4.08d Volumes of revolution: about x and y axes

2. The finite region bounded by the \(x\)-axis, the curve with equation \(y = 2 \mathrm { e } ^ { x }\), the \(y\)-axis and the line \(x = 1\) is rotated through one complete revolution about the \(x\)-axis to form a uniform solid. Use algebraic integration to

1. show that the volume of the solid is \(2 \pi \left( \mathrm { e } ^ { 2 } - 1 \right)\),
2. find, in terms of e, the \(x\) coordinate of the centre of mass of the solid.

3 Find the exact volume generated when the region enclosed between the \(x\)-axis and the portion of the curve \(y = \sin x\) between \(x = 0\) and \(x = \pi\) is rotated completely about the \(x\)-axis.

4 Fig. 4 shows a sketch of the region enclosed by the curve \(\sqrt { 1 + \mathrm { e } ^ { - 2 x } }\), the \(x\)-axis, the \(y\)-axis and the line \(x = 1\). \begin{figure}[h] \end{figure} Find the volume of the solid generated when this region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Give your answer in an exact form.

7 Fig. 7 shows the curve with parametric equations $$x = \cos \theta , y = \sin \theta - \frac { 1 } { 8 } \sin 2 \theta , 0 \leqslant \theta < 2 \pi$$ The curve crosses the \(x\)-axis at points \(\mathrm { A } ( 1,0 )\) and \(\mathrm { B } ( - 1,0 )\), and the positive \(y\)-axis at C . D is the maximum point of the curve, and E is the minimum point. The solid of revolution formed when this curve is rotated through \(360 ^ { \circ }\) about the \(x\)-axis is used to model the shape of an egg. \begin{figure}[h] \end{figure}

1. Show that, at the point \(\mathrm { A } , \theta = 0\). Write down the value of \(\theta\) at the point B , and find the coordinates of C .
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Hence show that, at the point D, $$2 \cos ^ { 2 } \theta - 4 \cos \theta - 1 = 0 .$$
3. Solve this equation, and hence find the \(y\)-coordinate of D , giving your answer correct to 2 decimal places. The cartesian equation of the curve (for \(0 \leqslant \theta \leqslant \pi\) ) is $$y = \frac { 1 } { 4 } ( 4 - x ) \sqrt { 1 - x ^ { 2 } } .$$
4. Show that the volume of the solid of revolution of this curve about the \(x\)-axis is given by $$\frac { 1 } { 16 } \pi \int _ { - 1 } ^ { 1 } \left( 16 - 8 x - 15 x ^ { 2 } + 8 x ^ { 3 } - x ^ { 4 } \right) \mathrm { d } x .$$ Evaluate this integral.

3 Fig. 3 shows part of the curve \(y = 1 + x ^ { 2 }\), together with the line \(y = 2\). \begin{figure}[h] \end{figure} The region enclosed by the curve, the \(y\)-axis and the line \(y = 2\) is rotated through \(360 ^ { \circ }\) about the \(y\)-axis. Find the volume of the solid generated, giving your answer in terms of \(\pi\).

3 Fig. 3 shows the curve \(y = \ln x\) and part of the line \(y = 2\). \begin{figure}[h] \end{figure} The shaded region is rotated through \(360 ^ { \circ }\) about the \(y\)-axis.

1. Show that the volume of the solid of revolution formed is given by \(\int _ { 0 } ^ { 2 } \pi \mathrm { e } ^ { 2 y } \mathrm {~d} y\).
2. Evaluate this, leaving your answer in an exact form.

2 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.

3 Fig. 3 shows the curve \(y = x ^ { 4 }\) and the line \(y = 4\). \begin{figure}[h] \end{figure} The finite region enclosed by the curve and the line is rotated through \(180 ^ { \circ }\) about the \(y\)-axis. Find the exact volume of revolution generated.

2 The graph shows part of the curve \(y = x ^ { 2 } + 1\). \includegraphics[max width=\textwidth, alt={}, center]{62dbc58e-f498-483f-a9aa-05cb5aa44881-2_380_876_715_575} Find the volume when the area between this curve, the axes and the line \(x = 2\) is rotated through \(360 ^ { 0 }\) about the \(x\)-axis.

3 The curve \(y ^ { 2 } = x - 1\) for \(1 \leq x \leq 3\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Find the volume of the solid formed.

3 The graph shows part of the curve \(y ^ { 2 } = ( x - 1 )\). \includegraphics[max width=\textwidth, alt={}, center]{73112db3-7b05-48db-9fff-fdbac7dbd564-2_428_860_973_616} Find the volume when the area between this curve, the \(x\)-axis and the line \(x = 5\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.

3. The diagram shows the curve with equation \(y = 2 \sin x + \operatorname { cosec } x , 0 < x < \pi\).
The shaded region bounded by the curve, the \(x\)-axis and the lines \(x = \frac { \pi } { 6 }\) and \(x = \frac { \pi } { 2 }\) is rotated through four right angles about the \(x\)-axis. Show that the volume of the solid formed is \(\frac { 1 } { 2 } \pi ( 4 \pi + 3 \sqrt { 3 } )\).

5. \includegraphics[max width=\textwidth, alt={}, center]{825f6c7d-5399-4e7f-bacd-b7c0831aab06-1_408_858_1893_488} The diagram shows the curve with equation \(y = 4 x ^ { \frac { 1 } { 2 } } \mathrm { e } ^ { - x }\).
The shaded region bounded by the curve, the \(x\)-axis and the line \(x = 2\) is rotated through four right angles about the \(x\)-axis. Find, in terms of \(\pi\) and e, the exact volume of the solid formed.

2 Fig. 2 shows the curve \(y = \overline { 1 + x ^ { 2 } }\). \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.

3 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.

1 Fig. 6 shows the region enclosed by the curve \(y = \left( 1 + 2 x ^ { 2 } \right) ^ { \frac { 1 } { 3 } }\) and the line \(y = 2\). \begin{figure}[h] \end{figure} This region is rotated about the \(y\)-axis. Find the volume of revolution formed, giving your answer as a multiple of \(\pi\).

2 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]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-2_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.

3 Fig. 6 shows the region enclosed by part of the curve \(y = 2 x ^ { 2 }\), the straight line \(x + y = 3\), and the \(y\)-axis. The curve and the straight line meet at \(\mathrm { P } ( 1,2 )\). \begin{figure}[h] \end{figure} The shaded region is rotated through \(360 ^ { \circ }\) about the \(y\)-axis. Find, in terms of \(\pi\), the volume of the solid of revolution formed.
[0pt] [You may use the formula \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) for the volume of a cone.]

4 The part of the curve \(y = 4 - x ^ { 2 }\) that is above the \(x\)-axis is rotated about the \(y\)-axis. This is shown in Fig. 4. Find the volume of revolution produced, giving your answer in terms of \(\pi\). \begin{figure}[h] \end{figure}

5 Fig. 2 shows the curve \(y = \sqrt { } 1 + \mathrm { e } ^ { 2 x }\). \begin{figure}[h] \end{figure} The region bounded by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 1\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Show that the volume of the solid of revolution produced is \(\frac { 1 } { 2 } \pi \left( 1 + \mathrm { e } ^ { 2 } \right)\).

6 Fig. 3 shows the curve \(y = \ln x\) and part of the line \(y = 2\). \begin{figure}[h] \end{figure} The shaded region is rotated through \(360 ^ { \circ }\) about the \(y\)-axis.

1. Show that the volume of the solid of revolution formed is given by \(\int _ { 0 } ^ { 2 } \pi \mathrm { e } ^ { 2 y } \mathrm {~d} y\).
2. Evaluate this, leaving your answer in an exact form.

7

1. Show that \(\int x \mathrm { e } ^ { - 2 x } \mathrm {~d} x = - \frac { 1 } { 4 } \mathrm { e } ^ { - 2 x } ( 1 + 2 x ) + c\). A vase is made in the shape of the volume of revolution of the curve \(y = x ^ { 1 / 2 } \mathrm { e } ^ { - x }\) about the \(x\)-axis between \(x = 0\) and \(x = 2\) (see Fig. 5). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-5_718_751_638_654} \captionsetup{labelformat=empty} \caption{Fig. 5}
   \end{figure}
2. Show that this volume of revolution is \(\frac { 1 } { 4 } \pi \left( 1 \frac { 5 } { \mathrm { e } ^ { 4 } } \right)\).

8 Fig. 4 shows a sketch of the region enclosed by the curve . \(\mathrm { J } 1 + \mathrm { e } - 2 \mathrm { x }\), the x -axis, the y -axis and the line \(\boldsymbol { x } = 1\). \begin{figure}[h] \end{figure} Find the volume of the solid generated when this region is rotated through \(360 ^ { \circ }\) about the \(\boldsymbol { x }\)-axis. Give your answer in an exact form.

1 Fig. 8 shows a cross-section of a car headlight whose inside reflective surface is modelled, in suitable units, by the curve $$x = 2 t ^ { 2 } , y = 4 t , \quad - \sqrt { 2 } \leqslant t \leqslant \sqrt { 2 } .$$ \(\mathrm { P } \left( 2 t ^ { 2 } , 4 t \right)\) is a point on the curve with parameter \(t\). TS is the tangent to the curve at P , and PR is the line through P parallel to the \(x\)-axis. Q is the point \(( 2,0 )\). The angles that PS and QP make with the positive \(x\)-direction are \(\theta\) and \(\phi\) respectively. \begin{figure}[h] \end{figure}

1. By considering the gradient of the tangent TS, show that \(\tan \theta = \frac { 1 } { t }\).
2. Find the gradient of the line QP in terms of \(t\). Hence show that \(\phi = 2 \theta\), and that angle TPQ is equal to \(\theta\).
   [0pt] [The above result shows that if a lamp bulb is placed at Q , then the light from the bulb is reflected to produce a parallel beam of light.] The inside surface of the headlight has the shape produced by rotating the curve about the \(x\)-axis.
3. Show that the curve has cartesian equation \(y ^ { 2 } = 8 x\). Hence find the volume of revolution of the curve, giving your answer as a multiple of \(\pi\).

2 Fig. 3 shows part of the curve \(y = 1 + x ^ { 2 }\), together with the line \(y = 2\). \begin{figure}[h] \end{figure} The region enclosed by the curve, the \(y\)-axis and the line \(y = 2\) is rotated through \(360 ^ { \circ }\) about the \(y\)-axis. Find the volume of the solid generated, giving your answer in terms of \(\pi\).