Multi-part: volume and area

11 \includegraphics[max width=\textwidth, alt={}, center]{4d8fcc3d-a2da-4d98-8500-075d10847be3-4_636_951_255_596} The diagram shows the line \(y = 1\) and part of the curve \(y = \frac { 2 } { \sqrt { } ( x + 1 ) }\).

1. Show that the equation \(y = \frac { 2 } { \sqrt { } ( x + 1 ) }\) can be written in the form \(x = \frac { 4 } { y ^ { 2 } } - 1\).
2. Find \(\int \left( \frac { 4 } { y ^ { 2 } } - 1 \right) \mathrm { d } y\). Hence find the area of the shaded region.
3. The shaded region is rotated through \(360 ^ { \circ }\) about the \(\boldsymbol { y }\)-axis. Find the exact value of the volume of revolution obtained.

10 The function f is defined by \(\mathrm { f } ( x ) = 2 x + ( x + 1 ) ^ { - 2 }\) for \(x > - 1\).

1. Find \(\mathrm { f } ^ { \prime } ( x )\) and \(\mathrm { f } ^ { \prime \prime } ( x )\) and hence verify that the function f has a minimum value at \(x = 0\). \includegraphics[max width=\textwidth, alt={}, center]{5c1ab2aa-3609-4245-b87a-98ecedc83a11-4_515_920_959_609} The points \(A \left( - \frac { 1 } { 2 } , 3 \right)\) and \(B \left( 1,2 \frac { 1 } { 4 } \right)\) lie on the curve \(y = 2 x + ( x + 1 ) ^ { - 2 }\), as shown in the diagram.
2. Find the distance \(A B\).
3. Find, showing all necessary working, the area of the shaded region. \footnotetext{Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity.
   To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at \href{http://www.cie.org.uk}{www.cie.org.uk} after the live examination series. Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. }

10 \includegraphics[max width=\textwidth, alt={}, center]{5201a3d5-7733-4d10-9de5-0c2255e3ad60-18_401_584_264_776} The diagram shows part of the curve \(y = \frac { 1 } { 2 } \left( x ^ { 4 } - 1 \right)\), defined for \(x \geqslant 0\).

1. Find, showing all necessary working, the area of the shaded region.
2. Find, showing all necessary working, the volume obtained when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
3. Find, showing all necessary working, the volume obtained when the shaded region is rotated through \(360 ^ { \circ }\) about the \(y\)-axis.

9. \begin{figure}[h] \end{figure} Figure 1 is a sketch of the curve \(C\) with equation $$y = 2 x ^ { \frac { 3 } { 2 } } ( 4 - x ) \quad x \geqslant 0$$ The point \(P\) is the stationary point of \(C\).

1. Find, using calculus, the \(x\) coordinate of \(P\). The region \(R _ { 1 }\), shown shaded in Figure 1, is bounded by \(C\) and the \(x\)-axis.
   The region \(R _ { 2 }\), also shown shaded in Figure 1, is bounded by \(C\), the \(x\)-axis and the line with equation \(x = k\), where \(k\) is a constant. Given that the area of \(R _ { 1 }\) is equal to the area of \(R _ { 2 }\)
2. find, using calculus, the exact value of \(k\).

3. \begin{figure}[h] \end{figure} The curve with equation \(y = 3 \sin \frac { x } { 2 } , 0 \leqslant x \leqslant 2 \pi\), is shown in Figure 1. The finite region enclosed by the curve and the \(x\)-axis is shaded.

1. Find, by integration, the area of the shaded region. This region is rotated through \(2 \pi\) radians about the \(x\)-axis.
2. Find the volume of the solid generated.

5. \begin{figure}[h] \end{figure} Figure 1 shows the curve with equation \(y = 4 x ^ { \frac { 1 } { 2 } } \mathrm { e } ^ { - x }\).
The shaded region is bounded by the curve, the \(x\)-axis and the line \(x = 2\).

1. Use the trapezium rule with four intervals of equal width to estimate the area of the shaded region. The shaded region is rotated through \(2 \pi\) radians about the \(x\)-axis.
2. Find, in terms of \(\pi\) and e, the exact volume of the solid formed.
   5. continued

5. \begin{figure}[h] \end{figure} Figure 1 shows the curve with equation \(y = \frac { 1 } { \sqrt { 3 x + 1 } }\).
The shaded region is bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 5\).

1. Find the area of the shaded region. The shaded region is rotated completely about the \(x\)-axis.
2. Find the volume of the solid formed, giving your answer in the form \(k \pi \ln 2\), where \(k\) is a simplified fraction.
   5. continued

6. \begin{figure}[h] \end{figure} Figure 1 shows the curve with equation \(y = x \sqrt { 1 - x } , 0 \leq x \leq 1\).

1. Use the substitution \(u ^ { 2 } = 1 - x\) to show that the area of the region bounded by the curve and the \(x\)-axis is \(\frac { 4 } { 15 }\).
2. Find, in terms of \(\pi\), the volume of the solid formed when the region bounded by the curve and the \(x\)-axis is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
   6. continued

8. \begin{figure}[h] \end{figure} Figure 2 shows the curve with equation \(y = \sqrt { \frac { x } { x + 1 } }\).
The shaded region is bounded by the curve, the \(x\)-axis and the line \(x = 3\).

1. 1. Use the trapezium rule with three strips to find an estimate for the area of the shaded region.
   2. Use the trapezium rule with six strips to find an improved estimate for the area of the shaded region. The shaded region is rotated through \(2 \pi\) radians about the \(x\)-axis.
2. Show that the volume of the solid formed is \(\pi ( 3 - \ln 4 )\).
   8. continued
   8. continued