OCR MEI C4 (Core Mathematics 4)

1 Fig. 6 shows the region enclosed by the curve \(y = \left( 1 + 2 x ^ { 2 } \right) ^ { \frac { 1 } { 3 } }\) and the line \(y = 2\). \begin{figure}[h] \end{figure} This region is rotated about the \(y\)-axis. Find the volume of revolution formed, giving your answer as a multiple of \(\pi\).

2 Fig. 7a shows the curve with the parametric equations $$x = 2 \cos \theta , \quad y = \sin 2 \theta , \quad - \frac { \pi } { 2 } \leqslant \theta \leqslant \frac { \pi } { 2 } .$$ The curve meets the \(x\)-axis at O and P . Q and R are turning points on the curve. The scales on the axes are the same. \begin{figure}[h] \end{figure}

1. State, with their coordinates, the points on the curve for which \(\theta = - \frac { \pi } { 2 } , \theta = 0\) and \(\theta = \frac { \pi } { 2 }\).
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Hence find the gradient of the curve when \(\theta = \frac { \pi } { 2 }\), and verify that the two tangents to the curve at the origin meet at right angles.
3. Find the exact coordinates of the turning point Q . When the curve is rotated about the \(x\)-axis, it forms a paperweight shape, as shown in Fig. 7b. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-2_324_389_1692_857} \captionsetup{labelformat=empty} \caption{Fig. 7b}
   \end{figure}
4. Express \(\sin ^ { 2 } \theta\) in terms of \(x\). Hence show that the cartesian equation of the curve is \(y ^ { 2 } = x ^ { 2 } \left( 1 - \frac { 1 } { 4 } x ^ { 2 } \right)\).
5. Find the volume of the paperweight shape.

3 Fig. 6 shows the region enclosed by part of the curve \(y = 2 x ^ { 2 }\), the straight line \(x + y = 3\), and the \(y\)-axis. The curve and the straight line meet at \(\mathrm { P } ( 1,2 )\). \begin{figure}[h] \end{figure} The shaded region is rotated through \(360 ^ { \circ }\) about the \(y\)-axis. Find, in terms of \(\pi\), the volume of the solid of revolution formed.
[0pt] [You may use the formula \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) for the volume of a cone.]

4 The part of the curve \(y = 4 - x ^ { 2 }\) that is above the \(x\)-axis is rotated about the \(y\)-axis. This is shown in Fig. 4. Find the volume of revolution produced, giving your answer in terms of \(\pi\). \begin{figure}[h] \end{figure}

5 Fig. 2 shows the curve \(y = \sqrt { } 1 + \mathrm { e } ^ { 2 x }\). \begin{figure}[h] \end{figure} The region bounded by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 1\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Show that the volume of the solid of revolution produced is \(\frac { 1 } { 2 } \pi \left( 1 + \mathrm { e } ^ { 2 } \right)\).

6 Fig. 3 shows the curve \(y = \ln x\) and part of the line \(y = 2\). \begin{figure}[h] \end{figure} The shaded region is rotated through \(360 ^ { \circ }\) about the \(y\)-axis.

1. Show that the volume of the solid of revolution formed is given by \(\int _ { 0 } ^ { 2 } \pi \mathrm { e } ^ { 2 y } \mathrm {~d} y\).
2. Evaluate this, leaving your answer in an exact form.

7

1. Show that \(\int x \mathrm { e } ^ { - 2 x } \mathrm {~d} x = - \frac { 1 } { 4 } \mathrm { e } ^ { - 2 x } ( 1 + 2 x ) + c\). A vase is made in the shape of the volume of revolution of the curve \(y = x ^ { 1 / 2 } \mathrm { e } ^ { - x }\) about the \(x\)-axis between \(x = 0\) and \(x = 2\) (see Fig. 5). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-5_718_751_638_654} \captionsetup{labelformat=empty} \caption{Fig. 5}
   \end{figure}
2. Show that this volume of revolution is \(\frac { 1 } { 4 } \pi \left( 1 \frac { 5 } { \mathrm { e } ^ { 4 } } \right)\).

8 Fig. 4 shows a sketch of the region enclosed by the curve . \(\mathrm { J } 1 + \mathrm { e } - 2 \mathrm { x }\), the x -axis, the y -axis and the line \(\boldsymbol { x } = 1\). \begin{figure}[h] \end{figure} Find the volume of the solid generated when this region is rotated through \(360 ^ { \circ }\) about the \(\boldsymbol { x }\)-axis. Give your answer in an exact form.