Questão 4(\mathbf{3 , 0}) : Considere as curvas:
  • \mathcal{C}_{3} : o arco da hélice \vec{r}_{3}(t)=(\cos t, \operatorname{sen} t, t) entre os pontos (-1,0,-\pi) e (1,0,0);
  • \mathcal{C}_{4} : segmento que une (1,0,0) a (1,2,-2);
  • \mathcal{C}_{5}=\mathcal{C}_{3} \cup \mathcal{C}_{4}
Calcule \int_{\mathcal{C}_{5}}\left(x^{2}+y^{2}+6 z^{2}-x^{2} y\right) d s

Passo 1

O enunciado pede o integral de linha de
f(x,y,z)=x^{2}+y^{2}+6z^{2}-x^{2}y
sobre a curva \mathcal{C}_{5}=\mathcal{C}_{3}\cup\mathcal{C}_{4}.
O conteúdo aqui é integral de função escalar em linha, ou seja
\int_{C}f\,ds=\int_{a}^{b}f\bigl(r(t)\bigr)\,\bigl\|r'(t)\bigr\|\,dt.
Vamos dividir em duas partes: primeiro sobre \mathcal{C}_{3} (hélice) e depois sobre \mathcal{C}_{4} (segmento de reta).

Passo 2

Para \mathcal{C}_{3}, use
r_{3}(t)=(\cos t,\;\sin t,\;t),\quad t\in[-\pi,0],
pois em t=-\pi temos (-1,0,-\pi) e em t=0 temos (1,0,0).
Calcule
r_{3}'(t)=(-\sin t,\;\cos t,\;1), \quad \|r_{3}'(t)\|=\sqrt{\sin^{2}t+\cos^{2}t+1}=\sqrt{2},
logo ds=\sqrt{2}\,dt. Em seguida,
f\bigl(r_{3}(t)\bigr) =\cos^{2}t+\sin^{2}t+6t^{2}-\cos^{2}t\;\sin t =1+6t^{2}-\cos^{2}t\,\sin t.
Portanto
\int_{\mathcal{C}_{3}}f\,ds =\sqrt{2}\int_{-\pi}^{0}\bigl(1+6t^{2}-\cos^{2}t\,\sin t\bigr)\,dt.
Separe em três integrais:
I_{1}=\sqrt{2}\int_{-\pi}^{0}1\,dt =\sqrt{2}\,\bigl[\,t\,\bigr]_{-\pi}^{0} =\pi\sqrt{2},
I_{2}=\sqrt{2}\int_{-\pi}^{0}6t^{2}\,dt =6\sqrt{2}\,\bigl[\tfrac{t^{3}}{3}\bigr]_{-\pi}^{0} =2\pi^{3}\sqrt{2},
I_{3}=-\sqrt{2}\int_{-\pi}^{0}\cos^{2}t\,\sin t\,dt.
Para I_{3} faça u=\cos t, du=-\sin t\,dt, de modo que
\int\cos^{2}t\,\sin t\,dt =-\int u^{2}\,du=-\tfrac{u^{3}}{3}.
Logo
I_{3} =-\sqrt{2}\Bigl[-\tfrac{\cos^{3}t}{3}\Bigr]_{-\pi}^{0} =\frac{\sqrt{2}}{3}\bigl[\cos^{3}t\bigr]_{-\pi}^{0} =\frac{2\sqrt{2}}{3}.
Somando I_{1}+I_{2}+I_{3}
\int_{\mathcal{C}_{3}}f\,ds =\sqrt{2}\,\pi+2\pi^{3}\sqrt{2}+\frac{2\sqrt{2}}{3} =\sqrt{2}\Bigl(\pi+2\pi^{3}+\tfrac{2}{3}\Bigr).

Passo 3

Para \mathcal{C}_{4}, o segmento de (1,0,0) a (1,2,-2), use
r_{4}(u)=(1,\;2u,\;-2u),\quad u\in[0,1].
Então
r_{4}'(u)=(0,2,-2), \quad \|r_{4}'(u)\|=\sqrt{0+4+4}=2\sqrt{2},
ou seja ds=2\sqrt{2}\,du. Avaliando f em r_{4}(u):
f\bigl(r_{4}(u)\bigr) =1^{2}+(2u)^{2}+6(-2u)^{2}-1^{2}\,(2u) =1+4u^{2}+24u^{2}-2u =1+28u^{2}-2u.
Portanto
\int_{\mathcal{C}_{4}}f\,ds =2\sqrt{2}\int_{0}^{1}(1+28u^{2}-2u)\,du =2\sqrt{2}\Bigl[u+\tfrac{28u^{3}}{3}-u^{2}\Bigr]_{0}^{1} =2\sqrt{2}\,\tfrac{28}{3} =\frac{56\sqrt{2}}{3}.

Passo 4

Agora some as duas partes:
\int_{\mathcal{C}_{5}}f\,ds =\int_{\mathcal{C}_{3}}f\,ds+\int_{\mathcal{C}_{4}}f\,ds =\sqrt{2}\Bigl(\pi+2\pi^{3}+\tfrac{2}{3}\Bigr) +\frac{56\sqrt{2}}{3} =\sqrt{2}\Bigl(\pi+2\pi^{3}+\tfrac{58}{3}\Bigr).

Resposta

\boxed{ \int_{\mathcal{C}_{5}}\bigl(x^{2}+y^{2}+6z^{2}-x^{2}y\bigr)\,ds =\sqrt{2}\Bigl(\pi+2\pi^{3}+\tfrac{58}{3}\Bigr). }

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