4.08d Volumes of revolution: about x and y axes

2. The diagram shows the curve with equation \(y = x \sqrt { 2 - x } , 0 \leq x \leq 2\).
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.

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 \(\pi\).

2. \includegraphics[max width=\textwidth, alt={}, center]{b124d427-1f9b-4770-95bb-ed79bae5b4fb-1_460_805_587_486} The diagram shows the curve with equation \(y = \frac { 1 } { 2 } \ln 3 x\).

1. Express the equation of the curve in the form \(x = \mathrm { f } ( y )\). The shaded region is bounded by the curve, the coordinate axes and the line \(y = 1\).
2. Find, in terms of \(\pi\) and e, the volume of the solid formed when the shaded region is rotated through four right angles about the \(y\)-axis.

4. \includegraphics[max width=\textwidth, alt={}, center]{c0b79c3c-9537-4c71-903b-01434dfb5d26-1_492_803_1562_452} The diagram shows the curves \(y = ( x - 1 ) ^ { 2 }\) and \(y = 2 - \frac { 2 } { x } , x > 0\).

1. Verify that the two curves meet at the points where \(x = 1\) and where \(x = 2\). The shaded region bounded by the two curves is rotated completely about the \(x\)-axis.
2. Find the exact volume of the solid formed.

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.

5

1. Find \(\int ( 3 x + 7 ) ^ { 9 } \mathrm {~d} x\).
2. \includegraphics[max width=\textwidth, alt={}, center]{32f90420-e1eb-47ab-b588-e3806b64813f-3_537_881_402_671} The diagram shows the curve \(y = \frac { 1 } { 2 \sqrt { x } }\). The shaded region is bounded by the curve and the lines \(x = 3 , x = 6\) and \(y = 0\). The shaded region is rotated completely about the \(x\)-axis. Find the exact volume of the solid produced, simplifying your answer.

4

1. \includegraphics[max width=\textwidth, alt={}, center]{e0e2a26b-d4d6-46ea-ac12-a882f3465e5e-2_579_785_1279_721} The diagram shows the curve \(y = \frac { 2 } { \sqrt { } x }\). The region \(R\), shaded in the diagram, is bounded by the curve and by the lines \(x = 1 , x = 5\) and \(y = 0\). The region \(R\) is rotated completely about the \(x\)-axis. Find the exact volume of the solid formed.
2. Use Simpson's rule, with 4 strips, to find an approximate value for $$\int _ { 1 } ^ { 5 } \sqrt { } \left( x ^ { 2 } + 1 \right) \mathrm { d } x ,$$ giving your answer correct to 3 decimal places.

9 \includegraphics[max width=\textwidth, alt={}, center]{ebfdf170-99c6-4785-b9d7-201c3425b4c9-4_556_720_676_715} The diagram shows the curve with equation \(y = 2 \ln ( x - 1 )\). The point \(P\) has coordinates ( \(0 , p\) ). The region \(R\), shaded in the diagram, is bounded by the curve and the lines \(x = 0 , y = 0\) and \(y = p\). The units on the axes are centimetres. The region \(R\) is rotated completely about the \(\boldsymbol { y }\)-axis to form a solid.

1. Show that the volume, \(V \mathrm {~cm} ^ { 3 }\), of the solid is given by $$V = \pi \left( \mathrm { e } ^ { p } + 4 \mathrm { e } ^ { \frac { 1 } { 2 } p } + p - 5 \right) .$$
2. It is given that the point \(P\) is moving in the positive direction along the \(y\)-axis at a constant rate of \(0.2 \mathrm {~cm} \mathrm {~min} ^ { - 1 }\). Find the rate at which the volume of the solid is increasing at the instant when \(p = 4\), giving your answer correct to 2 significant figures.

8

1. Given that \(\mathrm { y } = \frac { 4 \ln \mathrm { x } - 3 } { 4 \ln \mathrm { x } + 3 }\), show that \(\frac { \mathrm { dy } } { \mathrm { dx } } = \frac { 24 } { \mathrm { x } ( 4 \ln \mathrm { x } + 3 ) ^ { 2 } }\).
2. Find the exact value of the gradient of the curve \(y = \frac { 4 \ln x - 3 } { 4 \ln x + 3 }\) at the point where it crosses the \(x\)-axis.
3. \includegraphics[max width=\textwidth, alt={}, center]{133c38fb-307f-4f20-86cb-1bd57cc4f870-3_524_830_941_699} The diagram shows part of the curve with equation $$\mathrm { y } = \frac { 2 } { \mathrm { x } ^ { \frac { 1 } { 2 } } ( 4 \ln \mathrm { x } + 3 ) }$$ The region shaded in the diagram is bounded by the curve and the lines \(x = 1 , x = e\) and \(y = 0\). Find the exact volume of the solid produced when this shaded region is rotated completely about the x -axis.

6 \includegraphics[max width=\textwidth, alt={}, center]{5c501214-b41c-43a8-b9c6-986758e83e7d-3_586_798_267_676} The diagram shows the curves \(y = \mathrm { e } ^ { 3 x }\) and \(y = ( 2 x - 1 ) ^ { 4 }\). The shaded region is bounded by the two curves and the line \(x = \frac { 1 } { 2 }\). The shaded region is rotated completely about the \(x\)-axis. Find the exact volume of the solid produced.

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.

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the region \(R\) bounded by the curve with equation \(y = \mathrm { e } ^ { - x }\), the line \(x = 1\), the \(x\)-axis and the \(y\)-axis. A uniform solid \(S\) is formed by rotating \(R\) through \(2 \pi\) radians about the \(x\)-axis.

1. Show that the volume of \(S\) is \(\frac { \pi } { 2 } \left( 1 - \mathrm { e } ^ { - 2 } \right)\).
2. Find, in terms of e, the distance of the centre of mass of \(S\) from \(O\).

1. \begin{figure}[h] \end{figure} The shaded region \(R\) is bounded by the curve with equation \(y ^ { 2 } = 9 ( 4 - x )\), the positive \(x\)-axis and the positive \(y\)-axis, as shown in Figure 1. A uniform solid \(S\) is formed by rotating \(R\) through \(360 ^ { \circ }\) about the \(x\)-axis.
Use algebraic integration to find the \(x\) coordinate of the centre of mass of \(S\).

5. \begin{figure}[h] \end{figure} Figure 3 shows the finite region \(R\) which is bounded by part of the curve with equation \(y = \sin x\), the \(x\)-axis and the line with equation \(x = \frac { \pi } { 2 }\). A uniform solid \(S\) is formed by rotating \(R\) through \(2 \pi\) radians about the \(x\)-axis. Using algebraic integration,

1. show that the volume of \(S\) is \(\frac { \pi ^ { 2 } } { 4 }\)
2. find, in terms of \(\pi\), the \(x\) coordinate of the centre of mass of \(S\).

1. \begin{figure}[h] \end{figure} The region \(R\), shown shaded in Figure 1, is bounded by the curve with equation \(y = \frac { 1 } { x }\), the line with equation \(x = 1\), the positive \(x\)-axis and the line with equation \(x = a\) where \(a > 1\) A uniform solid \(S\) is formed by rotating \(R\) through \(2 \pi\) radians about the \(x\)-axis.

1. Show that the volume of \(S\) is $$\pi \left( 1 - \frac { 1 } { a } \right)$$
2. Find the \(x\) coordinate of the centre of mass of \(S\).

3. \begin{figure}[h] \end{figure} The shaded region in Figure 2 is bounded by the \(x\)-axis, the line with equation \(x = 2\) and the curve with equation \(y = \frac { 1 } { 4 } x ( 3 - x )\).
This region is rotated through \(2 \pi\) radians about the \(x\)-axis, to form a solid of revolution which is used to model a uniform solid \(S\). The volume of \(S\) is \(\frac { 2 } { 5 } \pi\)

1. Use the model and algebraic integration to show that the \(x\) coordinate of the centre of mass of \(S\) is \(\frac { 31 } { 24 }\) The solid \(S\) is placed with its circular face on a rough plane which is inclined at \(\alpha ^ { \circ }\) to the horizontal. The plane is sufficiently rough to prevent \(S\) from sliding. The solid \(S\) is on the point of toppling.
2. Find the value of \(\alpha\)

6. \begin{figure}[h] \end{figure} The shaded region \(R\) is bounded by part of the curve with equation \(y = x ^ { 2 } + 3\), the \(x\)-axis, the \(y\)-axis and the line with equation \(x = 2\), as shown in Figure 4. The unit of length on each axis is one centimetre. The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid \(S\).
Using algebraic integration,

1. show that the volume of \(S\) is \(\frac { 202 } { 5 } \pi \mathrm {~cm} ^ { 3 }\),
2. show that, to 2 decimal places, the centre of mass of \(S\) is 1.30 cm from \(O\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{b7cfcf0a-8f54-4350-8e07-a3b51d94d0f2-11_478_472_1407_762} \captionsetup{labelformat=empty} \caption{Figure 5}
   \end{figure} A uniform right circular solid cone, of base radius 7 cm and height 6 cm , is joined to \(S\) to form a solid \(T\). The base of the cone coincides with the larger plane face of \(S\), as shown in Figure 5. The vertex of the cone is \(V\).
   The mass per unit volume of \(S\) is twice the mass per unit volume of the cone.
3. Find the distance from \(V\) to the centre of mass of \(T\). The point \(A\) lies on the circumference of the base of the cone. The solid \(T\) is suspended from \(A\) and hangs freely in equilibrium.
4. Find the size of the angle between \(V A\) and the vertical.

1. The region enclosed by the curve with equation \(y = \frac { 1 } { 2 } \sqrt { x }\), the \(x\)-axis and the lines \(x = 2\) and \(x = 4\), is rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid \(S\). Use algebraic integration to find the exact value of the \(x\) coordinate of the centre of mass of \(S\).
   (6)

"都 D \(\_\_\_\_\) 1

1. The finite region enclosed by the curve with equation \(y = 3 - \sqrt { x }\) and the lines \(x = 0\) and \(y = 0\) is rotated through \(2 \pi\) radians about the \(x\)-axis, to form a uniform solid \(S\).

Use algebraic integration to

1. show that the volume of \(S\) is \(\frac { 27 } { 2 } \pi\)
2. find the \(x\) coordinate of the centre of mass of \(S\).

6. \begin{figure}[h] \end{figure} The shaded region \(R\) is bounded by part of the curve with equation \(y = x ^ { 2 } + 3\), the \(x\)-axis, the \(y\)-axis and the line with equation \(x = 2\), as shown in Figure 4. The unit of length on each axis is one centimetre. The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid \(S\).
Using algebraic integration,

1. show that the volume of \(S\) is \(\frac { 202 } { 5 } \pi \mathrm {~cm} ^ { 3 }\),
2. show that, to 2 decimal places, the centre of mass of \(S\) is 1.30 cm from \(O\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{bb73b211-7629-4ed7-9b71-91841c29bb85-20_483_469_1402_767} \captionsetup{labelformat=empty} \caption{Figure 5}
   \end{figure} A uniform right circular solid cone, of base radius 7 cm and height 6 cm , is joined to \(S\) to form a solid \(T\). The base of the cone coincides with the larger plane face of \(S\), as shown in Figure 5. The vertex of the cone is \(V\).
   The mass per unit volume of \(S\) is twice the mass per unit volume of the cone.
3. Find the distance from \(V\) to the centre of mass of \(T\). The point \(A\) lies on the circumference of the base of the cone. The solid \(T\) is suspended from \(A\) and hangs freely in equilibrium.
4. Find the size of the angle between \(V A\) and the vertical.

   Leave
   blank
   Q6

   VIIIV SIHI NI JAIIM ION OC

   VIIIV SIHI NI JIHMM ION OO

   VI4V SIHI NI JIIYM IONOO

6. \begin{figure}[h] \end{figure} The shaded region \(R\) is bounded by the curve with equation \(y = \frac { 1 } { 2 x ^ { 2 } }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\), as shown in Figure 4. The unit of length on each axis is 1 m . A uniform solid \(S\) has the shape made by rotating \(R\) through \(360 ^ { \circ }\) about the \(x\)-axis.

1. Show that the centre of mass of \(S\) is \(\frac { 2 } { 7 } \mathrm {~m}\) from its larger plane face. \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 5} \includegraphics[alt={},max width=\textwidth]{ab85ec29-b1fc-45a9-9343-09feb33ab6c5-010_616_431_1420_778}
   \end{figure} A sporting trophy \(T\) is a uniform solid hemisphere \(H\) joined to the solid \(S\). The hemisphere has radius \(\frac { 1 } { 2 } \mathrm {~m}\) and its plane face coincides with the larger plane face of \(S\), as shown in Figure 5. Both \(H\) and \(S\) are made of the same material.
2. Find the distance of the centre of mass of \(T\) from its plane face.

6. \begin{figure}[h] \end{figure} The region \(R\) is bounded by part of the curve with equation \(y = 4 - x ^ { 2 }\), the positive \(x\)-axis and the positive \(y\)-axis, as shown in Figure 3. The unit of length on both axes is one metre. A uniform solid \(S\) is formed by rotating \(R\) through \(360 ^ { \circ }\) about the \(x\)-axis.

1. Show that the centre of mass of \(S\) is \(\frac { 5 } { 8 } \mathrm {~m}\) from \(O\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{8374fa0f-cb28-497f-8696-877d7d0762f1-09_702_584_1138_676} \captionsetup{labelformat=empty} \caption{Figure 4}
   \end{figure} Figure 4 shows a cross section of a uniform solid \(P\) consisting of two components, a solid cylinder \(C\) and the solid \(S\). The cylinder \(C\) has radius 4 m and length \(l\) metres. One end of \(C\) coincides with the plane circular face of \(S\). The point \(A\) is on the circumference of the circular face common to \(C\) and \(S\). When the solid \(P\) is freely suspended from \(A\), the solid \(P\) hangs with its axis of symmetry horizontal.
2. Find the value of \(l\).

5. \begin{figure}[h] \end{figure} The shaded region \(R\) is bounded by the curve with equation \(y = ( x + 1 ) ^ { 2 }\), the \(x\)-axis, the \(y\)-axis and the line with equation \(x = 2\), as shown in Figure 3. The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid \(S\).

1. Use algebraic integration to find the \(x\) coordinate of the centre of mass of \(S\).
   (8) \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{f6ab162c-8473-4464-ad62-87a359d85ab3-08_558_492_1263_703} \captionsetup{labelformat=empty} \caption{Figure 4}
   \end{figure} A uniform solid hemisphere is fixed to \(S\) to form a solid \(T\). The hemisphere has the same radius as the smaller plane face of \(S\) and its plane face coincides with the smaller plane face of \(S\), as shown in Figure 4. The mass per unit volume of the hemisphere is 10 times the mass per unit volume of \(S\). The centre of the circular plane face of \(T\) is \(A\). All lengths are measured in centimetres.
2. Find the distance of the centre of mass of \(T\) from \(A\).

5. \begin{figure}[h] \end{figure} Figure 4 shows the region \(R\) bounded by part of the curve with equation \(y = \cos x\), the \(x\)-axis and the \(y\)-axis. A uniform solid \(S\) is formed by rotating \(R\) through \(2 \pi\) radians about the \(x\)-axis.

1. Show that the volume of \(S\) is \(\frac { \pi ^ { 2 } } { 4 }\)
2. Find, using algebraic integration, the \(x\) coordinate of the centre of mass of \(S\).