Beach & Coconut

To evaluate the volume of the sphere $V$ , we can integrate the volume of revolution of $y^2 + x^2 = R^2$ :

$V = \int \pi (y^2) \ dx = \pi \int (R^2 - x^2) \ dx = \pi[R^2(x) - \dfrac{x^3}{3}]$

However, this sphere is not fully filled and has its boundary from $-R$ to $kR$ for some constant $k$ as shown below:

Hence, $V = \pi[ R^2(x) - \dfrac{x^3}{3} |_{-R}^{kR}] = \pi[(R^2(kR) - \dfrac{k^3 R^3}{3}) - (-R^3 + \dfrac{R^3}{3})] = \pi R^3[k - \dfrac{k^3}{3} + \dfrac{2}{3}]$

Now the volume $V$ also equals to the full cylindrical can, so $V = \pi r^2 h$ .

And $2r = R + kR = R(1 + k)$ ; $r = \dfrac{R(k+1)}{2}$ .

$h = 2R$ .

Therefore, $V = \pi (\dfrac{R(k+1)}{2})^2 (2R) = (\pi R^3)(\dfrac{(k+1)^2}{2})$ .

Putting both terms in the same equation:

$V = (\pi R^3)(\dfrac{(k+1)^2}{2}) = \pi R^3[k - \dfrac{k^3}{3} + \dfrac{2}{3}]$

$\dfrac{(k+1)^2}{2} = k - \dfrac{k^3}{3} + \dfrac{2}{3}$

$3(k^2 + 2k +1) = 4 + 6k -2k^3$

$2k^3 +3k^2 -1 = 0$

$(2k -1)(k + 1)^2 = 0$

Since $k$ can't be negative, $k = \dfrac{1}{2}$ .

Thus, $2r = R(1 + k) = \dfrac{3R}{2}$ ; $4r = 3R$ .

Since $R$ & $r$ are co-prime integers, $R = 4$ and $r = 3$ .

As a result, $R + r = \boxed{7}$ .

Note: $V = \pi R^3[k - \dfrac{k^3}{3} + \dfrac{2}{3}] = \pi R^3[\dfrac{1}{2} - \dfrac{1}{24} + \dfrac{2}{3}] = (\dfrac{9}{8})\pi R^3$ for the partial sphere.

And $V = (\pi R^3)(\dfrac{(k+1)^2}{2} = (\dfrac{9}{8})\pi R^3$ for the full cylinder.

$Volume~ of~ Cylinder~~~~=2R\pi r^2.\\ Volume~ of~Sphere~~~~~=\dfrac 4 3 \pi R^3.\\ Volume~ of~empty~cap~~= \dfrac 1 3 \pi h^2(3R-h).\\ h=2R-2r.\\ \therefore~ 2R\pi r^2 = \dfrac 4 3 \pi R^3 - \dfrac 1 3 \pi \{2(R-r)\}^2\{3R-2(R-r)\}\\ Multiplying ~both~ sides~~by~\dfrac 3{2 \pi},\\ 3Rr^2=2R^3-2R^3+4R^2r-2Rr^2-4R^2r-4r^3.\\ Simplifying~~~3R=4r.\\ But ~ R~and~r~must~be~minimum~co-prime~integers.\\ So~R=4 ~and~r=3.\\ R+r=\color{#D61F06}{ \Large 7}$