Date | May 2012 | Marks available | 9 | Reference code | 12M.2.hl.TZ1.8 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Deduce | Question number | 8 | Adapted from | N/A |
Question
A cone has height h and base radius r . Deduce the formula for the volume of this cone by rotating the triangular region, enclosed by the line y=h−hrxy=h−hrx and the coordinate axes, through 2π2π about the y-axis.
Markscheme
x=r−rhy or x=rh(h−y) (or equivalent)x=r−rhy or x=rh(h−y) (or equivalent) (A1)
∫πx2dy∫πx2dy
=π∫h0(r−rhy)2dy=π∫h0(r−rhy)2dy M1A1
Note: Award M1 for ∫x2dy∫x2dy and A1 for correct expression.
Accept π∫h0(rhy−r)2dy and π∫h0(±(r−rhx))2dxπ∫h0(rhy−r)2dy and π∫h0(±(r−rhx))2dx
=π∫h0(r2−2r2hy+r2h2y2)dy=π∫h0(r2−2r2hy+r2h2y2)dy A1
Note: Accept substitution method and apply markscheme to corresponding steps.
=π[r2y−r2y2h+r2y33h2]h0=π[r2y−r2y2h+r2y33h2]h0 M1A1
Note: Award M1 for attempted integration of any quadratic trinomial.
=π(r2h−r2h+13r2h)=π(r2h−r2h+13r2h) M1A1
Note: Award M1 for attempted substitution of limits in a trinomial.
=13πr2h=13πr2h A1
Note: Throughout the question do not penalize missing dx/dy as long as the integrations are done with respect to correct variable.
[9 marks]
Examiners report
Most candidates attempted this question using either the formula given in the information booklet or the disk method. However, many were not successful, either because they started off with the incorrect expression or incorrect integration limits or even attempted to integrate the correct expression with respect to the incorrect variable.