| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2018 |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Surface area of revolution: parametric curve |
| Difficulty | Challenging +1.2 This is a Further Maths question on surface area of revolution with parametric equations, requiring the formula S = 2π∫y√((dx/dt)² + (dy/dt)²)dt, followed by a guided substitution. While the topic is advanced and involves multiple steps, the question provides the target form in part (a) and explicitly gives the substitution in part (b), making it more structured than open-ended. The algebra is moderately involved but routine for Further Maths students. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.08h Integration by substitution4.08d Volumes of revolution: about x and y axes |
| Answer | Marks |
|---|---|
| 7(a) | x(cid:32)3t4, y(cid:32)4t3 |
| Answer | Marks | Guidance |
|---|---|---|
| dt dt | Correct derivatives | B1 |
| Answer | Marks |
|---|---|
| M1: Substitutes their derivatives into a correct formula (2(cid:83) not required) | M1 |
| S (cid:32)(cid:11)2(cid:83)(cid:12)(cid:179) 4t3(cid:11) 144t4(cid:12)1 2 (cid:11) t2 (cid:14)1 (cid:12)1 2 dt | Attempt to factor out at least t4 - |
| numerical factor may be left | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0 | Correct completion | A1 |
| Answer | Marks |
|---|---|
| (b) | du dt |
| Answer | Marks | Guidance |
|---|---|---|
| dt du | Correct differentiation | B1 |
| t (cid:32)0(cid:159)u(cid:32)1, t (cid:32)1(cid:159)u(cid:32) 2 | Correct limits |
| Answer | Marks |
|---|---|
| award later when work seen) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| S (cid:32)(cid:11)96(cid:83)(cid:12)(cid:179) (cid:11) u2 (cid:16)1 (cid:12)2 (cid:117)u2du | M1: Complete substitution | M1 A1 |
| Answer | Marks |
|---|---|
| M1: Expands and attempts to integrate | dM1 |
| Answer | Marks |
|---|---|
| Alternative: If sub reversed, substitute the original limits | ddM1 |
| Answer | Marks |
|---|---|
| 105 | 64(cid:83) |
| Answer | Marks |
|---|---|
| 35 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Scheme | Marks |
Question 7:
--- 7(a) ---
7(a) | x(cid:32)3t4, y(cid:32)4t3
dx dy
(cid:32)12t3, (cid:32)12t2
dt dt | Correct derivatives | B1
1
S (cid:32)(cid:11)2(cid:83)(cid:12) (cid:179) y(cid:168) (cid:167) (cid:168) (cid:167)dx (cid:184) (cid:183) 2 (cid:14) (cid:168) (cid:167)dy (cid:184) (cid:183) 2(cid:183) (cid:184) 2 dt (cid:32)(cid:11)2(cid:83)(cid:12) (cid:179) 4t3 (cid:11) 12t3(cid:12)2 (cid:14) (cid:11) 12t2(cid:12)2 dt
(cid:168) (cid:169) dt (cid:185) (cid:169) dt (cid:185) (cid:184)
(cid:169) (cid:185)
(cid:167) (cid:183)
(cid:168)(cid:32)(cid:11)2(cid:83)(cid:12) (cid:179) 4t3(cid:11) 144t6 (cid:14)144t4(cid:12)1 2 dt(cid:184)
(cid:168) (cid:184)
(cid:169) (cid:185)
M1: Substitutes their derivatives into a correct formula (2(cid:83) not required) | M1
S (cid:32)(cid:11)2(cid:83)(cid:12)(cid:179) 4t3(cid:11) 144t4(cid:12)1 2 (cid:11) t2 (cid:14)1 (cid:12)1 2 dt | Attempt to factor out at least t4 -
numerical factor may be left | M1
S (cid:32)96(cid:83)(cid:179) 1 t5(cid:11) t2 (cid:14)1 (cid:12)1 2 dt
0 | Correct completion | A1
(4)
(b) | du dt
u2 (cid:32)t2 (cid:14)1(cid:159)2u (cid:32)2t or 2u (cid:32)2t
dt du | Correct differentiation | B1
t (cid:32)0(cid:159)u(cid:32)1, t (cid:32)1(cid:159)u(cid:32) 2 | Correct limits
Alternative:
Reverse the substitution later.
(Treat as M1 in this case and
award later when work seen) | B1
u
S (cid:32)(cid:11)96(cid:83)(cid:12)(cid:179) t5(cid:117)u(cid:117) du
t
S (cid:32)(cid:11)96(cid:83)(cid:12)(cid:179) (cid:11) u2 (cid:16)1 (cid:12)2 (cid:117)u2du | M1: Complete substitution | M1 A1
A1: Correct integral in terms of u.
Ignore limits, need not be
simplified
(cid:170)u7 2u5 u3(cid:186)
S (cid:32)(cid:11)96(cid:83)(cid:12)(cid:179) (cid:11) u6 (cid:16)2u4 (cid:14) u2(cid:12) du (cid:32)(cid:11)96(cid:83)(cid:12) (cid:16) (cid:14)
(cid:171) (cid:187)
7 5 3
(cid:172) (cid:188)
M1: Expands and attempts to integrate | dM1
(cid:170)u7 2u5 u3(cid:186) 2 (cid:173) (cid:176) (cid:167) 2 7 2 2 5 2 3 (cid:183) (cid:167)1 2 1(cid:183) (cid:189) (cid:176)
S (cid:32)96(cid:83) (cid:171) (cid:16) (cid:14) (cid:187) (cid:32)96(cid:83)(cid:174)(cid:168) (cid:16) (cid:14) (cid:184)(cid:16) (cid:168) (cid:16) (cid:14) (cid:184)(cid:190)
7 5 3 (cid:168) 7 5 3 (cid:184) (cid:169)7 5 3(cid:185)(cid:176)
(cid:172) (cid:188) 1 (cid:176) (cid:175) (cid:169) (cid:185) (cid:191)
M1: Correct use of their changed limits (both to be changed)
Alternative: If sub reversed, substitute the original limits | ddM1
192(cid:83)(cid:11) (cid:12)
S (cid:32) 11 2(cid:16)4
105 | 64(cid:83)
Cao eg
35 | A1
(7)
(11 marks)
Question | Scheme | MarksThe curve $C$ has parametric equations
$$x = 3t^4, \quad y = 4t^3, \quad 0 \leq t \leq 1$$
The curve $C$ is rotated through $2\pi$ radians about the $x$-axis. The area of the curved surface generated is $S$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$S = k\pi \int_{0}^{1} t^2(t^2 + 1)^{\frac{1}{2}} dt$$
where $k$ is a constant to be found.
[4]
\item Use the substitution $u^2 = t^2 + 1$ to find the value of $S$, giving your answer in the form $p\pi\left(11\sqrt{2} - 4\right)$ where $p$ is a rational number to be found.
[7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2018 Q7 [11]}}