21N.1.AHL.TZ0.17

attempt to use $V=\mathrm{\pi }{\int }_a^b{x}^2 dy$             (M1)

$x={\text{e}}^{\frac{y}{6}}$ or any reasonable attempt to find $x$ in terms of $y$             (M1)

$V=\mathrm{\pi }{\int }_0^h{\text{e}}^{\frac{y}{3}} dy$             A1

Note: Correct limits must be seen for the A1 to be awarded.

$=\mathrm{\pi }{\left[3{\text{e}}^{\frac{y}{3}}\right]}_0^h$             (A1)

Note: Condone the absence of limits for this A1 mark.

$=3\mathrm{\pi }\left[{\text{e}}^{\frac{h}{3}}-{\text{e}}^0\right]$             A1

$=3\mathrm{\pi }\left[{\text{e}}^{\frac{h}{3}}-1\right]$             AG

Note: If the variable used in the integral is $x$ instead of $y$ (i.e. $V=\mathrm{\pi }{\int }_0^h{\text{e}}^{\frac{x}{3}} dx$) and the candidate has not stated that they are interchanging $x$ and $y$ then award at most M1M1A0A1A1AG.

 

[5 marks]

maximum volume when $h=9 \text{cm}$             (M1)

max volume $=180 {\text{cm}}^3$             A1

 

[2 marks]

A number of candidates switched variables so that $y=6\ln x$ and then used $\mathrm{\pi }\int y^2\mathrm{d}x$. Other candidates who correctly found $x$ in terms of $y$ failed to use the limits $0$ and $h$, using $0$ and $9$ instead. As part (a) was to show that the volume was equal to the final expression it was necessary for examiners to see steps in obtaining the result. It was common to miss out any expression involving ${\mathrm{e}}^0$. Since the value $1$ could be written from the answer given, where this value came from needed to be shown. It was encouraging to see correct answers to (b), even when candidates had failed to gain marks for (a). Some candidates successfully used their GDC to calculate the value of the definite integral numerically.

A number of candidates switched variables so that $y=6\ln x$ and then used $\mathrm{\pi }\int y^2\mathrm{d}x$. Other candidates who correctly found $x$ in terms of $y$ failed to use the limits $0$ and $h$, using $0$ and $9$ instead. As part (a) was to show that the volume was equal to the final expression it was necessary for examiners to see steps in obtaining the result. It was common to miss out any expression involving ${\mathrm{e}}^0$. Since the value $1$ could be written from the answer given, where this value came from needed to be shown. It was encouraging to see correct answers to (b), even when candidates had failed to gain marks for (a). Some candidates successfully used their GDC to calculate the value of the definite integral numerically.