OCR C3 2007 January — Question 6 9 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2007
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVolumes of Revolution
TypeMulti-part: volume and area
DifficultyModerate -0.3 This is a straightforward volumes of revolution question requiring standard integration techniques. Part (i) involves integrating (3x+2)^{-1/2} using the reverse chain rule, and part (ii) uses the standard formula V = π∫y² dx which simplifies to a linear integration. Both parts are routine applications of C3 techniques with no problem-solving insight required, making it slightly easier than average.
Spec1.08b Integrate x^n: where n != -1 and sums4.08d Volumes of revolution: about x and y axes

6 \includegraphics[max width=\textwidth, alt={}, center]{1216a06e-7e14-48d7-a7ca-7acd8d71af5f-3_483_956_264_593} The diagram shows the curve with equation \(y = \frac { 1 } { \sqrt { 3 x + 2 } }\). The shaded region is bounded by the curve and the lines \(x = 0 , x = 2\) and \(y = 0\).
  1. Find the exact area of the shaded region.
  2. The shaded region is rotated completely about the \(x\)-axis. Find the exact volume of the solid formed, simplifying your answer.

Question 6:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Obtain integral of form \(k(3x+2)^{\frac{1}{2}}\)M1 any constant \(k\)
Obtain correct \(\frac{2}{3}(3x+2)^{\frac{3}{2}}\)A1 or equiv
Substitute limits 0 and 2 and attempt evaluationM1 for integral of form \(k(3x+2)^n\)
Obtain \(\frac{2}{3}(8^{\frac{3}{2}} - 2^{\frac{3}{2}})\)A1 4 or exact equiv suitably simplified
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
State or imply \(\pi\int\frac{1}{3x+2}\,dx\) or unsimplified versionB1 allow if \(dx\) absent or wrong
Obtain integral of form \(k\ln(3x+2)\)M1 any constant \(k\) involving \(\pi\) or not
Obtain \(\frac{1}{3}\pi\ln(3x+2)\) or \(\frac{1}{3}\ln(3x+2)\)A1
Show correct use of \(\ln a - \ln b\) propertyM1
Obtain \(\frac{1}{3}\pi\ln 4\)A1 5 or (similarly simplified) equiv
# Question 6:

## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Obtain integral of form $k(3x+2)^{\frac{1}{2}}$ | M1 | any constant $k$ |
| Obtain correct $\frac{2}{3}(3x+2)^{\frac{3}{2}}$ | A1 | or equiv |
| Substitute limits 0 and 2 and attempt evaluation | M1 | for integral of form $k(3x+2)^n$ |
| Obtain $\frac{2}{3}(8^{\frac{3}{2}} - 2^{\frac{3}{2}})$ | A1 | **4** or exact equiv suitably simplified |

## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply $\pi\int\frac{1}{3x+2}\,dx$ or unsimplified version | B1 | allow if $dx$ absent or wrong |
| Obtain integral of form $k\ln(3x+2)$ | M1 | any constant $k$ involving $\pi$ or not |
| Obtain $\frac{1}{3}\pi\ln(3x+2)$ or $\frac{1}{3}\ln(3x+2)$ | A1 | |
| Show correct use of $\ln a - \ln b$ property | M1 | |
| Obtain $\frac{1}{3}\pi\ln 4$ | A1 | **5** or (similarly simplified) equiv |

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6\\
\includegraphics[max width=\textwidth, alt={}, center]{1216a06e-7e14-48d7-a7ca-7acd8d71af5f-3_483_956_264_593}

The diagram shows the curve with equation $y = \frac { 1 } { \sqrt { 3 x + 2 } }$. The shaded region is bounded by the curve and the lines $x = 0 , x = 2$ and $y = 0$.\\
(i) Find the exact area of the shaded region.\\
(ii) The shaded region is rotated completely about the $x$-axis. Find the exact volume of the solid formed, simplifying your answer.

\hfill \mbox{\textit{OCR C3 2007 Q6 [9]}}