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. {www.cie.org.uk} after the live examination series. }

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.

\begin{figure}[h] \end{figure} Figure 1 shows part of the curve \(y = \frac { 3 } { \sqrt { } ( 1 + 4 x ) }\). The region \(R\) is bounded by the curve, the \(x\)-axis, and the lines \(x = 0\) and \(x = 2\), as shown shaded in Figure 1.

1. Use integration to find the area of \(R\). The region \(R\) is rotated \(360 ^ { \circ }\) about the \(x\)-axis.
2. Use integration to find the exact value of the volume of the solid formed.

5. The diagram 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 through four right angles about the \(x\)-axis.
2. Find the volume of the solid formed, giving your answer in the form \(k \pi \ln 2\).

4. The finite region \(R\) is bounded by the curve with equation \(y = \frac { 1 } { 2 x - 1 }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\).

1. Find the exact area of \(R\).
2. Show that the volume of the solid formed when \(R\) is rotated through four right angles about the \(x\)-axis is \(\frac { 1 } { 3 } \pi\).

7. The finite region \(R\) is bounded by the curve with equation \(y = x + \frac { 2 } { x }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4\).

1. Find the exact area of \(R\). The region \(R\) is rotated completely about the \(x\)-axis.
2. Find the volume of the solid formed, giving your answer in terms of \(\pi\).

5. The finite region \(R\) is bounded by the curve with equation \(y = \sqrt [ 3 ] { 3 x - 1 }\), the \(x\)-axis and the lines \(x = \frac { 2 } { 3 }\) and \(x = 3\).

1. Find the area of \(R\).
2. Find, in terms of \(\pi\), the volume of the solid formed when \(R\) is rotated through four right angles about the \(x\)-axis.

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.

4 \includegraphics[max width=\textwidth, alt={}, center]{b6b6e55a-a5ba-466c-ac9f-b5ef5bca7a3c-2_419_707_1576_660} The diagram shows the curve $$y = \frac { 1 } { \sqrt { } ( 4 x + 1 ) }$$ The region \(R\) (shaded in the diagram) is enclosed by the curve, the axes and the line \(x = 2\).

1. Show that the exact area of \(R\) is 1 .
2. The region \(R\) is rotated completely about the \(x\)-axis. Find the exact volume of the solid formed.

7 \includegraphics[max width=\textwidth, alt={}, center]{33a2b09d-0df9-48d6-9ee9-e0a1ec345f41-3_547_851_1749_605} The diagram shows the curve \(y = \sqrt { \frac { 3 } { 4 x + 1 } }\) for \(0 \leqslant x \leqslant 20\). The point \(P\) on the curve has coordinates \(\left( 20 , \frac { 1 } { 9 } \sqrt { 3 } \right)\). The shaded region \(R\) is enclosed by the curve and the lines \(x = 0\) and \(y = \frac { 1 } { 9 } \sqrt { 3 }\).

1. Find the exact area of \(R\).
2. Find the exact volume of the solid obtained when \(R\) is rotated completely about the \(x\)-axis.

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

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

\includegraphics{figure_2} Figure 2 shows part of the curve with equation \(y = x^2 + 2\). The finite region \(R\) is bounded by the curve, the \(x\)-axis and the lines \(x = 0\) and \(x = 2\).

1. Use the trapezium rule with 4 strips of equal width to estimate the area of \(R\). [5]
2. State, with a reason, whether your answer in part \((a)\) is an under-estimate or over-estimate of the area of \(R\). [1]
3. Using integration, find the volume of the solid generated when \(R\) is rotated through \(360°\) about the \(x\)-axis, giving your answer in terms of \(\pi\). [6]