OCR MEI Further Mechanics Major 2019 June — Question 7 8 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2019
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeCentre of mass of solid of revolution
DifficultyChallenging +1.8 This is a demanding Further Maths mechanics question requiring: (1) finding the volume of revolution about the y-axis using integration, (2) finding the first moment about the origin, (3) applying the centre of mass formula. Students must invert y = (2/3)ln x to get x = e^(3y/2), set up integrals with correct limits (0 to ln 2), and handle exponential integrals carefully. The 8-mark allocation and 'show detailed reasoning' requirement indicate extended multi-step work with potential for algebraic errors, placing it well above average difficulty.
Spec4.08d Volumes of revolution: about x and y axes6.04d Integration: for centre of mass of laminas/solids

In this question you must show detailed reasoning. \includegraphics{figure_7} Fig. 7 shows the curve with equation \(y = \frac{2}{3}\ln x\). The region R, shown shaded in Fig. 7, is bounded by the curve and the lines \(x = 0\), \(y = 0\) and \(y = \ln 2\). A uniform solid of revolution is formed by rotating the region R completely about the \(y\)-axis. Find the exact \(x\)-coordinate of the centre of mass of the solid. [8]

Question 7:
AnswerMarks
7DR
 2
V  ln2 e 3 2 y dy
0
ln2
1 
 e3y
 
3 
0
1   7
  e3ln2e0  
3 3
2
ln2  3y 
Vy  ye2  dy
0  
 ln2 
     1 ye3y    1  ln2 e3ydy  
 3  0 3 0 
ln2
1 1 
 ye3y  e3y
 
3 9 
0
8 7
ln2
3 9
y 
7
3
8 1
y  ln2
AnswerMarks
7 3M1*
A1
A1
M1*
A2
M1dep*
A1
AnswerMarks
[8]1.1
1.1
1.1
2.1
1.1
1.1
1.1
AnswerMarks
2.2a 2  2
For  e 3 2 y dy or  e3 2y dy
1e3y
For
3
For correct substitution of limits and
removing of logs
For yx2dy leading to
ye3y e3ydyor consistent with
 2
2y
their V if using  e3 dy
Both terms integrated correctly (A1
for one error)
Vy
M1 for y  - dependent on both
V
previous M marks
AnswerMarks
oeLimits not required for
M and first A mark
Must see one line of
working from
integrated expression to
answer
Clear indication of
integrating exponential
term and differentiating
𝑦 term
Limits not required for
M mark and both A
marks
Must have used correct
limits correctly
Allow absence of 
throughout
Question 7:
7 | DR
 2
V  ln2 e 3 2 y dy
0
ln2
1 
 e3y
 
3 
0
1   7
  e3ln2e0  
3 3
2
ln2  3y 
Vy  ye2  dy
0  
 ln2 
     1 ye3y    1  ln2 e3ydy  
 3  0 3 0 
ln2
1 1 
 ye3y  e3y
 
3 9 
0
8 7
ln2
3 9
y 
7
3
8 1
y  ln2
7 3 | M1*
A1
A1
M1*
A2
M1dep*
A1
[8] | 1.1
1.1
1.1
2.1
1.1
1.1
1.1
2.2a |  2  2
For  e 3 2 y dy or  e3 2y dy
1e3y
For
3
For correct substitution of limits and
removing of logs
For yx2dy leading to
ye3y e3ydyor consistent with
 2
2y
their V if using  e3 dy
Both terms integrated correctly (A1
for one error)
Vy
M1 for y  - dependent on both
V
previous M marks
oe | Limits not required for
M and first A mark
Must see one line of
working from
integrated expression to
answer
Clear indication of
integrating exponential
term and differentiating
𝑦 term
Limits not required for
M mark and both A
marks
Must have used correct
limits correctly
Allow absence of 
throughout
\textbf{In this question you must show detailed reasoning.}

\includegraphics{figure_7}

Fig. 7 shows the curve with equation $y = \frac{2}{3}\ln x$. The region R, shown shaded in Fig. 7, is bounded by the curve and the lines $x = 0$, $y = 0$ and $y = \ln 2$. A uniform solid of revolution is formed by rotating the region R completely about the $y$-axis.

Find the exact $x$-coordinate of the centre of mass of the solid. [8]

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2019 Q7 [8]}}