Sue Pemberton Solutions for Chapter: Integration, Exercise 9: EXERCISE 9I

The volume obtained when the shaded region is rotated through $360°$ about the $y-$axis is of the form of $\frac{k\pi }{2}unit{s}^3$, then $k=$

The volume obtained when the shaded region is rotated through $360°$ about the $y-$axis is of the form of $\frac{k\mathrm{\pi }}{15}$, then $k=$

The diagram shows part of the curve $y=\frac{a}{x}$, where $a>0$. The volume obtained when the shaded region is rotated through $360°$ about the $x-$axis is $18\pi$. Find the value of $a$.

The diagram shows part of the curve $y=\sqrt{x^3+4x^2+3x+2}$. Find the volume obtained when the shaded region is rotated through $360°$ about the $x-$axis is of the form of $\frac{k\mathrm{\pi }}{4}$, then $k=$

Sketch the graph of $y=(x-2{)}^2$.

Find the volume of the solid formed when the enclosed region bounded by the curve $y=(x-2{)}^2$, the $x-$axis and the $y-$axis is rotated through $360°$ about the $x-$axis is of the form of $\frac{k\mathrm{\pi }}{5}$, then $k=$

The diagram shows part of the curve $y=\frac{2}{2x+1}$. The shaded area is rotated through $360°$ about the $x-$axis between $x=0$ and $x=p$. Show that as $p\to \infty$, the volume approaches the value $2\pi$.

Use integration to prove that the volume, $V{ \mathrm{cm}}^3$, of a sphere with radius $r \mathrm{cm}$ is given by the formula $V=\frac{4}{3}\pi r^3$.