Volume of revolution with parts

9

1. Use integration by parts to find \(\int x \ln x \mathrm {~d} x\).
2. Given that \(y = ( \ln x ) ^ { 2 }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   (2 marks)
3. The diagram shows part of the curve with equation \(y = \sqrt { x } \ln x\). \includegraphics[max width=\textwidth, alt={}, center]{7148f43d-dc7d-43e2-b96e-ed1fb94073bf-5_406_645_696_719} The shaded region \(R\) is bounded by the curve \(y = \sqrt { x } \ln x\), the line \(x = \mathrm { e }\) and the \(x\)-axis from \(x = 1\) to \(x = \mathrm { e }\). Find the volume of the solid generated when the region \(R\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis, giving your answer in an exact form.
   (6 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)

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]

6. (a) Use integration by parts to show that $$\int _ { 0 } ^ { \frac { \pi } { 4 } } x \sec ^ { 2 } x \mathrm {~d} x = \frac { 1 } { 4 } \pi - \frac { 1 } { 2 } \ln 2 .$$ \begin{figure}[h] \end{figure} The finite region \(R\), bounded by the equation \(y = x ^ { \frac { 1 } { 2 } } \sec x\), the line \(x = \frac { \pi } { 4 }\) and the \(x\)-axis is shown in Fig. 1. The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
(b) Find the volume of the solid of revolution generated.
(c) Find the gradient of the curve with equation \(y = x ^ { \frac { 1 } { 2 } } \sec x\) at the point where \(x = \frac { \pi } { 4 }\).

5. (a) Use integration by parts to show that $$\int _ { 0 } ^ { \frac { \pi } { 4 } } x \sec ^ { 2 } x \mathrm {~d} x = \frac { 1 } { 4 } \pi - \frac { 1 } { 2 } \ln 2$$ \begin{figure}[h] \end{figure} The finite region \(R\), bounded by the equation \(y = x ^ { \frac { 1 } { 2 } } \sec x\), the line \(x = \frac { \pi } { 4 }\) and the \(x\)-axis is shown in Fig. 1. The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
(b) Find the volume of the solid of revolution generated.
(c) Find the gradient of the curve with equation \(y = x ^ { \frac { 1 } { 2 } } \sec x\) at the point where \(x = \frac { \pi } { 4 }\).