4.08d Volumes of revolution: about x and y axes

3 Fig. 7 shows the curve BC defined by the parametric equations $$x = 5 \ln u , y = u + \frac { 1 } { u } , \quad 1 \leqslant u \leqslant 10$$ The point A lies on the \(x\)-axis and AC is parallel to the \(y\)-axis. The tangent to the curve at C makes an angle \(\theta\) with AC, as shown. \begin{figure}[h] \end{figure}

1. Find the lengths \(\mathrm { OA } , \mathrm { OB }\) and AC .
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(u\). Hence find the angle \(\theta\).
3. Show that the cartesian equation of the curve is \(y = \mathrm { e } ^ { \frac { 1 } { 5 } x } + \mathrm { e } ^ { - \frac { 1 } { 5 } x }\). An object is formed by rotating the region OACB through \(360 ^ { \circ }\) about \(\mathrm { O } x\).
4. Find the volume of the object.

4 Fig. 2 shows the curve \(y = \sqrt { 1 + x ^ { 2 } }\). \begin{figure}[h] \end{figure}

1. The following table gives some values of \(x\) and \(y\).

   \(x\)	0	0.25	0.5	0.75	1
   \(y\)	1	1.0308		1.25	1.4142

   Find the missing value of \(y\), giving your answer correct to 4 decimal places. Hence show that, using the trapezium rule with four strips, the shaded area is approximately 1.151 square units.
2. Jenny uses a trapezium rule with 8 strips, and obtains a value of 1.158 square units. Explain why she must have made a mistake.
3. The shaded area is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Find the exact volume of the solid of revolution formed.

5 Fig. 4 shows the curve \(y = \sqrt { 1 + \mathrm { e } ^ { 2 x } }\), and the region between the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 2\). \begin{figure}[h] \end{figure}

1. Find the exact volume of revolution when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
2. 1. Complete the table of values, and use the trapezium rule with 4 strips to estimate the area of the shaded region.

      \(x\)	0	0.5	1	1.5	2
      \(y\)		1.9283	2.8964	4.5919

   2. The trapezium rule for \(\int _ { 0 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { 2 x } } \mathrm {~d} x\) with 8 and 16 strips gives 6.797 and 6.823, although not necessarily in that order. Without doing the calculations, say which result is which, explaining your reasoning.

6 Fig. 8 illustrates a hot air balloon on its side. The balloon is modelled by the volume of revolution about the \(x\)-axis of the curve with parametric equations $$x = 2 + 2 \sin \theta , \quad y = 2 \cos \theta + \sin 2 \theta , \quad ( 0 \leqslant \theta \leqslant 2 \pi ) .$$ The curve crosses the \(x\)-axis at the point \(\mathrm { A } ( 4,0 )\). B and C are maximum and minimum points on the curve. Units on the axes are metres. \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\).
2. Verify that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(\theta = \frac { 1 } { 6 } \pi\), and find the exact coordinates of B . Hence find the maximum width BC of the balloon.
3. (A) Show that \(y = x \cos \theta\).
   (B) Find \(\sin \theta\) in terms of \(x\) and show that \(\cos ^ { 2 } \theta = x - \frac { 1 } { 4 } x ^ { 2 }\).
   (C) Hence show that the cartesian equation of the curve is \(y ^ { 2 } = x ^ { 3 } - \frac { 1 } { 4 } x ^ { 4 }\).
4. Find the volume of the balloon.

2 Fig. 7 shows the curve BC defined by the parametric equations $$x = 5 \ln u , y = u + \frac { 1 } { u } , \quad 1 \leqslant u \leqslant 10$$ The point A lies on the \(x\)-axis and AC is parallel to the \(y\)-axis. The tangent to the curve at C makes an angle \(\theta\) with AC, as shown. \begin{figure}[h] \end{figure}

1. Find the lengths \(\mathrm { OA } , \mathrm { OB }\) and AC .
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(u\). Hence find the angle \(\theta\).
3. Show that the cartesian equation of the curve is \(y = \mathrm { e } ^ { \frac { 1 } { 5 } x } + \mathrm { e } ^ { - \frac { 1 } { 5 } x }\). An object is formed by rotating the region OACB through \(360 ^ { \circ }\) about \(\mathrm { O } x\).
4. Find the volume of the object.

1 Fig. 8 illustrates a hot air balloon on its side. The balloon is modelled by the volume of revolution about the \(x\)-axis of the curve with parametric equations $$x = 2 + 2 \sin \theta , \quad y = 2 \cos \theta + \sin 2 \theta , \quad ( 0 \leqslant \theta \leqslant 2 \pi ) .$$ The curve crosses the \(x\)-axis at the point \(\mathrm { A } ( 4,0 )\). B and C are maximum and minimum points on the curve. Units on the axes are metres. \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\).
2. Verify that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(\theta = \frac { 1 } { 6 } \pi\), and find the exact coordinates of B . Hence find the maximum width BC of the balloon.
3. (A) Show that \(y = x \cos \theta\).
   (B) Find \(\sin \theta\) in terms of \(x\) and show that \(\cos ^ { 2 } \theta = x - \frac { 1 } { 4 } x ^ { 2 }\).
   (C) Hence show that the cartesian equation of the curve is \(y ^ { 2 } = x ^ { 3 } - \frac { 1 } { 4 } x ^ { 4 }\).
4. Find the volume of the balloon.

4 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]{1601927c-74d7-4cc2-a7f2-2c2a2e8c2c4c-4_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.
6. Express \(\frac { 3 } { ( y - 2 ) ( y + 1 ) }\) in partial fractions.
7. Hence, given that \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = x ^ { 2 } ( y - 2 ) ( y + 1 )$$ show that \(\frac { y - 2 } { y + 1 } = A \mathrm { e } ^ { x ^ { 3 } }\), where \(A\) is a constant.

1 A curve has parametric equations \(x = \sec \theta , y = 2 \tan \theta\).

1. Given that the derivative of \(\sec \theta\) is \(\sec \theta \tan \theta\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \operatorname { cosec } \theta\).
2. Verify that the cartesian equation of the curve is \(y ^ { 2 } = 4 x ^ { 2 } - 4\). Fig. 5 shows the region enclosed by the curve and the line \(x = 2\). This region is rotated through \(180 ^ { \circ }\) about the \(x\)-axis. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1e788cb0-36b0-42a9-9e0c-077022d410ae-1_556_867_580_588} \captionsetup{labelformat=empty} \caption{Fig. 5}
   \end{figure}
3. Find the volume of revolution produced, giving your answer in exact form.

9

1. Prove that \(\int _ { 0 } ^ { N } \ln ( 1 + x ) \mathrm { d } x = ( N + 1 ) \ln ( N + 1 ) - N\), where \(N\) is a positive constant.
2. \includegraphics[max width=\textwidth, alt={}, center]{63a316f6-1c18-4224-930f-0b58112c9f71-4_616_1261_406_482} The diagram shows the curve \(y = \ln ( 1 + x )\), for \(0 \leqslant x \leqslant 70\), together with a set of rectangles of unit width.
   1. By considering the areas of these rectangles, explain why $$\ln 2 + \ln 3 + \ln 4 + \ldots + \ln 70 < \int _ { 0 } ^ { 70 } \ln ( 1 + x ) d x$$
   2. By considering the areas of another set of rectangles, show that $$\ln 2 + \ln 3 + \ln 4 + \ldots + \ln 70 > \int _ { 0 } ^ { 69 } \ln ( 1 + x ) d x$$
   3. Hence find bounds between which \(\ln ( 70 ! )\) lies. Give the answers correct to 1 decimal place.

8

1. Use the substitution \(x = \cosh ^ { 2 } u\) to find \(\int \sqrt { \frac { x } { x - 1 } } \mathrm {~d} x\), giving your answer in the form \(\mathrm { f } ( x ) + \ln ( \mathrm { g } ( x ) )\). \includegraphics[max width=\textwidth, alt={}, center]{d25d17c4-a87c-4dcf-900c-400086af6610-3_693_1041_927_593}
2. Hence calculate the exact area of the region between the curve \(y = \sqrt { \frac { x } { x - 1 } }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4\) (see diagram).
3. What can you say about the volume of the solid of revolution obtained when the region defined in part (ii) is rotated completely about the \(x\)-axis? Justify your answer.

7

1. By using a set of rectangles of unit width to approximate an area under the curve \(y = \frac { 1 } { x }\), show that \(\sum _ { x = 1 } ^ { \infty } \frac { 1 } { x }\) is infinite.
2. By using a set of rectangles of unit width to approximate an area under the curve \(y = \frac { 1 } { x ^ { 2 } }\), find an upper limit for the series \(\sum _ { x = 1 } ^ { \infty } \frac { 1 } { x ^ { 2 } }\).

5 \includegraphics[max width=\textwidth, alt={}, center]{e4e1c424-8dd5-4d18-9950-e902de0301b0-3_444_999_1258_539} The diagram shows the curve \(y = \frac { 1 } { x + 1 }\) together with four rectangles of unit width.

1. Explain how the diagram shows that $$\frac { 1 } { 2 } + \frac { 1 } { 3 } + \frac { 1 } { 4 } + \frac { 1 } { 5 } < \int _ { 0 } ^ { 4 } \frac { 1 } { x + 1 } \mathrm {~d} x$$ The curve \(y = \frac { 1 } { x + 2 }\) passes through the top left-hand corner of each of the four rectangles shown.
2. By considering the rectangles in relation to this curve, write down a second inequality involving \(\frac { 1 } { 2 } + \frac { 1 } { 3 } + \frac { 1 } { 4 } + \frac { 1 } { 5 }\) and a definite integral.
3. By considering a suitable range of integration and corresponding rectangles, show that $$\ln ( 500.5 ) < \sum _ { r = 2 } ^ { 1000 } \frac { 1 } { r } < \ln ( 1000 ) .$$

1. \hspace{0pt} [In this question you may assume the following formulae for the volume and curved] surface area of a cone of base radius \(r\) and height \(h\) and of a sphere of radius \(r\).

Cone: volume \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) and curved surface area \(S = \pi r \sqrt { h ^ { 2 } + r ^ { 2 } }\) Sphere: volume \(V = \frac { 4 \pi } { 3 } r ^ { 3 }\) and curved surface area \(S = 4 \pi r ^ { 2 }\) \includegraphics[max width=\textwidth, alt={}, center]{78ba3acc-4cca-4d15-8362-a27e425c5859-22_782_755_637_657} Figure 3
Figure 3 shows the design for a garden ornament.
The ornament is made of a hemisphere on top of a truncated cone.
The truncated cone has base radius \(2 r \mathrm {~cm}\), top radius \(r \mathrm {~cm}\) and height \(4 r \mathrm {~cm}\).
The hemisphere has radius \(R \mathrm {~cm}\).
Given that the volume of the ornament is \(2100 \pi \mathrm {~cm} ^ { 3 }\)

1. show that $$R ^ { 3 } = 3150 - 14 r ^ { 3 }$$
2. Find an expression involving \(\frac { \mathrm { d } R } { \mathrm {~d} r }\) in terms of \(r\) and/or \(R\). The base of the truncated cone of the ornament is fixed to the ground.
3. Show that the visible surface area of the ornament, \(A \mathrm {~cm} ^ { 2 }\), is given by $$A = ( 3 \sqrt { 17 } - 1 ) \pi r ^ { 2 } + 3 \pi R ^ { 2 }$$
4. Hence show that $$\frac { \mathrm { d } A } { \mathrm {~d} r } = \gamma \pi r - \frac { \delta \pi r ^ { 2 } } { R }$$ where \(\gamma\) and \(\delta\) are real numbers to be determined. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{78ba3acc-4cca-4d15-8362-a27e425c5859-23_705_803_625_630} \captionsetup{labelformat=empty} \caption{Figure 4}
   \end{figure} Figure 4 shows a sketch of \(A\) against \(r\), for \(r \geqslant 0\) There is a local minimum at \(r = 0\) and a local maximum at the point \(M\). The overall minimum point is at the point \(N\), where the gradient of the curve is undefined.
5. 1. Determine the \(r\) coordinate of the point \(N\).
   2. Explain why, for the ornament, \(r\) must be less than this value.
6. Show that the \(r\) coordinate of the point \(M\) is $$\sqrt [ 3 ] { \frac { p ( 3 \sqrt { 17 } - 1 ) ^ { 3 } } { 3 q ^ { 2 } + ( 3 \sqrt { 17 } - 1 ) ^ { 3 } } }$$ where \(p\) and \(q\) are integers to be determined.

12. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the curve with parametric equations $$x = 3 \sin t , \quad y = 2 \sin 2 t , \quad 0 \leqslant t \leqslant \frac { \pi } { 2 }$$ The finite region \(S\), shown shaded in Figure 3, is bounded by the curve, the \(x\)-axis and the line with equation \(x = \frac { 3 } { 2 }\) The shaded region \(S\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.

1. Show that the volume of the solid of revolution is given by $$k \int _ { 0 } ^ { a } \sin ^ { 2 } t \cos ^ { 3 } t \mathrm {~d} t$$ where \(k\) and \(a\) are constants to be given in terms of \(\pi\).
2. Use the substitution \(u = \sin t\), or otherwise, to find the exact value of this volume, giving your answer in the form \(\frac { p \pi } { q }\) where \(p\) and \(q\) are integers. (Solutions based entirely on graphical or numerical methods are not acceptable.)

5 \includegraphics[max width=\textwidth, alt={}, center]{774bb427-5392-45d3-8e4e-47d08fb8a792-02_559_1191_1749_479} The diagram shows the curve with equation \(y = \frac { 6 } { \sqrt { 3 x - 2 } }\). The region \(R\), shaded in the diagram, is bounded by the curve and the lines \(x = 1 , x = a\) and \(y = 0\), where \(a\) is a constant greater than 1 . It is given that the area of \(R\) is 16 square units. Find the value of \(a\) and hence find the exact volume of the solid formed when \(R\) is rotated completely about the \(x\)-axis.
[0pt] [9]

2 \includegraphics[max width=\textwidth, alt={}, center]{89e54367-bb83-483a-add5-0527b71a5cac-2_490_713_447_660} The diagram shows part of the curve \(y = \frac { 6 } { ( 2 x + 1 ) ^ { 2 } }\). The shaded region is bounded by the curve and the lines \(x = 0 , x = 1\) and \(y = 0\). Find the exact volume of the solid produced when this shaded region is rotated completely about the \(x\)-axis.

9

1. Show that the derivative with respect to \(y\) of $$y \ln ( 2 y ) - y$$ is \(\ln ( 2 y )\).
2. 
   
   [diagram]
   The diagram shows the curve with equation \(y = \frac { 1 } { 2 } \mathrm { e } ^ { x ^ { 2 } }\). The point \(P \left( 2 , \frac { 1 } { 2 } \mathrm { e } ^ { 4 } \right)\) lies on the curve. The shaded region is bounded by the curve and the lines \(x = 0\) and \(y = \frac { 1 } { 2 } e ^ { 4 }\). Find the exact volume of the solid produced when the shaded region is rotated completely about the \(y\)-axis.
3. Hence find the volume of the solid produced when the region bounded by the curve and the lines \(x = 0\), \(x = 2\) and \(y = 0\) is rotated completely about the \(y\)-axis.

6 \includegraphics[max width=\textwidth, alt={}, center]{cf154c94-6248-4dda-91e8-61349cc10482-3_606_846_251_614} The diagram shows the curves \(y = \cos x\) and \(y = \sin x\), for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). The region \(R\) is bounded by the curves and the \(x\)-axis. Find the volume of the solid of revolution formed when \(R\) is rotated completely about the \(x\)-axis, giving your answer in terms of \(\pi\).

9

1. Find \(\int ( x + \cos 2 x ) ^ { 2 } \mathrm {~d} x\).
2. \includegraphics[max width=\textwidth, alt={}, center]{80f94db1-39be-46f5-896e-277c93cbe4b8-3_538_935_383_646} The diagram shows the part of the curve \(y = x + \cos 2 x\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). The shaded region bounded by the curve, the axes and the line \(x = \frac { 1 } { 2 } \pi\) is rotated completely about the \(x\)-axis to form a solid of revolution of volume \(V\). Find \(V\), giving your answer in an exact form.

9

1. Show that \(\frac { \mathrm { d } } { \mathrm { d } x } ( x \ln x - x ) = \ln x\).
2. \includegraphics[max width=\textwidth, alt={}, center]{8492b214-aaac-4354-8649-e317bf7b3535-3_481_725_1064_751} In the diagram, \(C\) is the curve \(y = \ln x\). The region \(R\) is bounded by \(C\), the \(x\)-axis and the line \(x = \mathrm { e }\).
   1. Find the exact volume of the solid of revolution formed by rotating \(R\) completely about the \(x\)-axis.
   2. The region \(R\) is rotated completely about the \(y\)-axis. Explain why the volume of the solid of revolution formed is given by $$\pi \mathrm { e } ^ { 2 } - \pi \int _ { 0 } ^ { 1 } \mathrm { e } ^ { 2 y } \mathrm {~d} y ,$$ and find this volume.

8 Fig. 8 illustrates a hot air balloon on its side. The balloon is modelled by the volume of revolution about the \(x\)-axis of the curve with parametric equations $$x = 2 + 2 \sin \theta , \quad y = 2 \cos \theta + \sin 2 \theta , \quad ( 0 \leqslant \theta \leqslant 2 \pi ) .$$ The curve crosses the \(x\)-axis at the point \(\mathrm { A } ( 4,0 )\). B and C are maximum and minimum points on the curve. Units on the axes are metres. \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\).
2. Verify that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(\theta = \frac { 1 } { 6 } \pi\), and find the exact coordinates of B . Hence find the maximum width BC of the balloon.
3. (A) Show that \(y = x \cos \theta\).
   (B) Find \(\sin \theta\) in terms of \(x\) and show that \(\cos ^ { 2 } \theta = x - \frac { 1 } { 4 } x ^ { 2 }\).
   (C) Hence show that the cartesian equation of the curve is \(y ^ { 2 } = x ^ { 3 } - \frac { 1 } { 4 } x ^ { 4 }\).
4. Find the volume of the balloon.

4 Fig. 4 shows the curve \(y = \sqrt { 1 + \mathrm { e } ^ { 2 x } }\), and the region between the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 2\). \begin{figure}[h] \end{figure}

1. Find the exact volume of revolution when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
2. 1. Complete the table of values, and use the trapezium rule with 4 strips to estimate the area of the shaded region.

      \(x\)	0	0.5	1	1.5	2
      \(y\)		1.9283	2.8964	4.5919

   2. The trapezium rule for \(\int _ { 0 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { 2 x } } \mathrm {~d} x\) with 8 and 16 strips gives 6.797 and 6.823, although not necessarily in that order. Without doing the calculations, say which result is which, explaining your reasoning.

5 A curve has parametric equations \(x = \sec \theta , y = 2 \tan \theta\).

1. Given that the derivative of \(\sec \theta\) is \(\sec \theta \tan \theta\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \operatorname { cosec } \theta\).
2. Verify that the cartesian equation of the curve is \(y ^ { 2 } = 4 x ^ { 2 } - 4\). Fig. 5 shows the region enclosed by the curve and the line \(x = 2\). This region is rotated through \(180 ^ { \circ }\) about the \(x\)-axis. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{132ae754-bd4c-4819-80ef-4823ac2ead4f-02_545_853_1738_607} \captionsetup{labelformat=empty} \caption{Fig. 5}
   \end{figure}
3. Find the volume of revolution produced, giving your answer in exact form.

1 The finite region enclosed by the line \(y = k x\), where \(k\) is a positive constant, the \(x\)-axis for \(0 \leqslant x \leqslant h\), and the line \(x = h\) is rotated through 1 complete revolution about the \(x\)-axis. Prove by integration that the centroid of the resulting cone is at a distance \(\frac { 3 } { 4 } h\) from the origin \(O\).
[0pt] [The volume of a cone of height \(h\) and base radius \(r\) is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\).]

3 The equation of a curve is \(y = \lambda x ^ { 2 }\), where \(\lambda > 0\). The region bounded by the curve, the \(x\)-axis and the line \(x = a\), where \(a > 0\), is denoted by \(R\). The \(y\)-coordinate of the centroid of \(R\) is \(a\). Show that \(\lambda = \frac { 10 } { 3 a }\).