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Area Enclosed by Parametric Curves

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

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

So, A=∫abF(x)dx=∫αβF(f(t))f′(t)dt=∫αβg(t)f′(t)dt{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=∫αβg(t)f′(t)dt{A}={\int_{{\alpha}}^{{\beta}}}{g{{\left({t}\right)}}}{f{'}}{\left({t}\right)}{d}{t} (or ∫βαg(t)f′(t)dt{\int_{\beta}^{\alpha}}{g{{\left({t}\right)}}}{f{'}}{\left({t}\right)}{d}{t}).

Example. Find the area under one arch of cycloid. x=r(t−sin⁡(t)){x}={r}{\left({t}-{\sin{{\left({t}\right)}}}\right)}, y=r(1−cos⁡(t)){y}={r}{\left({1}-{\cos{{\left({t}\right)}}}\right)}.area enclosed by parametric curve: cycloid

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

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

Therefore, A=∫02πr(1−cos⁡(t))r(1−cos⁡(t))dt={A}={\int_{{0}}^{{{2}\pi}}}{r}{\left({1}-{\cos{{\left({t}\right)}}}\right)}{r}{\left({1}-{\cos{{\left({t}\right)}}}\right)}{d}{t}=

=r2∫02π(1−cos⁡(t))2dt=r2∫02π(1−2cos⁡(t)+cos⁡2(t))dt={{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:

Ar2∫02π(1−2cos⁡(t)+12(1+cos⁡(2t)))dt=r2∫02π(32−2cos⁡(t)+12cos⁡(2t))dt={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}=

=r2(32t−2sin⁡(t)+14sin⁡(2t))∣02π=={{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}}}=

=r2((322π−2sin⁡(2π)+14sin⁡(2⋅2π))−(32⋅0−2sin⁡(0)+14sin⁡(2⋅0)))=={{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)}=

=r2((3π−2⋅0+14⋅0)−(0−2⋅0+14⋅0))=3πr2={{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}}.