# Find the exact length of the polar curve. r=\theta^{2},\ 0\le\theta\le2

Find the exact length of the polar curve.
$$\displaystyle{r}=\theta^{{{2}}},\ {0}\le\theta\le{2}\pi$$

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Opeance1951
Step 1
It is given that
$$\displaystyle{r}=\theta^{{{2}}}$$
Differentiate using the power rule
$$\displaystyle{\frac{{{d}{r}}}{{{d}\theta}}}={2}\cdot\theta^{{{2}-{1}}}$$
$$\displaystyle{\frac{{{d}{r}}}{{{d}\theta}}}={2}\theta$$
Formula 6 from this section:
Length of a curve with polar equation $$\displaystyle{r}={f{{\left(\theta\right)}}},\ {a}\le\theta\le{b}$$ is given by
$$\displaystyle{L}={\int_{{{a}}}^{{{b}}}}\sqrt{{{r}^{{{2}}}+{\left[{\frac{{{d}{r}}}{{{d}\theta}}}\right]}^{{{2}}}}}{d}\theta$$
$$\displaystyle={\int_{{{0}}}^{{{2}\pi}}}\sqrt{{{\left(\theta^{{{2}}}\right)}^{{{2}}}+{\left({2}\theta\right)}^{{{2}}}}}{d}\theta$$
$$\displaystyle={\int_{{{0}}}^{{{2}\pi}}}\sqrt{{\theta^{{{4}}}+{4}\theta^{{{2}}}}}{d}\theta$$
Note that: $$\displaystyle\sqrt{{\theta^{{{4}}}+{4}\theta^{{{2}}}}}=\sqrt{{\theta^{{{2}}}{\left(\theta^{{{2}}}+{4}\right)}}}=\sqrt{{\theta^{{{2}}}}}\cdot\sqrt{{\theta^{{{2}}}+{4}}}=\theta\cdot\sqrt{{\theta^{{{2}}}+{4}}}$$
Since we are integrating over $$\displaystyle{\left[{0},\ {2}\pi\right]},\ \theta$$ is positive and $$\displaystyle\sqrt{{\theta^{{{2}}}={\left|\theta\right|}=\theta}}$$
$$\displaystyle={\int_{{{0}}}^{{{2}\pi}}}\theta\cdot\sqrt{{\theta^{{{2}}}+{4}}}{d}\theta$$
$$\displaystyle={\int_{{{0}}}^{{{2}\pi}}}\sqrt{{\theta^{{{2}}}+{4}}}{\left[\theta\ {d}\theta\right]}$$
We will now substitute
$$\displaystyle\theta^{{{2}}}+\={u}$$
Step 2
On differentiation, we get
$$\displaystyle{2}\theta,{d}\theta={d}{u}\Rightarrow\theta{d}\theta={\frac{{{1}}}{{{2}}}}{d}{u}$$
The limits of integration will change from $$\displaystyle{\int_{{{0}}}^{{{2}\pi}}}$$ to $$\displaystyle{\int_{{{0}^{{{2}}}+{4}}}^{{{4}\pi^{{{2}}}+{4}}}}$$
$$\displaystyle={\int_{{{4}}}^{{{4}\pi^{{{2}}}+{4}}}}\sqrt{{{u}}}{\left[{\frac{{{1}}}{{{2}}}}{d}{u}\right]}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}{\int_{{{4}}}^{{{4}\pi^{{{2}}}+{4}}}}{u}^{{\frac{{1}}{{2}}}}{d}{u}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}{{\left[{\frac{{{u}^{{{\left({\frac{{{1}}}{{{2}}}}\right)}+{1}}}}}{{{\left({\frac{{{1}}}{{{2}}}}\right)}+{1}}}}\right]}_{{{4}}}^{{{4}\pi^{{{2}}}+{4}}}}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}{\left[{u}^{{{\frac{{{3}}}{{{2}}}}}}\right\rbrace}{\left\lbrace{\frac{{{3}}}{{{2}}}}\right\rbrace}{{]}_{{{4}}}^{{{4}\pi^{{{2}}}+{4}}}}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}{{\left[{\frac{{{2}}}{{{3}}}}\cdot{u}^{{{\frac{{{3}}}{{{2}}}}}}\right]}_{{{4}}}^{{{4}\pi^{{{2}}}+{4}}}}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}{\left[{\frac{{{2}}}{{{3}}}}\cdot{\left({4}\pi^{{{2}}}+{4}\right)}^{{{\frac{{{3}}}{{{2}}}}}}-{\frac{{{2}}}{{{3}}}}\cdot{4}^{{{\frac{{{3}}}{{{2}}}}}}\right]}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}\cdot{\left({2}^{{{2}}}\right)}^{{{\frac{{{3}}}{{{2}}}}}}\cdot{\left(\pi^{{{2}}}+{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}-{\frac{{{1}}}{{{3}}}}\cdot{\left({2}^{{{2}}}\right)}^{{{\frac{{{3}}}{{{2}}}}}}$$
$$\displaystyle={\frac{{{1}}}{{{3}}}}\cdot{2}^{{{3}}}\cdot{\left(\pi^{{{2}}}+{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}-{\frac{{{1}}}{{{3}}}}\cdot{2}^{{{3}}}$$
$$\displaystyle={\frac{{{8}}}{{{3}}}}\cdot{\left[{\left(\pi^{{{2}}}+{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}-{1}\right]}$$
$$\displaystyle{L}={\frac{{{8}}}{{{3}}}}\cdot{\left[{\left(\pi^{{{2}}}+{1}\right)}^{{{\frac{{{3}}}{{{2}}}}}}-{1}\right]}$$
###### Have a similar question?
huckelig75
Step 1
Given: $$\displaystyle{r}=\theta^{{{2}}}$$
$$\displaystyle{l}={\int_{{{a}}}^{{{b}}}}\sqrt{{{r}^{{{2}}}+{\left({\frac{{{d}{r}}}{{{d}\theta}}}\right)}^{{{2}}}{d}\theta}}$$ - formula for length of a curve.
$$\displaystyle{\frac{{{d}{r}}}{{{d}\theta}}}={2}\theta$$
$$\displaystyle{L}={\int_{{{0}}}^{{{2}\pi}}}\sqrt{{{\left(\theta^{{{2}}}\right)}^{{{2}}}+{\left({2}\theta\right)}^{{{2}}}}}{d}\theta$$
$$\displaystyle={\int_{{{0}}}^{{{2}\pi}}}\sqrt{{\theta^{{{4}}}+{4}\theta^{{{2}}}}}{d}\theta$$
$$\displaystyle={\int_{{{0}}}^{{{2}\pi}}}\sqrt{{\theta^{{{2}}}{\left(\theta^{{{2}}}+{4}\right)}}}{d}\theta$$
$$\displaystyle={\int_{{{0}}}^{{{2}\pi}}}\theta\sqrt{{\theta^{{{2}}}+{4}}}{d}\theta$$
Substitution:
$$\displaystyle{t}=\theta^{{{2}}}+{4}$$
$$\displaystyle{\left.{d}{t}\right.}={2}\theta{d}\theta\rightarrow\theta{d}\theta={\frac{{{\left.{d}{t}\right.}}}{{{2}}}}$$
$$\displaystyle{0}\rightarrow{4}$$
$$\displaystyle{2}\pi\rightarrow{4}\pi^{{{2}}}+{4}$$
$$\displaystyle{L}={\frac{{{1}}}{{{2}}}}{\int_{{{4}}}^{{{4}\pi^{{{2}}}+{4}}}}\sqrt{{{t}{\left.{d}{t}\right.}}}={\frac{{{1}}}{{{2}}}}{\left({\frac{{{2}}}{{{3}}}}{t}^{{{\frac{{{3}}}{{{2}}}}}}\right)}{{\mid}_{{{4}}}^{{{4}\pi^{{{2}}}+{4}}}}$$
$$\displaystyle={\frac{{{1}}}{{{3}}}}{\left({\left({4}\pi^{{{2}}}+{4}\right)}^{{{\frac{{{3}}}{{{2}}}}}}-{4}^{{{\frac{{{3}}}{{{2}}}}}}\right)}\approx{92.89}$$
$$\displaystyle{L}\approx{92.89}$$