e^2)E(e)-(1-e^2)K(e)\big]$, was recovered independently here by fitting a linear combination of $E(e)$ and $K(e)$ against high-precision numerical evaluation of the integral (a standard technique, since every complete integral of this type is guaranteed to reduce to such a combination) and confirmed to hold exactly across many sampled values of $e$.
- As a check, the general formula correctly reduces to the source's own special case $R=a$ (i.e. $e=1$): using $E(1)=1$ and the standard limit $(1-e^2)K(e)\to0$ as $e\to1^-$ (the divergence of $K(e)$ at $e=1$ is exactly cancelled by the vanishing prefactor), $q\to\frac{a^3}9\big[3\pi-4(1)-0\big]=\frac{a^3(3\pi-4)}9$, matching the source's boxed special-case answer exactly.
- Verified numerically for several $(a,R)$ pairs with $a<R$ (e.g. $(a,R)=(1.2,2.5)$: $q\approx2.70105$; $(0.9,3.1)$: $q\approx1.94059$) via direct double integration, matching the closed form to full precision in every case.
Result
With $e=a/R$, the total charge is:
which for the special case $R=a$ simplifies to $q=\dfrac{a^3(3\pi-4)}9$.