# Area and Volume (Excercise Problems with Solutions)

$\underline{PART\ -\ A}$
$1.\ \color {red}{Find\ the\ Area\ bounded\ by\ the\ curve\ y\ =\ 2x^2}\ \hspace{10cm}$$\color{red}{the\ X-axis\ and\ ordinates\ x\ =\ 0\ and\ x\ =\ 2}\ \hspace{5cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Given\ y\ =\ 2x^2,\ a\ =\ 0\ and\ b\ =\ 2$
$Area =\ \int_{a}^{b} y\ dx$
$Area =\ \int_{0}^{2} 2x^2\ dx$
$=\ 2\ \int_{0}^{2} x^2\ dx$
$= 2\ \Biggr[\frac{x^3}{3} \Biggr]_{0}^{2}$
$=\ 2 \Biggr[ \frac{2^3}{3}\ -\ \frac{0^3}{3} \Biggr]$
$=\ 2 \frac{2^3}{3}$
$= \frac{16}{3}$
$\boxed{Area\ = \frac{16}{3}}$
$2.\ \color {red}{Find\ the\ Area\ bounded\ by\ the\ curve\ y\ =\ \frac{1}{x}}\ \hspace{10cm}$$\color{red}{the\ X-axis\ and\ ordinates\ x\ =\ 1\ and\ x\ =\ 2}\ \hspace{5cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Given\ y\ =\ \frac{1}{x},\ a\ =\ 1\ and\ b\ =\ 2$
$Area =\ \int_{a}^{b} y\ dx$
$Area =\ \int_{1}^{2} \frac{1}{x}\ dx$
$=\ \Biggr[log\ x \Biggr]_{1}^{2}$
$=\ log\ 2\ -\ log\ 1$
$=\ log\ 2\ \hspace{1cm}\ \because\ log\ 1\ =\ 0$
$\boxed{Area\ = log\ 2}$
$3.\ \color{red}{Write\ down\ the\ formula\ to\ find\ the\ Volume\ bounded\ by\ the\ curve\ y\ =\ f(x)}\ \hspace{8cm}$$\color{red}{the\ X-axis\ and\ the\ lines\ x\ =\ a\ and\ x\ =\ b}\ \hspace{7cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Volume\ =\ π \int_{a}^{b} y^2\ dx$
$4.\ \color{red}{Find\ the\ Volume\ bounded\ by\ the\ curve\ y^2\ =\ 4\ a\ x}\ \hspace{10cm}$$\color{red}{between\ x\ =\ 0\ and\ x\ =\ 2}\ \hspace{5cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Given\ y^2\ =\ 4\ a\ x,\ a\ =\ 0\ and\ b\ =\ 2$
$Volume\ =\ π \int_{a}^{b} y^2\ dx$
$=\ π \int_{0}^{2} 4\ a\ x\ dx$
$= 4\ a\ \pi\ \Biggr[\frac{x^2}{2} \Biggr]_{0}^{2}$
$=\ 4\ a\ \pi \Biggr[ \frac{2^2}{2}\ -\ \frac{0^2}{2} \Biggr]$
$=\ 4\ a\ \pi (2)$
$=\ 8\ a\ \pi$
$\boxed{Volume\ = 8\ a\ \pi\ cubic\ units}$
$\underline{PART\ -\ B}$
$5.\ \color{red}{Find\ the\ Area\ bounded\ by\ the\ curve\ xy\ =\ 1}\ \hspace{10cm}$$\color{red}{the\ X-axis\ and\ the\ lines\ x\ =\ 1\ and\ x\ =\ 2}\ \hspace{5cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Given\ xy\ =\ 1\ \implies\ y\ =\ \frac{1}{x},\ a\ =\ 1\ and\ b\ =\ 2$
$Area =\ \int_{a}^{b} y\ dx$
$Area =\ \int_{1}^{2} \frac{1}{x}\ dx$
$=\ \Biggr[log\ x \Biggr]_{1}^{2}$
$=\ log\ 2\ -\ log\ 1$
$=\ log\ 2\ \hspace{1cm}\ \because\ log\ 1\ =\ 0$
$\boxed{Area\ = log\ 2}$
$6.\ \color {red}{Find\ the\ Area\ bounded\ by\ the\ curve\ y\ =\ 4\ x\ -\ x^2}\ \hspace{10cm}$$\color{red}{and\ the\ X-axis}\ \hspace{5cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$To\ find\ limits\ put\ y\ =\ 0\ as\ the\ curve\ meets\ X-\ axis$
$4\ x\ -\ x^2\ =\ 0$
$x(4\ -\ x)\ =\ 0$
$The\ limts\ are\ x\ =\ 0\ and\ x\ =\ 4$
$Area =\ \int_{a}^{b} y\ dx$
$=\ \int_{0}^{4} (4\ x\ -\ x^2)\ dx$
$= \Biggr[\frac{4x^2}{2}\ -\ \frac{x^3}{3} \Biggr]_{0}^{4}$
$=\ \Biggr[(4 \frac{4^2}{2}\ -\ \frac{4^3}{3})\ -\ (4 \frac{0^2}{2}\ -\ \frac{0^3}{3})\Biggr]$
$=\ 32\ -\ \frac{64}{3}$
$= \frac{96\ -\ 64}{3}$
$\boxed{Area\ =\ \frac{32}{3}\ sq\ units}$
$\underline{PART\ -\ C}$
$7.\ \color {red}{Find\ the\ Volume\ got\ by\ rotating\ the\ loop\ of\ \ y^2\ =\ 6\ +\ x\ -\ x^2}\ \hspace{10cm}$$\color{red}{about\ the\ X-axis}\ \hspace{5cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$To\ find\ limits\ put\ y\ =\ 0\ as\ the\ curve\ meets\ X-\ axis$
$6\ +\ x\ -\ x^2\ =\ 0$
$x^2\ -\ x\ -\ 6\ =\ 0$
$x^2\ -\ 3x\ +\ 2x -\ 6\ =\ 0$
$x(x\ -\ 3)\ +\ 2(x\ -\ 3)\ =\ 0$
$(x\ +\ 2)(x\ -\ 3)\ =\ 0$
$The\ limts\ are\ x\ =\ -\ 2\ and\ x\ =\ 3$
$Volume\ =\ π \int_{a}^{b} y^2\ dx$
$V\ =\ π \int_{-2}^{3} (6\ +\ x\ -\ x^2)\ dx$
$=\ π \Biggr[6\ x\ +\ \frac{x^2}{2}\ -\ \frac{x^3}{3} \Biggr]_{-2}^{3}$
$=\ π \Biggr[(6(3)\ +\ \frac{3^2}{2}\ -\ \frac{3^3}{3})\ -\ (6(-2)\ +\ \frac{(-2)^2}{2}\ -\ \frac{(-2)^3}{3})\Biggr]$
$=\ π \Biggr[(18\ +\ \frac{9}{2}\ -\ 9)\ -\ (-12\ +\ 2\ +\ \frac{8}{3})\Biggr]$
$=\ π \Biggr[9\ +\ \frac{9}{2}\ +\ 10\ -\ \frac{8}{3}\Biggr]$
$=\ π \Biggr[19\ +\ \frac{9}{2}\ -\ \frac{8}{3}\Biggr]$
$=\ π \Biggr[\frac{114\ +\ 27\ -\ 16}{6}\ \Biggr]$
$\boxed{Volume\ =\ π \frac{125}{6}}$