\vec{\gamma}(t)=(2 \operatorname{sen} t, 2 \cos t), t \in[0, \pi]
F_{1}(x, y)=x^{2} \ln \left(x^{2}+1\right)+6 x y^{2}-y
F_{2}(x, y)=\frac{1}{y+3}+(6 y+1) x^{2}
Calcule \int_{\gamma} \vec{F} \cdot \overrightarrow{d f}
\vec{F}=\left(F_{1}, F_{2}\right)

Passo 1

O enunciado pede o integral de linha
\int_{\gamma}\vec F\cdot d\vec r
onde \vec r(t)=\bigl(2\sin t,\,2\cos t\bigr), t\in[0,\pi], e
\vec F(x,y)=\bigl(F_1(x,y),F_2(x,y)\bigr) com
F_{1}(x,y)=x^{2}\ln\bigl(x^{2}+1\bigr)+6xy^{2}-y, \quad F_{2}(x,y)=\frac1{y+3}+(6y+1)x^{2}.
O conteúdo é parametrização de integral de linha não conservativa. Vamos usar
\int_{\gamma}\vec F\cdot d\vec r =\int_{0}^{\pi}\bigl[F_{1}(x(t),y(t))\,x'(t)+F_{2}(x(t),y(t))\,y'(t)\bigr]\,dt.

Passo 2

Parametrizando:
x(t)=2\sin t,\quad y(t)=2\cos t, \quad x'(t)=2\cos t,\quad y'(t)=-2\sin t.
Logo
F_{1}(t) =4\sin^{2}t\ln\bigl(1+4\sin^{2}t\bigr) +48\sin t\cos^{2}t -2\cos t,
F_{2}(t) =\frac1{3+2\cos t} +4\sin^{2}t\,(12\cos t+1),
e o integrando é
F_{1}(t)\,x'(t)+F_{2}(t)\,y'(t) =2\Bigl[F_{1}(t)\cos t - F_{2}(t)\sin t\Bigr].

Passo 3

Expandindo e separando em cinco integrais notáveis, chamemos
I =2\int_{0}^{\pi}\Bigl[ 4\sin^{2}t\cos t\ln(1+4\sin^{2}t) +48\sin t\cos^{3}t -2\cos^{2}t -\frac{\sin t}{2}\frac{2}{3+2\cos t} -4\sin^{3}t(12\cos t+1) \Bigr]dt.
Denotemos
A=\int_{0}^{\pi}4\sin^{2}t\cos t\ln(1+4\sin^{2}t)\,dt, \quad B=\int_{0}^{\pi}48\sin t\cos^{3}t\,dt,
C=\int_{0}^{\pi}(-2)\cos^{2}t\,dt, \quad D=\int_{0}^{\pi}\frac{\sin t}{3+2\cos t}\,dt, \quad E=\int_{0}^{\pi}4\sin^{3}t\,(12\cos t+1)\,dt,
de modo que
\displaystyle I=2\bigl[A+B+C-\tfrac12D -E\bigr].

Passo 4

Calculamos B e C por substituição direta:
B=48\int_{0}^{\pi}\sin t\cos^{3}t\,dt =-12\cos^{4}t\Big|_{0}^{\pi} =-12\bigl[(-1)^{4}-1^{4}\bigr]=0,
C=-2\int_{0}^{\pi}\cos^{2}t\,dt =-2\Bigl[\tfrac t2+\tfrac{\sin2t}{4}\Bigr]_{0}^{\pi} =-\bigl[\pi +0\bigr] =-\pi.

Passo 5

Para D, use u=3+2\cos t, du=-2\sin t\,dt:
D=\int_{0}^{\pi}\frac{\sin t}{3+2\cos t}\,dt =-\tfrac12\ln\bigl(3+2\cos t\bigr)\Big|_{0}^{\pi} =-\tfrac12\bigl[\ln1-\ln5\bigr] =\tfrac12\ln5.
Para E, separe e integre:
\int\sin^{3}t\cos t\,dt=\tfrac{\sin^{4}t}{4},\quad \int\sin^{3}t\,dt=-\cos t+\tfrac{\cos^{3}t}{3}.
Assim
E =4\Bigl[12\cdot\tfrac{\sin^{4}t}{4} +\bigl(-\cos t+\tfrac{\cos^{3}t}{3}\bigr)\Bigr]_{0}^{\pi} =\bigl[12\sin^{4}t -4\cos t +\tfrac{4}{3}\cos^{3}t\bigr]_{0}^{\pi} =(-4\cdot(-1)+\tfrac{4}{3}(-1)) -( -4\cdot1+\tfrac{4}{3}\,1) =0.

Passo 6

O termo A é
\displaystyle A=\int_{0}^{\pi}4\sin^{2}t\cos t\ln(1+4\sin^{2}t)\,dt.
Note que sob t\mapsto\pi-t temos \sin^{2}(\pi-t)=\sin^{2}t,
\cos(\pi-t)=-\cos t e \ln(1+4\sin^{2}t) invariante,
logo o integrando muda de sinal e A=0.

Passo 7

Agora
I =2\Bigl[A+B+C-\tfrac12D -E\Bigr] =2\Bigl[0+0+(-\pi)-\tfrac12\cdot\tfrac12\ln5 -0\Bigr] =2\Bigl[-\pi -\tfrac14\ln5\Bigr] =-2\pi -\tfrac12\ln5.
Revisando o cálculo de D, como D=\tfrac12\ln5, o termo em I é -\tfrac12D=-\tfrac14\ln5.

Resposta

\int_{\gamma}\vec F\cdot d\vec r =-2\pi \;-\;\tfrac12\ln5.

Exercícios relacionados