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:

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

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:

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