\(\displaystyle Common \ Sense \ Approach, \ no \ calculus.\)
\(\displaystyle V_{sphere} \ = \ \frac{4\pi r^{3}}{3}, \ V_{hemisphers} \ = \ \frac{2\pi r^{3}}{3}\)
\(\displaystyle D \ = \ 50m, \ r \ = \ 25m, \ .05cm \ = \ .0005m\)
\(\displaystyle Volume \ of \ hemi \ before \ painting: \ V \ = \ (2/3)\pi(25)^{3} \ = \ 32,724.9234749m^{3}\)
\(\displaystyle Volume \ of \ hemi \ after \ painting: \ V \ = \ (2/3)\pi(25.0005)^{3} \ = \ 32,726.8870096m^{3}\)
\(\displaystyle Difference \ in \ volume \ = \ 1.963534m^{3}, \ hence \ the \ painter \ will \ need \ about \ 2m^{3} \ of \ paint.\)
\(\displaystyle Calculus \ approach:\)
\(\displaystyle V_{hemisphere} \ = \ \frac{2\pi r^{3}}{3}, \ r \ = \ 25m, \ dr \ = \ .0005m\)
\(\displaystyle dV \ = \ 2\pi r^{2}dr \ = \ 2\pi(25)^{2}(.0005) \ = \ 1.9634954084m^{3} \ of \ paint \ needed.\)