# Area Enclosed by Parametric Curves

We know that area under the curve ${y}={F}{\left({x}\right)}$ is ${A}={\int_{{a}}^{{b}}}{F}{\left({x}\right)}{d}{x}$ where ${f{{\left({x}\right)}}}\ge{0}$.

If curve is given by parametric equations ${x}={f{{\left({t}\right)}}}$ and ${y}={g{{\left({t}\right)}}}$ then using substitution rule with ${x}={f{{\left({t}\right)}}}$ we have that ${d}{x}={f{'}}{\left({t}\right)}{d}{t}$ and since ${x}$ is changing from ${a}$ to ${b}$ then ${t}$ is changing from $\alpha={{f}}^{{-{1}}}{\left({a}\right)}$ to $\beta={{f}}^{{-{1}}}{\left({b}\right)}$. It is not always a case, sometimes ${t}$ is changing from $\beta$ to $\alpha$.

So, ${A}={\int_{{a}}^{{b}}}{F}{\left({x}\right)}{d}{x}={\int_{\alpha}^{\beta}}{F}{\left({f{{\left({t}\right)}}}\right)}{f{'}}{\left({t}\right)}{d}{t}={\int_{\alpha}^{\beta}}{g{{\left({t}\right)}}}{f{'}}{\left({t}\right)}{d}{t}$.

Area Enclosed by Parametric Curves: ${A}={\int_{{\alpha}}^{{\beta}}}{g{{\left({t}\right)}}}{f{'}}{\left({t}\right)}{d}{t}$ (or ${\int_{\beta}^{\alpha}}{g{{\left({t}\right)}}}{f{'}}{\left({t}\right)}{d}{t}$).

Example. Find the area under one arch of cycloid. ${x}={r}{\left({t}-{\sin{{\left({t}\right)}}}\right)}$, ${y}={r}{\left({1}-{\cos{{\left({t}\right)}}}\right)}$.

One arch of the cycloid is given by ${0}\le{t}\le{2}\pi$.

We have that ${x}_{{t}}'={r}{\left({1}-{\cos{{\left({t}\right)}}}\right)}$.

Therefore, ${A}={\int_{{0}}^{{{2}\pi}}}{r}{\left({1}-{\cos{{\left({t}\right)}}}\right)}{r}{\left({1}-{\cos{{\left({t}\right)}}}\right)}{d}{t}=$

$={{r}}^{{2}}{\int_{{0}}^{{{2}\pi}}}{{\left({1}-{\cos{{\left({t}\right)}}}\right)}}^{{2}}{d}{t}={{r}}^{{2}}{\int_{{0}}^{{{2}\pi}}}{\left({1}-{2}{\cos{{\left({t}\right)}}}+{{\cos}}^{{2}}{\left({t}\right)}\right)}{d}{t}$.

Now, we need to use double angle formula and integral becomes:

${A}{{r}}^{{2}}{\int_{{0}}^{{{2}\pi}}}{\left({1}-{2}{\cos{{\left({t}\right)}}}+\frac{{1}}{{2}}{\left({1}+{\cos{{\left({2}{t}\right)}}}\right)}\right)}{d}{t}={{r}}^{{2}}{\int_{{0}}^{{{2}\pi}}}{\left(\frac{{3}}{{2}}-{2}{\cos{{\left({t}\right)}}}+\frac{{1}}{{2}}{\cos{{\left({2}{t}\right)}}}\right)}{d}{t}=$

$={{r}}^{{2}}{\left(\frac{{3}}{{2}}{t}-{2}{\sin{{\left({t}\right)}}}+\frac{{1}}{{4}}{\sin{{\left({2}{t}\right)}}}\right)}{{\mid}_{{0}}^{{{2}\pi}}}=$

$={{r}}^{{2}}{\left({\left(\frac{{3}}{{2}}{2}\pi-{2}{\sin{{\left({2}\pi\right)}}}+\frac{{1}}{{4}}{\sin{{\left({2}\cdot{2}\pi\right)}}}\right)}-{\left(\frac{{3}}{{2}}\cdot{0}-{2}{\sin{{\left({0}\right)}}}+\frac{{1}}{{4}}{\sin{{\left({2}\cdot{0}\right)}}}\right)}\right)}=$

$={{r}}^{{2}}{\left({\left({3}\pi-{2}\cdot{0}+\frac{{1}}{{4}}\cdot{0}\right)}-{\left({0}-{2}\cdot{0}+\frac{{1}}{{4}}\cdot{0}\right)}\right)}={3}\pi{{r}}^{{2}}$.