Ache o comprimento do arco dado. Quando aparecer a, a>0.
r=a \operatorname{sen}^{3} \frac{1}{3} \theta; de \theta=0 a \theta=\theta_{1}.

Passo 1

Resumo do enunciado: queremos o comprimento de arco da curva dada em coordenadas polares por
r = a\sin^{3}\!\bigl(\tfrac{\theta}{3}\bigr),
para \theta variando de 0 até \theta_{1}.
Conteúdo utilizado: comprimento de arco em coordenadas polares, cuja fórmula é
L = \int_{\theta=0}^{\theta=\theta_{1}} \sqrt{r^{2} + \bigl(\tfrac{dr}{d\theta}\bigr)^{2}}\;d\theta.

Passo 2

Calcular a derivada dr/d\theta. Escrevendo u=\sin(\theta/3), temos r=a\,u^{3}. Então
\frac{dr}{d\theta} = a\cdot 3u^{2}\,\frac{du}{d\theta} =3a\sin^{2}\!\bigl(\tfrac{\theta}{3}\bigr)\cdot\Bigl(\cos\!\bigl(\tfrac{\theta}{3}\bigr)\cdot\tfrac{1}{3}\Bigr) = a\,\sin^{2}\!\bigl(\tfrac{\theta}{3}\bigr)\cos\!\bigl(\tfrac{\theta}{3}\bigr).

Passo 3

Montar o integrando \sqrt{r^{2} + (dr/d\theta)^{2}}. Temos
r^{2} = a^{2}\sin^{6}\!\bigl(\tfrac{\theta}{3}\bigr), \quad \bigl(\tfrac{dr}{d\theta}\bigr)^{2} = a^{2}\sin^{4}\!\bigl(\tfrac{\theta}{3}\bigr)\cos^{2}\!\bigl(\tfrac{\theta}{3}\bigr).
Somando,
r^{2} + \bigl(\tfrac{dr}{d\theta}\bigr)^{2} = a^{2}\sin^{4}\!\bigl(\tfrac{\theta}{3}\bigr)\bigl[\sin^{2}\!\bigl(\tfrac{\theta}{3}\bigr)+\cos^{2}\!\bigl(\tfrac{\theta}{3}\bigr)\bigr] = a^{2}\sin^{4}\!\bigl(\tfrac{\theta}{3}\bigr),
logo
\sqrt{r^{2} + \bigl(\tfrac{dr}{d\theta}\bigr)^{2}} = a\,\sin^{2}\!\bigl(\tfrac{\theta}{3}\bigr).

Passo 4

Substituir no integral de comprimento:
L = \int_{0}^{\theta_{1}} a\,\sin^{2}\!\bigl(\tfrac{\theta}{3}\bigr)\,d\theta = a\int_{0}^{\theta_{1}}\sin^{2}\!\bigl(\tfrac{\theta}{3}\bigr)\,d\theta.
Usamos a identidade \sin^{2}x=(1-\cos(2x))/2, com x=\theta/3:
\sin^{2}\!\bigl(\tfrac{\theta}{3}\bigr) =\frac{1-\cos\!\bigl(\tfrac{2\theta}{3}\bigr)}{2}.
Então
L = a\int_{0}^{\theta_{1}}\frac{1-\cos\!\bigl(\tfrac{2\theta}{3}\bigr)}{2}\,d\theta =\frac{a}{2}\Bigl[\int_{0}^{\theta_{1}}1\,d\theta -\int_{0}^{\theta_{1}}\cos\!\bigl(\tfrac{2\theta}{3}\bigr)\,d\theta\Bigr].

Passo 5

Calcular as integrais:
\int_{0}^{\theta_{1}}1\,d\theta = \theta_{1}, \qquad \int_{0}^{\theta_{1}}\cos\!\bigl(\tfrac{2\theta}{3}\bigr)\,d\theta = \frac{3}{2}\,\sin\!\bigl(\tfrac{2\theta}{3}\bigr)\Big|_{0}^{\theta_{1}} = \frac{3}{2}\,\sin\!\bigl(\tfrac{2\theta_{1}}{3}\bigr).
Logo
L = \frac{a}{2}\Bigl[\theta_{1} - \frac{3}{2}\sin\!\bigl(\tfrac{2\theta_{1}}{3}\bigr)\Bigr].

Resposta

L = \frac{a\,\theta_{1}}{2}-\frac{3a}{4}\,\sin\!\bigl(\tfrac{2\theta_{1}}{3}\bigr).

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