Show integral then evaluate volume

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

1. Show that the volume of the solid of revolution formed is given by $$4 \pi \int _ { 0 } ^ { \frac { \pi } { 3 } } \left( \tan ^ { 2 } t - \sin ^ { 2 } t \right) \mathrm { d } t$$
2. Hence use integration to find the exact value for this volume.

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.

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.

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

1. Show that the volume of the solid of revolution is given by $$k \int _ { 0 } ^ { a } \sin ^ { 2 } t \cos ^ { 3 } t \mathrm {~d} t$$ where \(k\) and \(a\) are constants to be given in terms of \(\pi\).
2. Use the substitution \(u = \sin t\), or otherwise, to find the exact value of this volume, giving your answer in the form \(\frac { p \pi } { q }\) where \(p\) and \(q\) are integers. (Solutions based entirely on graphical or numerical methods are not acceptable.)

\includegraphics{figure_3} The curve \(C\), shown in Figure 3, has equation $$y = \frac{x^{-\frac{1}{4}}}{\sqrt{1+x}\left(\arctan\sqrt{x}\right)}$$ The region \(R\), shown shaded in Figure 3, is bounded by \(C\), the line with equation \(x = 3\), the \(x\)-axis and the line with equation \(x = \frac{1}{3}\) The region \(R\) is rotated through \(360°\) about the \(x\)-axis to form a solid. Using the substitution \(\tan u = \sqrt{x}\)

1. show that the volume \(V\) of the solid formed is given by $$k \int_a^b \frac{1}{u^2} du$$ where \(k\), \(a\) and \(b\) are constants to be found. [6]
2. Hence, using algebraic integration, find the value of \(V\) in simplest form. [3]