# Area and Volume

$We\ apply\ the\ concept\ of\ definite\ integral\ to\ find\ the\ area\ and\ volume$

## Area

$The\ area\ under\ the\ curve\ y\ =\ f(x)\ between\ the\ x-axis\ and$$the\ ordinates\ x\ =\ a\ and\ x\ =\ b\ is\ given\ by$$\boxed{Area =\ \int_{a}^{b} y\ dx}$

Volume

$The\ volume\ of\ the\ solid\ obtained\ by\ rotating\ the\ area\ bounded\ by\ the\ curve\ y\ =\ f(x)\, the\ x – axis$$and\ the\ ordinates\ is\ given\ by$$\boxed{Volume\ =\ π \int_{a}^{b} y^2\ dx}$

Example  1:

$\color{red}{Find\ the\ Area\ bounded\ by\ the\ curve\ y\ =\ x^2}\ \hspace{8cm}$$\color{red}{the\ X-axis\ and\ ordinates\ x\ =\ 0\ and\ x\ =\ 2}\ \hspace{7cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Given\ y\ =\ x^2,\ a\ =\ 0\ and\ b\ =\ 2$
$Area =\ \int_{a}^{b} y\ dx$
$Area =\ \int_{0}^{2} x^2\ dx$
$= \frac{x^3}{3} \Biggr]_{0}^{2}$
$= [\frac{2^3}{3} – \frac{0^3}{3}]$
$= \frac{8}{3} – \ 0$
$= \frac{8}{3}$
$\boxed{Area\ = \frac{8}{3}}$

#### Example  2:

$\color{red}{Find\ the\ Volume\ of\ the\ solid\ formed\ when\ the\ area\ bounded\ by\ the\ curve\ y^2\ =\ 4\ x}\ \hspace{8cm}$$\color{red}{between\ x\ =\ 0\ and\ x\ =\ 1\ is\ rotated\ about\ the\ X\ -\ axis.}\ \hspace{5cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Given\ y^2\ =\ 4\ x,\ a\ =\ 0\ and\ b\ =\ 1$
$Volume\ =\ π \int_{a}^{b} y^2\ dx$
$=\ π \int_{0}^{1} 4\ x\ dx$
$= 4\ \pi\ \Biggr[\frac{x^2}{2} \Biggr]_{0}^{1}$
$=\ 4\ \pi \Biggr[ \frac{1^2}{2}\ -\ \frac{0^2}{2} \Biggr]$
$=\ 4\ \pi (\frac{1} {2})$
$=\ 2\ \pi$
$\boxed{Volume\ =\ 2\ \pi\ cubic\ units}$

Example  3:

$\color{red}{Find\ the\ Volume\ bounded\ by\ the\ curve\ y^2\ =\ x^6,}\ \hspace{8cm}$$\color{red}{the\ X-axis\ and\ ordinates\ x\ =\ 0\ and\ x\ =\ 1}\ \hspace{7cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Given\ y^2\ =\ x^6,\ a\ =\ 0\ and\ b\ =\ 1$
$Volume\ =\ π \int_{a}^{b} y^2\ dx$
$=\ π \int_{0}^{1} x^6\ dx$
$=\ π \frac{x^7}{7} \Biggr]_{0}^{1}$
$=\ π [\frac{1^7}{7} – \frac{0^7}{7}]$
$= \frac{π}{7}$
$\boxed{Volume\ = \frac{π}{7}}$

Example  4:

$\color {red}{Find\ the\ Area\ bounded\ by\ the\ curve\ y\ =\ x^2\ -\ 6x\ +\ 8}\ \hspace{10cm}$$\color{red}{and\ the\ X-axis}\ \hspace{5cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$To\ find\ limits\ put\ y\ =\ 0\ as\ the\ curve\ meets\ X-\ axis$
$x^2\ -\ 6x\ +\ 8\ =\ 0$
$x^2\ -\ 2x\ -\ 4x +\ 8\ =\ 0$
$x(x\ -\ 2)\ -\ 4(x\ -\ 2)\ =\ 0$
$(x\ -\ 2)(x\ -\ 4)\ =\ 0$
$The\ limts\ are\ x\ =\ \ 2\ and\ x\ =\ 4$
$Area =\ \int_{a}^{b} y\ dx$
$Area\ =\ \int_{2}^{4} (x^2\ -\ 6x\ +\ 8)\ dx$
$=\ \Biggr[\frac{x^3}{3}\ -\ 6\ \frac{x^2}{2}\ +\ 8\ x \Biggr]_{2}^{4}$
$=\ \Biggr[(\frac{4^3}{3}\ -\ 6\frac{4^2}{2}\ +\ 8(4))\ -\ (\frac{(2)^3}{3}\ -\ 6\frac{2^2}{2}\ +\ 8(2))\Biggr]$
$=\ \Biggr[(\frac{64}{3}\ -\ 48\ +\ 32)\ -\ (\frac{8}{3}\ -\ 12\ +\ 16)\Biggr]$
$=\ \Biggr[(\frac{64}{3}\ -\ 16)\ -\ (\frac{8}{3}\ +\ 4)\Biggr]$
$=\ \Biggr[\frac{64}{3}\ -\ \frac{8}{3}\ -\ 20\Biggr]$
$=\ \Biggr[\frac{64\ -\ 8\ -\ 60}{3}\Biggr]$
$= \frac{-\ 4}{3}$
$\boxed{Area\ =\ \frac{4}{3}\ sq\ units (as\ Area\ is\ positive)}$

Example  4:

$\color{red}{Find\ the\ Area\ of\ the\ circle\ whose\ radius\ is\ ‘a’\ units}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Equation\ of\ circle\ is\ x^2\ +\ y^2\ =\ a^2$
$y^2\ =\ a^2\ -\ x^2$
$y\ =\ \sqrt{a^2\ -\ x^2}$
$Area\ of\ OAB\ =\ \int_{0}^{a} \sqrt{a^2\ -\ x^2}\ dx$
$put\ x\ =\ a\ sin\ \theta$
$\frac{dx}{d\theta}\ =\ a\ cos\ \theta$
$dx\ =\ a\ cos\ \theta\ d\theta$
$when\ x\ =\ 0\ \hspace{2cm}\ \theta\ =\ 0$
$when\ x\ =\ a\ \hspace{2cm}\ a\ =\ a\ sin\ \theta$
$\hspace{4cm}\ sin\ \theta\ =\ 1$
$\hspace{4cm}\ \theta\ =\ \frac{\pi}{2}$
$Area\ of\ OAB\ =\ \int_{0}^{\frac{\pi}{2}} \sqrt{a^2\ -\ a^2\ sin^2\ \theta}\ a\ cos\ \theta\ d \theta$
$=\ \int_{0}^{\frac{\pi}{2}}\ a \sqrt{1\ -\ sin^2\ \theta}\ a\ cos\ \theta\ d \theta$
$=\ \int_{0}^{\frac{\pi}{2}}\ a \ cos\ \theta\ a\ cos\ \theta\ d \theta$
$=\ \int_{0}^{\frac{\pi}{2}}\ a^2 \ cos^2\ \theta\ d \theta$
$=\ \int_{0}^{\frac{\pi}{2}}\ a^2 \ (\frac{1\ +\ cos\ 2\ \theta}{2})\ d \theta$
$=\ \frac{a^2}{2}[\int_{0}^{\frac{\pi}{2}} 1\ d \theta\ +\ \int_{0}^{\frac{\pi}{2}} cos\ 2\ \theta\ d \theta]$
$=\ \frac{a^2}{2}\ [ \Biggr[ \theta \Biggr]_{0}^{\frac{\pi}{2}}\ +\ \Biggr[\frac{sin\ 2\theta}{2} \Biggr]_{0}^{\frac{\pi}{2}}]$
$=\ \frac{a^2}{2}\ [(\frac{\pi}{2}\ -\ 0)\ +\ \frac{1}{2}(sin\ \pi\ -\ sin\ 0)]$
$=\ \frac{a^2}{2}\ [(\frac{\pi}{2})\ +\ \frac{1}{2}(0)]$
$Area\ of\ OAB\ =\ \frac{\pi\ a^2}{4}$
$Area\ of\ circle\ =\ 4\ \times\ Area\ of\ OAB$
$=\ 4\ \times\ \frac{\pi\ a^2}{4}$
$=\ \pi\ a^2$
$\boxed{Area\ of\ circle\ =\ \pi\ a^2}$

Example  5:

$\color{red}{Find\ the\ volume\ of\ sphere\ having\ radius\ ‘a’\ by\ using\ integration}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Rotate\ the\ area\ OAB\ (Quadrant\ of\ circle)\ about\ OA,\ the\ x- axis.$$Then\ we\ get\ a\ hemisphere.$
$Volume\ of\ hemisphere\ is$
$V\ =\ π \int_{0}^{a} y^2\ dx$
$\ =\ π \int_{0}^{a} (a^2\ -\ x^2)\ dx$
$=\ \pi \Biggr[a^2\ x\ -\ \frac{x^3}{3}\Biggr]_ {0}^{a}$
$=\ \pi \Biggr[a^2\ a\ -\ \frac{a^3}{3}\Biggr]$
$=\ \pi \Biggr[a^3\ -\ \frac{a^3}{3}\Biggr]$
$=\ \pi \Biggr[\frac{3a^3\ -\ a^3}{3}\Biggr]$
$=\ \pi \Biggr[\frac{2a^3}{3}\Biggr]$
$V\ =\ \frac{2\ \pi\ a^3}{3}$
$Volume\ of\ sphere\ =\ 2\ \times\ \frac{2\ \pi\ a^3}{3}$
$\boxed{Volume\ of\ sphere\ =\ \frac{4\ \pi\ a^3}{3}}$

Example  6:

$\color{red}{Find\ the\ volume\ of\ right\ circular\ cone\ \ of\ height\ ‘h’\ and\ base\ radius\ ‘r’\ by\ integration}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Equation\ of\ OB\ is\ y\ =\ mx$
$m\ =\ tan\ \theta\ =\ \frac{r}{h}$
$Volume\ of\ the\ cone\ is$
$V\ =\ π \int_{0}^{h} y^2\ dx$
$V\ =\ π \int_{0}^{h} m^2\ x^2\ dx$
$=\ π \int_{0}^{h} \frac{r^2}{h^2}\ x^2\ dx$
$=\ π \frac{r^2}{h^2} \Biggr[ \frac{x^3}{3}\ \Biggr]_{0}^{h}$
$=\ π \frac{r^2}{h^2} \Biggr[ \frac{h^3}{3}\ -\ \frac{0^3}{3} \Biggr]$
$=\ π \frac{r^2}{h^2} \frac{h^3}{3}$
$\boxed{Volume\ of\ cone\ =\ \frac{1}{3}\ \pi\ r^2\ h}$

#### Example  7:

$\color {red}{Find\ the\ Volume\ of\ the\ solid\ generated\ by\ the\ area\ enclosed\ by\ the\ curve\ y^2\ =\ x(x\ -\ 1)^2}\ \hspace{10cm}$$\color{red}{and\ the\ X-axis\ when\ rotated\ about\ X- axis}\ \hspace{5cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$y^2\ =\ x(x\ -\ 1)^2$
$To\ find\ limits\ put\ y\ =\ 0\ as\ the\ curve\ meets\ X-\ axis$
$x(x\ -\ 1)^2\ =\ 0$
$x\ =\ 0\ and\ (x\ -\ 1)^2\ =\ 0$
$The\ limts\ are\ x\ =\ 0\ and\ x\ =\ 1$
$Volume\ =\ π \int_{a}^{b} y^2\ dx$
$V\ =\ π \int_{0}^{1} x(x\ -\ 1)^2\ dx$
$=\ π \int_{0}^{1} x (x^2\ -\ 2x\ +\ 1)\ dx$
$=\ π \int_{0}^{1} (x^3\ -\ 2x^2\ +\ x)\ dx$
$=\ π \Biggr[\frac{x^4}{4}\ -\ 2\ \frac{x^3}{3}\ +\ \frac{x^2}{2} \Biggr]_{0}^{1}$
$=\ π \Biggr[(\frac{1^4}{4}\ -\ 2\ \frac{1^3}{3}\ +\ \frac{1^2}{2})\ -\ ((\frac{0^4}{4}\ -\ 2\ \frac{0^3}{3}\ +\ \frac{0^2}{2})\Biggr]$
$=\ π \Biggr[(\frac{1}{4}\ -\ \frac{2}{3}\ +\ \frac{1}{2})\ -\ (0\ -\ 2(0)\ +\ 0)\Biggr]$
$=\ π \Biggr[(\frac{3\ -\ 8\ +\ 6}{12})\ -\ (0) \Biggr]$
$=\ π \Biggr[\frac{1}{12}\ \Biggr]$
$\boxed{Volume\ =\ \frac{π}{12}\ cubic\ units}$