4.08d Volumes of revolution: about x and y axes

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

1. Show that the exact volume of this solid of revolution is given by $$\int _ { 0 } ^ { k } p ( 1 - \cos 2 \theta ) d \theta$$ where \(p\) and \(k\) are constants to be found.
2. Hence find, by algebraic integration, the exact volume of this solid of revolution.

7. \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} Figure 3 shows the design of a doorknob.
The shape of the doorknob is formed by rotating the curve shown in Figure 4 through \(360 ^ { \circ }\) about the \(x\)-axis, where the units are centimetres. The equation of the curve is given by $$\mathrm { f } ( x ) = \frac { 1 } { 4 } ( 4 - x ) \mathrm { e } ^ { x } \quad 0 \leqslant x \leqslant 4$$

1. Show that the volume, \(V \mathrm {~cm} ^ { 3 }\), of the doorknob is given by $$V = K \int _ { 0 } ^ { 4 } \left( x ^ { 2 } - 8 x + 16 \right) \mathrm { e } ^ { 2 x } \mathrm {~d} x$$ where \(K\) is a constant to be found.
2. Hence, find the exact value of the volume of the doorknob. Give your answer in the form \(p \pi \left( \mathrm { e } ^ { q } + r \right) \mathrm { cm } ^ { 3 }\) where \(p , q\) and \(r\) are simplified rational numbers to be found.

3. \begin{figure}[h] \end{figure} \section*{In this question you must show all stages of your working.} \section*{Solutions relying entirely on calculator technology are not acceptable.} Figure 1 shows a sketch of the curve with equation $$y = \sqrt { \frac { 3 x } { 3 x ^ { 2 } + 5 } } \quad x \geqslant 0$$ The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis and the lines with equations \(x = \sqrt { 5 }\) and \(x = 5\) The region \(R\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
Use integration to find the exact volume of the solid generated. Give your answer in the form \(a \ln b\), where \(a\) is an irrational number and \(b\) is a prime number.

1. (a) Using the substitution \(u = 4 x + 2 \sin 2 x\), or otherwise, show that

$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \mathrm { e } ^ { 4 x + 2 \sin 2 x } \cos ^ { 2 } x \mathrm {~d} x = \frac { 1 } { 8 } \left( \mathrm { e } ^ { 2 \pi } - 1 \right)$$ Figure 3 The curve shown in Figure 3, has equation $$y = 6 \mathrm { e } ^ { 2 x + \sin 2 x } \cos x$$ The region \(R\), shown shaded in Figure 3, is bounded by the positive \(x\)-axis, the positive \(y\)-axis and the curve. The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid.
(b) Use the answer to part (a) to find the volume of the solid formed, giving the answer in simplest form.

4. \begin{figure}[h] \end{figure} Figure 1 shows the finite shaded region, \(R\), which is bounded by the curve \(y = x \mathrm { e } ^ { x }\), the line \(x = 1\), the line \(x = 3\) and the \(x\)-axis. The region \(R\) is rotated through 360 degrees about the \(x\)-axis.
Use integration by parts to find an exact value for the volume of the solid generated.
(8)

2. \begin{figure}[h] \end{figure} The curve with equation \(y = \frac { 1 } { 3 ( 1 + 2 x ) } , x > - \frac { 1 } { 2 }\), is shown in Figure 1.
The region bounded by the lines \(x = - \frac { 1 } { 4 } , x = \frac { 1 } { 2 }\), the \(x\)-axis and the curve is shown shaded in Figure 1. This region is rotated through 360 degrees about the \(x\)-axis.

1. Use calculus to find the exact value of the volume of the solid generated. \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 2} \includegraphics[alt={},max width=\textwidth]{d366e541-15f6-4fb5-9afb-faf6120f1a1c-03_383_447_1411_753}
   \end{figure} Figure 2 shows a paperweight with axis of symmetry \(A B\) where \(A B = 3 \mathrm {~cm}\). \(A\) is a point on the top surface of the paperweight, and \(B\) is a point on the base of the paperweight. The paperweight is geometrically similar to the solid in part (a).
2. Find the volume of this paperweight.

3. \begin{figure}[h] \end{figure} The curve shown in Figure 2 has equation \(y = \frac { 1 } { ( 2 x + 1 ) }\). The finite region bounded by the curve, the \(x\)-axis and the lines \(x = a\) and \(x = b\) is shown shaded in Figure 2. This region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis to generate a solid of revolution. Find the volume of the solid generated. Express your answer as a single simplified fraction, in terms of \(a\) and \(b\).

8. (a) Using the substitution \(x = 2 \cos u\), or otherwise, find the exact value of $$\int _ { 1 } ^ { \sqrt { 2 } } \frac { 1 } { x ^ { 2 } \sqrt { } \left( 4 - x ^ { 2 } \right) } d x$$ \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve with equation \(y = \frac { 4 } { x \left( 4 - x ^ { 2 } \right) ^ { \frac { 1 } { 4 } } } , \quad 0 < x < 2\). The shaded region \(S\), shown in Figure 3, is bounded by the curve, the \(x\)-axis and the lines with equations \(x = 1\) and \(x = \sqrt { } 2\). The shaded region \(S\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.
(b) Using your answer to part (a), find the exact volume of the solid of revolution formed.

1. The curve \(C\) has parametric equations

$$x = \ln t , \quad y = t ^ { 2 } - 2 , \quad t > 0$$ Find

1. an equation of the normal to \(C\) at the point where \(t = 3\),
2. a cartesian equation of \(C\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{a3ece8a8-8107-4c3a-a6a9-c19b5e35ec5a-10_579_759_740_571} \captionsetup{labelformat=empty} \caption{Figure 1}
   \end{figure} The finite area \(R\), shown in Figure 1, is bounded by \(C\), the \(x\)-axis, the line \(x = \ln 2\) and the line \(x = \ln 4\). The area \(R\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
3. Use calculus to find the exact volume of the solid generated.

4. \begin{figure}[h] \end{figure} Figure 1 shows the curve with equation $$y = \sqrt { } \left( \frac { 2 x } { 3 x ^ { 2 } + 4 } \right) , x \geqslant 0$$ The finite region \(S\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis and the line \(x = 2\) The region \(S\) is rotated \(360 ^ { \circ }\) about the \(x\)-axis.
Use integration to find the exact value of the volume of the solid generated, giving your answer in the form \(k \ln a\), where \(k\) and \(a\) are constants.

6. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve with equation \(y = 1 - 2 \cos x\), where \(x\) is measured in radians. The curve crosses the \(x\)-axis at the point \(A\) and at the point \(B\).

1. Find, in terms of \(\pi\), the \(x\) coordinate of the point \(A\) and the \(x\) coordinate of the point \(B\). The finite region \(S\) enclosed by the curve and the \(x\)-axis is shown shaded in Figure 3. The region \(S\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
2. Find, by integration, the exact value of the volume of the solid generated.

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.

7. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with equation \(y = \sqrt { } ( \tan x )\). The finite region \(R\), which is bounded by the curve, the \(x\)-axis and the line \(x = \frac { \pi } { 4 }\), is shown shaded in Figure 1.

1. Given that \(y = \sqrt { } ( \tan x )\), complete the table with the values of \(y\) corresponding to \(x = \frac { \pi } { 16 } , \frac { \pi } { 8 }\) and \(\frac { 3 \pi } { 16 }\), giving your answers to 5 decimal places.

   \(x\)	0	\(\frac { \pi } { 16 }\)	\(\frac { \pi } { 8 }\)	\(\frac { 3 \pi } { 16 }\)	\(\frac { \pi } { 4 }\)
   \(y\)	0				1

2. Use the trapezium rule with all the values of \(y\) in the completed table to obtain an estimate for the area of the shaded region \(R\), giving your answer to 4 decimal places. The region \(R\) is rotated through \(2 \pi\) radians around the \(x\)-axis to generate a solid of revolution.
3. Use integration to find an exact value for the volume of the solid generated. \section*{LO}

8. (a) Using the identity \(\cos 2 \theta = 1 - 2 \sin ^ { 2 } \theta\), find \(\int \sin ^ { 2 } \theta \mathrm {~d} \theta\). \begin{figure}[h] \end{figure} Figure 4 shows part of the curve \(C\) with parametric equations $$x = \tan \theta , \quad y = 2 \sin 2 \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$ The finite shaded region \(S\) shown in Figure 4 is bounded by \(C\), the line \(x = \frac { 1 } { \sqrt { 3 } }\) and the \(x\)-axis. This shaded region is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.
(b) Show that the volume of the solid of revolution formed is given by the integral $$k \int _ { 0 } ^ { \frac { \pi } { 6 } } \sin ^ { 2 } \theta \mathrm {~d} \theta$$ where \(k\) is a constant.
(c) Hence find the exact value for this volume, giving your answer in the form \(p \pi ^ { 2 } + q \pi \sqrt { } 3\), where \(p\) and \(q\) are constants.

3. \begin{figure}[h] \end{figure} Figure 1 shows the finite region \(R\) bounded by the \(x\)-axis, the \(y\)-axis, the line \(x = \frac { \pi } { 2 }\) and the curve with equation $$y = \sec \left( \frac { 1 } { 2 } x \right) , \quad 0 \leqslant x \leqslant \frac { \pi } { 2 }$$ The table shows corresponding values of \(x\) and \(y\) for \(y = \sec \left( \frac { 1 } { 2 } x \right)\).

1. Complete the table above giving the missing value of \(y\) to 6 decimal places.
2. Using the trapezium rule, with all of the values of \(y\) from the completed table, find an approximation for the area of \(R\), giving your answer to 4 decimal places. Region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
3. Use calculus to find the exact volume of the solid formed.

4. (a) Express \(\frac { 25 } { x ^ { 2 } ( 2 x + 1 ) }\) in partial fractions. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve \(C\) with equation \(y = \frac { 5 } { x \sqrt { } ( 2 x + 1 ) } , x > 0\) The finite region \(R\) is bounded by the curve \(C\), the \(x\)-axis, the line with equation \(x = 1\) and the line with equation \(x = 4\) This region is shown shaded in Figure 2 The region \(R\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
(b) Use calculus to find the exact volume of the solid of revolution generated, giving your answer in the form \(a + b \ln c\), where \(a , b\) and \(c\) are constants.

7. (a) Find $$\int ( 2 x - 1 ) ^ { \frac { 3 } { 2 } } d x$$ giving your answer in its simplest form. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve \(C\) with equation $$y = ( 2 x - 1 ) ^ { \frac { 3 } { 4 } } , \quad x \geqslant \frac { 1 } { 2 }$$ The curve \(C\) cuts the line \(y = 8\) at the point \(P\) with coordinates \(( k , 8 )\), where \(k\) is a constant.
(b) Find the value of \(k\). The finite region \(S\), shown shaded in Figure 3, is bounded by the curve \(C\), the \(x\)-axis, the \(y\)-axis and the line \(y = 8\). This region is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.
(c) Find the exact value of the volume of the solid generated.

5. \begin{figure}[h] \end{figure} Diagram not drawn to scale The finite region \(S\), shown shaded in Figure 2, is bounded by the \(y\)-axis, the \(x\)-axis, the line with equation \(x = \ln 4\) and the curve with equation $$y = \mathrm { e } ^ { x } + 2 \mathrm { e } ^ { - x } , \quad x \geqslant 0$$ The region \(S\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
Use integration to find the exact value of the volume of the solid generated. Give your answer in its simplest form.
[0pt] [Solutions based entirely on graphical or numerical methods are not acceptable.]

8. \begin{figure}[h] \end{figure}

1. Find \(\int x \cos 4 x d x\) Figure 3 shows part of the curve with equation \(y = \sqrt { x } \sin 2 x , \quad x \geqslant 0\) The finite region \(R\), shown shaded in Figure 3, is bounded by the curve, the \(x\)-axis and the line with equation \(x = \frac { \pi } { 4 }\) The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.
2. Find the exact value of the volume of this solid of revolution, giving your answer in its simplest form.
   (Solutions based entirely on graphical or numerical methods are not acceptable.) \includegraphics[max width=\textwidth, alt={}, center]{0c4a3759-ecaa-47c3-a071-ce25fd11159f-32_2630_1828_121_121}

7. (a) Given that \(u = \frac { x } { 2 } - \frac { 1 } { 8 } \sin 4 x\), show that \(\frac { \mathrm { d } u } { \mathrm {~d} x } = \sin ^ { 2 } 2 x\). \begin{figure}[h] \end{figure} Figure 2 shows the finite region bounded by the curve \(y = x ^ { \frac { 1 } { 2 } } \sin 2 x\), the line \(x = \frac { \pi } { 4 }\) and the \(x\)-axis. This region is rotated through \(2 \pi\) radians about the \(x\)-axis.
(b) Using the result in part (a), or otherwise, find the exact value of the volume generated.
(8)

2. \includegraphics[max width=\textwidth, alt={}, center]{960fe82f-c180-422c-b409-a5cdc5fae924-06_974_1088_116_548} \section*{Figure 1} Figure 1 shows a sketch of part of the curve with equation $$y = \frac { 9 } { ( 2 x - 3 ) ^ { 1.25 } } \quad x > \frac { 3 } { 2 }$$ The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the line with equation \(y = 9\) and the line with equation \(x = 6\) This region is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution. Find, by algebraic integration, the exact volume of the solid generated.

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

Solutions relying entirely on calculator technology are not acceptable. \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} Figure 2 shows the curve with equation $$y = 10 x \mathrm { e } ^ { - \frac { 1 } { 2 } x } \quad 0 \leqslant x \leqslant 10$$ The finite region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis and the line with equation \(x = 10\) The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.

1. Show that the volume, \(V\), of this solid is given by $$V = k \int _ { 0 } ^ { 10 } x ^ { 2 } \mathrm { e } ^ { - x } \mathrm {~d} x$$ where \(k\) is a constant to be found.
2. Find \(\int x ^ { 2 } e ^ { - x } d x\) Figure 3 represents an exercise weight formed by joining two of these solids together.
   The exercise weight has mass 5 kg and is 20 cm long.
   Given that $$\text { density } = \frac { \text { mass } } { \text { volume } }$$ and using your answers to part (a) and part (b),
3. find the density of this exercise weight. Give your answer in grams per \(\mathrm { cm } ^ { 3 }\) to 3 significant figures.

8. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve \(C\) with parametric equations $$x = t + \frac { 1 } { t } \quad y = t - \frac { 1 } { t } \quad t > 0.7$$ The curve \(C\) intersects the \(x\)-axis at the point \(Q\).

1. Find the \(x\) coordinate of \(Q\). The line \(l\) is the normal to \(C\) at the point \(P\) as shown in Figure 2.
   Given that \(t = 2\) at \(P\)
2. write down the coordinates of \(P\)
3. Using calculus, show that an equation of \(l\) is $$3 x + 5 y = 15$$ The region, \(R\), shown shaded in Figure 2 is bounded by the curve \(C\), the line \(l\) and the \(x\)-axis.
4. Using algebraic integration, find the exact volume of the solid of revolution formed when the region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis.

3. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \mathrm { e } ^ { 0.5 x } - 2\) The region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis and the \(y\)-axis. The region \(R\) is rotated \(360 ^ { \circ }\) about the \(x\)-axis to form a solid of revolution.
Show that the volume of this solid can be written in the form \(a \ln 2 + b\), where \(a\) and \(b\) are constants to be found.

8. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the curve \(C\) with parametric equations $$x = 6 t - 3 \sin 2 t \quad y = 2 \cos t \quad 0 \leqslant t \leqslant \frac { \pi } { 2 }$$ The curve meets the \(y\)-axis at 2 and the \(x\)-axis at \(k\), where \(k\) is a constant.

1. State the value of \(k\).
2. Use parametric differentiation to show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \lambda \operatorname { cosec } t$$ where \(\lambda\) is a constant to be found. The point \(P\) with parameter \(\mathrm { t } = \frac { \pi } { 4 }\) lies on \(C\).
   The tangent to \(C\) at the point \(P\) cuts the \(y\)-axis at the point \(N\).
3. Find the exact \(y\) coordinate of \(N\), giving your answer in simplest form. The region bounded by the curve, the \(x\)-axis and the \(y\)-axis is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.
4. 1. Show that the volume of this solid is given by $$\int _ { 0 } ^ { \alpha } \beta ( 1 - \cos 4 t ) d t$$ where \(\alpha\) and \(\beta\) are constants to be found.
   2. Hence, using algebraic integration, find the exact volume of this solid.