$\tan{\theta}={{r}\over{l}}={{\Delta r}\over{\Delta l}}$
The used volume $V_1 = {1 \over 3}\pi {{\Delta l}^{3}} \tan ^ {2} \theta$.
The removed volume $V_2 = \pi \Delta l^3 \tan ^ 2 \theta + {1 \over 3} \pi l ^ 3 \tan ^ 2 \theta - {1 \over 3} \pi \Delta l^3 \tan ^ 2 \theta - {1 \over 3} \pi l^3 \tan ^ 2 \theta \\ = {2\over 3}\pi \Delta l^3 \tan ^2 \theta$
${{\Delta V_2}\over{\Delta V_1}} = 2$
${{\Delta V_2}\over{\Delta V_1 + \Delta V_2}} = {2\over 3}$
${{\Delta V_1}\over{\Delta V_1 + \Delta V_2}} = {2\over 3}$
This shows that only $1\over 3$ of pencil is used.
