Ache o comprimento do arco dado. Quando aparecer a, a>0.
r=\operatorname{sen}^{2} \frac{1}{2} \theta; de \theta=0 a \theta=\frac{1}{2} \pi.

Passo 1

O enunciado pede o comprimento de arco da curva polar
r = \operatorname{sen}^{2}\!\Bigl(\frac{\theta}{2}\Bigr)
para \theta variando de 0 a \tfrac{\pi}{2}. O conteúdo usado é a fórmula do comprimento de arco em polares:
L = \int_{\theta_1}^{\theta_2} \sqrt{r^2 + \Bigl(\frac{dr}{d\theta}\Bigr)^{2}}\\ d\theta.

Passo 2

Vamos achar \dfrac{dr}{d\theta} usando a regra da cadeia. Seja u=\tfrac{\theta}{2}, então r=\operatorname{sen}^{2}u e
\frac{dr}{d\theta} = 2\,\operatorname{sen} u\cos u\;\frac{du}{d\theta} = 2\,\operatorname{sen}\!\Bigl(\frac{\theta}{2}\Bigr)\cos\!\Bigl(\frac{\theta}{2}\Bigr)\cdot\frac12 = \operatorname{sen}\!\Bigl(\frac{\theta}{2}\Bigr)\cos\!\Bigl(\frac{\theta}{2}\Bigr).
Como \operatorname{sen}\theta=2\operatorname{sen}\!\bigl(\tfrac{\theta}{2}\bigr)\cos\!\bigl(\tfrac{\theta}{2}\bigr), podemos também escrever
\frac{dr}{d\theta} = \frac{\operatorname{sen}\theta}{2}.

Passo 3

Substituindo na fórmula de comprimento de arco, obtemos
L = \int_{0}^{\frac{\pi}{2}} \sqrt{\bigl(\operatorname{sen}^{2}\!\tfrac{\theta}{2}\bigr)^{2} + \Bigl(\tfrac{\operatorname{sen}\theta}{2}\Bigr)^{2}} \;d\theta.

Passo 4

Simplificamos o integrando:
\bigl(\operatorname{sen}^{2}\!\tfrac{\theta}{2}\bigr)^{2} + \Bigl(\tfrac{\operatorname{sen}\theta}{2}\Bigr)^{2} = \operatorname{sen}^{4}\!\tfrac{\theta}{2} + \tfrac{\operatorname{sen}^{2}\theta}{4} = \operatorname{sen}^{4}\!\tfrac{\theta}{2} + \operatorname{sen}^{2}\!\tfrac{\theta}{2}\cos^{2}\!\tfrac{\theta}{2} = \operatorname{sen}^{2}\!\tfrac{\theta}{2}\,\bigl(\operatorname{sen}^{2}\!\tfrac{\theta}{2}+\cos^{2}\!\tfrac{\theta}{2}\bigr) = \operatorname{sen}^{2}\!\tfrac{\theta}{2}.
Logo \sqrt{r^2 + (dr/d\theta)^2} = \operatorname{sen}\!\bigl(\tfrac{\theta}{2}\bigr), e
L = \int_{0}^{\frac{\pi}{2}} \operatorname{sen}\!\Bigl(\frac{\theta}{2}\Bigr)\,d\theta.

Passo 5

Para integrar, use u=\tfrac{\theta}{2}, então d\theta=2\,du, e quando \theta vai de 0 a \tfrac{\pi}{2}, u vai de 0 a \tfrac{\pi}{4}. Assim
L = \int_{0}^{\frac{\pi}{4}} \operatorname{sen}(u)\cdot 2\,du = 2\int_{0}^{\frac{\pi}{4}}\operatorname{sen} u\,du = 2\bigl[-\cos u\bigr]_{0}^{\frac{\pi}{4}} = 2\bigl(1 - \cos\tfrac{\pi}{4}\bigr).

Passo 6

Como \cos\!\bigl(\tfrac{\pi}{4}\bigr)=\tfrac{\sqrt{2}}{2}, obtemos
L = 2\Bigl(1 - \frac{\sqrt{2}}{2}\Bigr) = 2 - \sqrt{2}.

Resposta

O comprimento do arco é 2 - \sqrt{2}.

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