Calcule as integrais de linha a seguir
\int_{\gamma}\left(1+\frac{9}{4} z^{\frac{2}{3}}\right)^{\frac{1}{4}} d s. \gamma é a curva parametrizada por \vec{r}(t)=\cos t \vec{i}+\sin t \vec{j}+t^{\frac{3}{2}} \vec{k}, 0 \leq t \leq \frac{20}{3}.

Passo 1

O enunciado pede a integral de linha da função escalar
f(x,y,z)=\Bigl(1+\tfrac{9}{4}\,z^{2/3}\Bigr)^{1/4}
ao longo da curva \gamma dada por
\vec r(t)=\cos t\,\mathbf i+\sin t\,\mathbf j+t^{3/2}\,\mathbf k, \quad 0\le t\le \tfrac{20}{3}.
O conteúdo necessário é a definição de integral de linha para campo escalar, que usa
\int_\gamma f\,ds=\int_{a}^{b}f\bigl(\vec r(t)\bigr)\,\|\vec r\,'(t)\|\,dt.

Passo 2

Calculamos a derivada
\vec r\,'(t)=-\sin t\,\mathbf i+\cos t\,\mathbf j+\tfrac{3}{2}\,t^{1/2}\,\mathbf k
e seu módulo
\|\vec r\,'(t)\| =\sqrt{(-\sin t)^2+(\cos t)^2+\Bigl(\tfrac{3}{2}t^{1/2}\Bigr)^2} =\sqrt{1+\tfrac{9}{4}\,t}.

Passo 3

Na curva, a coordenada z=t^{3/2} implica
z^{2/3}=\bigl(t^{3/2}\bigr)^{2/3}=t.
Logo o integrando fica
f\bigl(\vec r(t)\bigr) =\Bigl(1+\tfrac{9}{4}\,t\Bigr)^{1/4}, \quad \|\vec r\,'(t)\|=\Bigl(1+\tfrac{9}{4}\,t\Bigr)^{1/2}.

Passo 4

A integral de linha é
\int_{0}^{20/3}\Bigl(1+\tfrac{9}{4}\,t\Bigr)^{1/4}\, \Bigl(1+\tfrac{9}{4}\,t\Bigr)^{1/2}\,dt =\int_{0}^{20/3}\Bigl(1+\tfrac{9}{4}\,t\Bigr)^{3/4}dt.

Passo 5

Fazemos a mudança de variável
u=1+\tfrac{9}{4}\,t,\quad du=\tfrac{9}{4}\,dt,
quando t=0\Rightarrow u=1 e t=\tfrac{20}{3}\Rightarrow u=16. Assim
dt=\tfrac{4}{9}du,
e a integral vira
\int_{1}^{16}u^{3/4}\,\tfrac{4}{9}\,du =\tfrac{4}{9}\int_{1}^{16}u^{3/4}du.

Passo 6

Integramos em u:
\tfrac{4}{9}\int u^{3/4}du =\tfrac{4}{9}\cdot\tfrac{u^{7/4}}{7/4} =\tfrac{4}{9}\cdot\tfrac{4}{7}\,u^{7/4} =\tfrac{16}{63}\,u^{7/4}.
Avaliaremos de 1 a 16:
u^{7/4}\Big|_{1}^{16} =16^{7/4}-1^{7/4} =(16^{1/4})^7-1 =2^7-1 =128-1=127.

Resposta

\int_{\gamma}\Bigl(1+\tfrac{9}{4}\,z^{2/3}\Bigr)^{1/4}ds =\tfrac{16}{63}\times127 =\tfrac{2032}{63}.

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