Rotation about x-axis: polynomial or root function

2 \includegraphics[max width=\textwidth, alt={}, center]{b24ed4c7-ab07-45f4-adf2-027734c36b62-2_633_787_402_680} The diagram shows the curve \(y = 3 x ^ { \frac { 1 } { 4 } }\). The shaded region is bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 4\). Find the volume of the solid obtained when this shaded region is rotated completely about the \(x\)-axis, giving your answer in terms of \(\pi\).

3

1. Sketch the curve \(y = ( x - 2 ) ^ { 2 }\).
2. The region enclosed by the curve, the \(x\)-axis and the \(y\)-axis is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Find the volume obtained, giving your answer in terms of \(\pi\).

1. The region bounded by the curve \(y = x ^ { 2 } - 2 x\) and the \(x\)-axis is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.

Find the volume of the solid formed, giving your answer in terms of \(\pi\).

2. The diagram shows the curve with equation \(y = x \sqrt { 2 - x } , 0 \leq x \leq 2\).
Find, in terms of \(\pi\), the volume of the solid formed when the region bounded by the curve and the \(x\)-axis is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.

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.

6

1. 1. Sketch the graph of \(y = 4 - x ^ { 2 }\), indicating the coordinates of the points where the graph crosses the coordinate axes.
   2. The region between the graph and the \(x\)-axis from \(x = 0\) to \(x = 2\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Find the exact value of the volume of the solid generated.
2. 1. Sketch the graph of \(y = \left| 4 - x ^ { 2 } \right|\).
   2. Solve \(\left| 4 - x ^ { 2 } \right| = 3\).
   3. Hence, or otherwise, solve the inequality \(\left| 4 - x ^ { 2 } \right| < 3\).

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.

2 The diagram shows the curve with equation \(y = \sqrt { ( x - 2 ) ^ { 5 } }\) for \(x \geqslant 2\). \includegraphics[max width=\textwidth, alt={}, center]{59b896ae-60ce-49ea-9c70-0f76fc5fffae-2_885_1125_854_461} The shaded region \(R\) is bounded by the curve \(y = \sqrt { ( x - 2 ) ^ { 5 } }\), the \(x\)-axis and the lines \(x = 3\) and \(x = 4\). Find the exact value of the volume of the solid formed when the region \(R\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.

2

1. Differentiate \(( x - 1 ) ^ { 4 }\) with respect to \(x\).
2. The diagram shows the curve with equation \(y = 2 \sqrt { ( x - 1 ) ^ { 3 } }\) for \(x \geqslant 1\). \includegraphics[max width=\textwidth, alt={}, center]{9fd9fa54-b0e6-413d-8645-de34b99b859a-02_789_1180_1190_431} The shaded region \(R\) is bounded by the curve \(y = 2 \sqrt { ( x - 1 ) ^ { 3 } }\), the lines \(x = 2\) and \(x = 4\), and the \(x\)-axis. Find the exact value of the volume of the solid formed when the region \(R\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
3. Describe a sequence of two geometrical transformations that maps the graph of \(y = \sqrt { x ^ { 3 } }\) onto the graph of \(y = 2 \sqrt { ( x - 1 ) ^ { 3 } }\).

2 The function f is defined by $$f ( x ) = 2 x + 3 \quad 0 \leq x \leq 5$$ The region \(R\) is enclosed by \(y = \mathrm { f } ( x ) , x = 5\), the \(x\)-axis and the \(y\)-axis.
The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
Give an expression for the volume of the solid formed.
Tick ( ✓ ) one box. \(\pi \int _ { 0 } ^ { 5 } ( 2 x + 3 ) d x\) \includegraphics[max width=\textwidth, alt={}, center]{47b12ae4-ca3f-472c-9d15-2ef17a2a4d87-02_113_108_1539_1000} \(\pi \int _ { 0 } ^ { 5 } ( 2 x + 3 ) ^ { 2 } \mathrm {~d} x\) \includegraphics[max width=\textwidth, alt={}, center]{47b12ae4-ca3f-472c-9d15-2ef17a2a4d87-02_115_108_1699_1000} \(2 \pi \int _ { 0 } ^ { 5 } ( 2 x + 3 ) d x\) □ \(2 \pi \int _ { 0 } ^ { 5 } ( 2 x + 3 ) ^ { 2 } \mathrm {~d} x\) □

5 The region bounded by the curve with equation \(y = 3 + \sqrt { x }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4\) is rotated through \(2 \pi\) radians about the \(x\)-axis. Use integration to show that the volume generated is \(\frac { 125 \pi } { 2 }\) [0pt] [5 marks]

4 Find the volume of the solid generated when the region bounded by the \(x\)-axis, \(x = 1 , x = 2\) and the curve given by \(y = x ^ { 3 }\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.

\includegraphics{figure_1} Figure 1 shows part of a curve \(C\) with equation \(y = x^2 + 3\). The shaded region is bounded by \(C\), the \(x\)-axis and the lines \(x = 1\) and \(x = 3\). The shaded region is rotated \(360°\) about the \(x\)-axis. Using calculus, calculate the volume of the solid generated. Give your answer as an exact multiple of \(\pi\). [7]

\includegraphics{figure_2} The diagram shows the curve with equation \(y = (2x - 3)^2\). The shaded region is bounded by the curve and the lines \(x = 0\) and \(y = 0\). Find the exact volume obtained when the shaded region is rotated completely about the \(x\)-axis. [5]

The region bounded by the curve \(y = x^2 - 2x\) and the \(x\)-axis is rotated through \(2\pi\) radians about the \(x\)-axis. Find the volume of the solid formed, giving your answer in terms of \(\pi\). [6]

Please remember to show detailed reasoning in your answer \includegraphics{figure_9} The diagram shows the curve with equation \(y = (2x - 3)^2\). The shaded region is bounded by the curve and the lines \(x = 0\) and \(y = 0\). Find the exact volume obtained when the shaded region is rotated completely about the \(x\)-axis. [5]

A finite region is bounded by the curve with equation \(y = x + x^{-\frac{3}{2}}\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\) This region is rotated through \(2\pi\) radians about the \(x\)-axis. Show that the volume generated is \(\pi\left(a\sqrt{2} + b\right)\), where \(a\) and \(b\) are rational numbers to be determined. [5 marks]

A finite region is bounded by the curve with equation \(y = x + x^{\frac{3}{2}}\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\) This region is rotated through \(2\pi\) radians about the \(x\)-axis. Show that the volume generated is \(\pi\left(a\sqrt{2} + b\right)\), where \(a\) and \(b\) are rational numbers to be determined. [5 marks]