4.08d Volumes of revolution: about x and y axes

7

1. Use integration by parts to find:
   1. \(\quad \int x \cos 4 x \mathrm {~d} x\);
      (4 marks)
   2. \(\int x ^ { 2 } \sin 4 x d x\).
      (4 marks)
2. The region bounded by the curve \(y = 8 x \sqrt { ( \sin 4 x ) }\) and the lines \(x = 0\) and \(x = 0.2\) is rotated through \(2 \pi\) radians about the \(x\)-axis. Find the value of the volume of the solid generated, giving your answer to three significant figures.
   (3 marks)

4

1. By using integration by parts, find \(\int x \mathrm { e } ^ { 6 x } \mathrm {~d} x\).
   (4 marks)
2. The diagram shows part of the curve with equation \(y = \sqrt { x } \mathrm { e } ^ { 3 x }\). \includegraphics[max width=\textwidth, alt={}, center]{d3c66c34-b09c-4223-8383-cf0a68419bf9-4_547_846_536_591} The shaded region \(R\) is bounded by the curve \(y = \sqrt { x } \mathrm { e } ^ { 3 x }\), the line \(x = 1\) and the \(x\)-axis from \(x = 0\) to \(x = 1\). Find the volume of the solid generated when the region \(R\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis, giving your answer in the form \(\pi \left( p \mathrm { e } ^ { 6 } + q \right)\), where \(p\) and \(q\) are rational numbers.
   (3 marks)

9 The shape of a vase can be modelled by rotating the curve with equation \(16 x ^ { 2 } - ( y - 8 ) ^ { 2 } = 32\) between \(y = 0\) and \(y = 16\) completely about the \(\boldsymbol { y }\)-axis. \includegraphics[max width=\textwidth, alt={}, center]{063bbfa5-df49-44a1-8143-5e076397f63f-09_890_1210_1555_424} The vase has a base.
Find the volume of water needed to fill the vase, giving your answer as an exact value.

6

1. By using integration by parts twice, find $$\int x ^ { 2 } \sin 2 x d x$$
2. A curve has equation \(y = x \sqrt { \sin 2 x }\), for \(0 \leqslant x \leqslant \frac { \pi } { 2 }\). The region bounded by the curve and the \(x\)-axis is rotated through \(2 \pi\) radians about the \(x\)-axis to generate a solid. Find the exact value of the volume of the solid generated.
   [0pt] [3 marks]

5. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with equation \(y = 1 + \frac { 1 } { 2 \sqrt { x } }\). The shaded region \(R\), bounded by the curve, that \(x\)-axis and the lines \(x = 1\) and \(x = 4\), is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Using integration, show that the volume of the solid generated is \(\pi \left( 5 + \frac { 1 } { 2 } \ln 2 \right)\).
(8)

5. \begin{figure}[h] \end{figure} Figure 1 shows a graph of \(y = x \sqrt { } \sin x , 0 < x < \pi\). The maximum point on the curve is \(A\).

1. Show that the \(x\)-coordinate of the point \(A\) satisfies the equation \(2 \tan x + x = 0\). The finite region enclosed by the curve and the \(x\)-axis is shaded as shown in Fig. 1.
   A solid body \(S\) is generated by rotating this region through \(2 \pi\) radians about the \(x\)-axis.
2. Find the exact value of the volume of \(S\).
   (7)

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 parametric equations $$x = 3 \sin t , \quad y = 2 \sin 2 t , \quad 0 \leq t < \pi .$$ The curve meets the \(x\)-axis at the origin, \(O\), and at the point \(A\).

1. Find the value of \(t\) at \(O\) and the value of \(t\) at \(A\). The region enclosed by the curve is rotated through \(\pi\) radians about the \(x\)-axis.
2. Show that the volume of the solid formed is given by $$\int _ { 0 } ^ { \frac { \pi } { 2 } } 12 \pi \sin ^ { 2 } 2 t \cos t \mathrm {~d} t$$
3. Using the substitution \(u = \sin t\), or otherwise, evaluate this integral, giving your answer as an exact multiple of \(\pi\).
   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

7. \begin{figure}[h] \end{figure} Figure 1 shows the curve with parametric equations $$x = t ^ { 3 } + 1 , \quad y = \frac { 2 } { t } , \quad t > 0 .$$ The shaded region is bounded by the curve, the \(x\)-axis and the lines \(x = 2\) and \(x = 9\).

1. Find the area of the shaded region.
2. Show that the volume of the solid formed when the shaded region is rotated through \(2 \pi\) radians about the \(x\)-axis is \(12 \pi\).
3. Find a cartesian equation for the curve in the form \(y = \mathrm { f } ( x )\).
   7. continued

1. \begin{figure}[h] \end{figure} Figure 1 shows the curve with equation \(y = \frac { 3 x + 1 } { \sqrt { x } } , x > 0\).
The shaded region is bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 3\).
Find the volume of the solid formed when the shaded region is rotated through \(2 \pi\) radians about the \(x\)-axis, giving your answer in the form \(\pi ( a + \ln b )\), where \(a\) and \(b\) are integers.
□

4 In this question, \(a\) is a constant with \(a > 1\).
Fig. 4 shows the region bounded by the curve \(y = \frac { 1 } { x ^ { 2 } }\) for \(1 \leqslant x \leqslant a\), the \(x\)-axis, and the lines \(x = 1\) and \(x = a\). \begin{figure}[h] \end{figure} This region is occupied by a uniform lamina ABCD , where A is \(( 1,1 ) , \mathrm { B }\) is \(( 1,0 ) , \mathrm { C }\) is \(( a , 0 )\) and D is \(\left( a , \frac { 1 } { a ^ { 2 } } \right)\). The centre of mass of this lamina is \(( \bar { x } , \bar { y } )\).

1. Find \(\bar { x }\) in terms of \(a\), and show that \(\bar { y } = \frac { a ^ { 3 } - 1 } { 6 \left( a ^ { 3 } - a ^ { 2 } \right) }\).
2. In the case \(a = 2\), the lamina is freely suspended from the point A , and hangs in equilibrium. Find the angle which AB makes with the vertical. The region shown in Fig. 4 is now rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid of revolution.
3. Find the \(x\)-coordinate of the centre of mass of this solid of revolution, in terms of \(a\), and show that it is less than 1.5.

4 Fig. 4.1 shows the region \(R\) bounded by the curve \(y = x ^ { - \frac { 1 } { 3 } }\) for \(1 \leqslant x \leqslant 8\), the \(x\)-axis, and the lines \(x = 1\) and \(x = 8\). \begin{figure}[h] \end{figure}

1. Find the \(x\)-coordinate of the centre of mass of a uniform solid of revolution obtained by rotating \(R\) through \(2 \pi\) radians about the \(x\)-axis.
2. Find the coordinates of the centre of mass of a uniform lamina in the shape of the region \(R\).
3. Using your answer to part (ii), or otherwise, find the coordinates of the centre of mass of a uniform lamina in the shape of the region (shown shaded in Fig. 4.2) bounded by the curve \(y = x ^ { - \frac { 1 } { 3 } }\) for \(1 \leqslant x \leqslant 8\), the line \(y = \frac { 1 } { 2 }\) and the line \(x = 1\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{c470e80e-b346-4335-9c08-beb5a46cc506-4_595_1015_1610_607} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
   \end{figure}

6. \begin{figure}[h] \end{figure} Figure 3 shows part of the curve \(y = x ^ { 2 } + 1\). The shaded region enclosed by the curve, the coordinate axes and the line \(x = 1\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.

1. Find the coordinates of the centre of mass of the solid obtained. The solid is suspended from a point on its larger circular rim and hangs in equilibrium.
2. Find, correct to the nearest degree, the acute angle which the plane surfaces of the solid make with the vertical.
   (3 marks)

3 The region \(R\) is bounded by the \(x\)-axis, the \(y\)-axis, the curve \(y = a \mathrm { e } ^ { \frac { x } { a } }\) and the line \(x = a \ln 2\) (where \(a\) is a positive constant). A uniform solid of revolution, of mass \(M\), is formed by rotating \(R\) through \(2 \pi\) radians about the \(x\)-axis. Find, in terms of \(M\) and \(a\), the moment of inertia about the \(x\)-axis of this solid of revolution.
[0pt] [8]

4 A curve \(C\) is given parametrically by the equations $$x = \frac { 1 } { 2 } \cosh 2 t , \quad y = 2 \sinh t$$

1. Express $$\left( \frac { \mathrm { d } x } { \mathrm {~d} t } \right) ^ { 2 } + \left( \frac { \mathrm { d } y } { \mathrm {~d} t } \right) ^ { 2 }$$ in terms of \(\cosh t\).
2. The arc of \(C\) from \(t = 0\) to \(t = 1\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
   1. Show that \(S\), the area of the curved surface generated, is given by $$S = 8 \pi \int _ { 0 } ^ { 1 } \sinh t \cosh ^ { 2 } t \mathrm {~d} t$$
   2. Find the exact value of \(S\).

3 Fig. 3.1 shows the curve with equation \(y = x ^ { 2 } + 3\). The region \(R\), shown shaded, is bounded by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 2\). A uniform solid of revolution S is formed by rotating the region R through \(2 \pi\) about the \(x\)-axis. The volume of \(S\) is \(\frac { 202 } { 5 } \pi\). \begin{figure}[h] \end{figure} \section*{(a) In this question you must show detailed reasoning.} Show that the \(x\)-coordinate of the centre of mass of S is \(\frac { 395 } { 303 }\). S is fixed to a cylinder of base radius 3 units and height 2 units to form the uniform solid D . The smaller circular face of S is joined to the top circular face of the cylinder, as shown in Fig. 3.2. \begin{figure}[h] \end{figure} (b) Find the distance of the centre of mass of D from its smaller circular face. D is placed with its smaller circular face in contact with a rough plane which is inclined at an angle of \(30 ^ { \circ }\) to the horizontal. It is given that D does not slip.
(c) Determine whether D topples.

4 The equation of a curve is \(\mathrm { y } = \frac { 1 } { \sqrt { \mathrm {~K} ^ { 2 } + \mathrm { x } ^ { 2 } } }\), where \(k\) is a positive constant. The region between the \(x\)-axis, the \(y\)-axis and the line \(x = k\) is rotated through \(2 \pi\) radians about the \(x\)-axis. Given that the volume of the solid of revolution formed is 1 unit \({ } ^ { 3 }\), find the exact value of \(k\).

9. \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} A mathematics student is modelling the profile of a glass bottle of water. Figure 1 shows a sketch of a central vertical cross-section \(A B C D E F G H A\) of the bottle with the measurements taken by the student. The horizontal cross-section between \(C F\) and \(D E\) is a circle of diameter 8 cm and the horizontal cross-section between \(B G\) and \(A H\) is a circle of diameter 2 cm . The student thinks that the curve \(G F\) could be modelled as a curve with equation $$y = a x ^ { 2 } + b \quad 1 \leqslant x \leqslant 4$$ where \(a\) and \(b\) are constants and \(O\) is the fixed origin, as shown in Figure 2.

1. Find the value of \(a\) and the value of \(b\) according to the model.
2. Use the model to find the volume of water that the bottle can contain.
3. State a limitation of the model. The label on the bottle states that the bottle holds approximately \(750 \mathrm {~cm} ^ { 3 }\) of water.
4. Use this information and your answer to part (b) to evaluate the model, explaining your reasoning.

9. $$\mathrm { f } ( x ) = 2 x ^ { \frac { 1 } { 3 } } + x ^ { - \frac { 2 } { 3 } } \quad x > 0$$ The finite region bounded by the curve \(y = \mathrm { f } ( x )\), the line \(x = \frac { 1 } { 8 }\), the \(x\)-axis and the line \(x = 8\) is rotated through \(\theta\) radians about the \(x\)-axis to form a solid of revolution. Given that the volume of the solid formed is \(\frac { 461 } { 2 }\) units cubed, use algebraic integration to find the angle \(\theta\) through which the region is rotated.

3. \begin{figure}[h] \end{figure} Figure 1 shows a circle with radius \(r\) and centre at the origin.
The region \(R\), shown shaded in Figure 1, is bounded by the \(x\)-axis and the part of the circle for which \(y > 0\) The region \(R\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis to create a sphere with volume \(V\) Use integration to show that \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\)

9. \begin{figure}[h] \end{figure} Figure 2 shows the vertical cross-section, \(A O B C D E\), through the centre of a wax candle.
In a model, the candle is formed by rotating the region bounded by the \(y\)-axis, the line \(O B\), the curve \(B C\), and the curve \(C D\) through \(360 ^ { \circ }\) about the \(y\)-axis. The point \(B\) has coordinates \(( 3,0 )\) and the point \(C\) has coordinates \(( 5,15 )\).
The units are in centimetres.
The curve \(B C\) is represented by the equation $$y = \frac { \sqrt { 225 x ^ { 2 } - 2025 } } { a } \quad 3 \leqslant x < 5$$ where \(a\) is a constant.

1. Determine the value of \(a\) according to this model. The curve \(C D\) is represented by the equation $$y = 16 - 0.04 x ^ { 2 } \quad 0 \leqslant x < 5$$
2. Using algebraic integration, determine, according to the model, the exact volume of wax that would be required to make the candle.
3. State a limitation of the model. When the candle was manufactured, \(700 \mathrm {~cm} ^ { 3 }\) of wax were required.
4. Use this information and your answer to part (b) to evaluate the model, explaining your reasoning.

8. \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} Figure 1 shows a sketch of a 16 cm tall vase which has a flat circular base with diameter 8 cm and a circular opening of diameter 8 cm at the top. A student measures the circular cross-section halfway up the vase to be 8 cm in diameter.
The student models the shape of the vase by rotating a curve, shown in Figure 2, through \(360 ^ { \circ }\) about the \(x\)-axis.

1. State the value of \(a\) that should be used when setting up the model. Two possible equations are suggested for the curve in the model. $$\begin{array} { l l } \text { Model A } & y = a - 2 \sin \left( \frac { 45 } { 2 } x \right) ^ { \circ } \\ \text { Model B } & y = a + \frac { x ( x - 8 ) ( x + 8 ) } { 100 } \end{array}$$ For each model,
2. 1. find the distance from the base at which the widest part of the vase occurs,
   2. find the diameter of the vase at this widest point. The widest part of the vase has diameter 12 cm and is just over 3 cm from the base.
3. Using this information and making your reasoning clear, suggest which model is more appropriate.
4. Using algebraic integration, find the volume for the vase predicted by Model B. You must make your method clear. The student pours water from a full one litre jug into the vase and finds that there is 100 ml left in the jug when the vase is full.
5. Comment on the suitability of Model B in light of this information.

1. In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable. \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} A large pile of concrete waste is created on a building site.
Figure 1 shows a central vertical cross-section of the concrete waste.
The curve \(C\), shown in Figure 2, has equation $$y + x ^ { 2 } = 2 \quad 0 \leqslant x \leqslant \sqrt { 2 }$$ The region \(R\), shown shaded in Figure 2, is bounded by the \(y\)-axis, the \(x\)-axis and the curve \(C\). The volume of concrete waste is modelled by the volume of revolution formed when \(R\) is rotated through \(360 ^ { \circ }\) about the \(y\)-axis. The units are metres. The density of the concrete waste is \(900 \mathrm { kgm } ^ { - 3 }\)

1. Use the model to estimate the mass of the concrete waste. Give your answer to 2 significant figures.
2. Give a limitation of the model. The mass of the concrete waste is approximately 5500 kg .
3. Use this information and your answer to part (a) to evaluate the model, giving a reason for your answer.