Standard +0.3 This is a standard two-part volumes of revolution question requiring integration of y² for volume and solving for a constant from a given area. The function simplifies nicely (y² = 36/(3x-2)), and both integrations are straightforward using standard techniques. Slightly easier than average due to the clean algebraic manipulation and routine application of formulas.
5
\includegraphics[max width=\textwidth, alt={}, center]{774bb427-5392-45d3-8e4e-47d08fb8a792-02_559_1191_1749_479}
The diagram shows the curve with equation \(y = \frac { 6 } { \sqrt { 3 x - 2 } }\). The region \(R\), shaded in the diagram, is bounded by the curve and the lines \(x = 1 , x = a\) and \(y = 0\), where \(a\) is a constant greater than 1 . It is given that the area of \(R\) is 16 square units. Find the value of \(a\) and hence find the exact volume of the solid formed when \(R\) is rotated completely about the \(x\)-axis. [0pt]
[9]
Integrate to obtain form \(k(3x - 2)^{\frac{1}{4}}\)
M1
any non-zero constant \(k\); or equiv involving substitution
Obtain correct \(4(3x - 2)^{\frac{1}{4}}\)
A1
or (unsimplified) equiv such as \(\frac{6(3x-2)^{\frac{1}{4}}}{3 \times \frac{1}{4}}\)
Apply limits and attempt solution for \(a\)
M1
assuming integral of form \(k(3x - 2)^n\); taking solution as far as removal of root; with subtraction the right way round; if sub'n used, limits must be appropriate
Obtain \(a = 9\)
A1
(this answer written down with no working scores 0/4 so far but all subsequent marks are available)
State or imply formula \(\int \frac{36\pi}{3x-2} dx\)
B1
or (unsimplified) equiv; condone absence of \(dx\); allow B1 retroactively if \(\pi\) absent here but inserted later
Integrate to obtain form \(k \ln(3x - 2)\)
*M1
any constant \(k\) including \(\pi\) or not; condone absence of brackets
Obtain \(12\pi \ln(3x - 2)\) or \(12 \ln(3x - 2)\)
A1
following their integral of form \(\int \frac{k}{3x-2} dx\)
Apply limits the correct way round
M1
dep *M; use of limit 1 is implied by absence of second term; allow use of limit \(a\)
Obtain \(12\pi \ln 25\) (or \(24\pi \ln 5\))
A1
9 or exact equiv but not with \(\ln 1\) remaining; condone answers such as \(\pi 12 \ln 25\) and \(12\ln 25\pi\)
Integrate to obtain form $k(3x - 2)^{\frac{1}{4}}$ | M1 | any non-zero constant $k$; or equiv involving substitution
Obtain correct $4(3x - 2)^{\frac{1}{4}}$ | A1 | or (unsimplified) equiv such as $\frac{6(3x-2)^{\frac{1}{4}}}{3 \times \frac{1}{4}}$
Apply limits and attempt solution for $a$ | M1 | assuming integral of form $k(3x - 2)^n$; taking solution as far as removal of root; with subtraction the right way round; if sub'n used, limits must be appropriate
Obtain $a = 9$ | A1 | (this answer written down with no working scores 0/4 so far but all subsequent marks are available)
State or imply formula $\int \frac{36\pi}{3x-2} dx$ | B1 | or (unsimplified) equiv; condone absence of $dx$; allow B1 retroactively if $\pi$ absent here but inserted later
Integrate to obtain form $k \ln(3x - 2)$ | *M1 | any constant $k$ including $\pi$ or not; condone absence of brackets
Obtain $12\pi \ln(3x - 2)$ or $12 \ln(3x - 2)$ | A1 | following their integral of form $\int \frac{k}{3x-2} dx$
Apply limits the correct way round | M1 | dep *M; use of limit 1 is implied by absence of second term; allow use of limit $a$
Obtain $12\pi \ln 25$ (or $24\pi \ln 5$) | A1 | 9 or exact equiv but not with $\ln 1$ remaining; condone answers such as $\pi 12 \ln 25$ and $12\ln 25\pi$
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5\\
\includegraphics[max width=\textwidth, alt={}, center]{774bb427-5392-45d3-8e4e-47d08fb8a792-02_559_1191_1749_479}
The diagram shows the curve with equation $y = \frac { 6 } { \sqrt { 3 x - 2 } }$. The region $R$, shaded in the diagram, is bounded by the curve and the lines $x = 1 , x = a$ and $y = 0$, where $a$ is a constant greater than 1 . It is given that the area of $R$ is 16 square units. Find the value of $a$ and hence find the exact volume of the solid formed when $R$ is rotated completely about the $x$-axis.\\[0pt]
[9]
\hfill \mbox{\textit{OCR C3 2011 Q5 [9]}}