1a Questao Integral de Linha

Considere a curva C dada por:
x=2 \operatorname{sen}(t), y=2 \cos (t), t \operatorname{em}[0, p i]
Calcule a integral de linha:
Integral_C F . dr
onde:
F(x, y)=\left(x^{\wedge} 2 \ln \left(x^{\wedge} 2+1\right)+6 x y^{\wedge} 2-y\right) i+\left[\left(1+(6 y+1) x^{\wedge} 2 y^{\wedge} 2\right) /(y+3)\right] j
Resultado da integral (forma final):
Integral_0^pi \left[2 \cos (t)^{*}\left(4 \operatorname{sen}^{\wedge} 2(t)^{*} \ln \left(4 \operatorname{sen}^{\wedge} 2(t)+1\right)+48 \operatorname{sen}(t) \cos ^{\wedge} 2(t)-2 \cos (t)\right)\right.
\left.-2 \operatorname{sen}(t)^{\star}\left\{\left(1+(12 \cos (t)+1)^{\star} 16 \operatorname{sen}^{\wedge} 2(t) \cos ^{\wedge} 2(t)\right) /(2 \cos (t)+3)\right\}\right] d t

Passo 1

O enunciado pede para calcular a integral de linha
\int_{C}\mathbf F\cdot d\mathbf r
onde a curva C é dada por
x=2\sin t,\quad y=2\cos t,\quad t\in[0,\pi],
e o campo vetorial é
\mathbf F(x,y)=\bigl(P(x,y),\,Q(x,y)\bigr),
com
P(x,y)=x^{2}\ln\bigl(x^{2}+1\bigr)+6\,x\,y^{2}-y,
Q(x,y)=\frac{1+(6y+1)\,x^{2}y^{2}}{y+3}.
Para resolver, vamos usar a definição de integral de linha por parametrização:
\int_{C}\mathbf F\cdot d\mathbf r =\int_{0}^{\pi}\mathbf F\bigl(r(t)\bigr)\cdot r'(t)\,dt.

Passo 2

Parametrizamos C por
r(t)=(x(t),y(t))=(2\sin t,\,2\cos t),
logo
r'(t)=\bigl(2\cos t,\,-2\sin t\bigr).

Passo 3

Substituímos na componente P:
P\bigl(r(t)\bigr) = (2\sin t)^{2}\,\ln\bigl((2\sin t)^{2}+1\bigr) +6\,(2\sin t)\,(2\cos t)^{2} -2\cos t
\quad=4\sin^{2}t\;\ln\bigl(4\sin^{2}t+1\bigr) +48\,\sin t\cos^{2}t -2\cos t.
E na componente Q:
Q\bigl(r(t)\bigr) =\frac{1+(6\cdot2\cos t+1)\,(2\sin t)^{2}(2\cos t)^{2}}{2\cos t+3}
\quad=\frac{1+(12\cos t+1)\,16\sin^{2}t\cos^{2}t}{2\cos t+3}.

Passo 4

O produto escalar é
\mathbf F\bigl(r(t)\bigr)\cdot r'(t) =P\bigl(r(t)\bigr)\,x'(t)+Q\bigl(r(t)\bigr)\,y'(t)
\quad=2\cos t\;\bigl[4\sin^{2}t\ln\bigl(4\sin^{2}t+1\bigr) +48\sin t\cos^{2}t-2\cos t\bigr] -2\sin t\;\frac{1+(12\cos t+1)\,16\sin^{2}t\cos^{2}t}{2\cos t+3}.

Passo 5

Portanto a integral de linha fica
\int_{C}\mathbf F\cdot d\mathbf r =\int_{0}^{\pi}\Bigl\{2\cos t\,[4\sin^{2}t\ln(4\sin^{2}t+1) +48\sin t\cos^{2}t-2\cos t]
\quad-2\sin t\;\frac{1+(12\cos t+1)\,16\sin^{2}t\cos^{2}t}{2\cos t+3}\Bigr\}\,dt.

Resposta

\int_{C}\mathbf F\cdot d\mathbf r =\int_{0}^{\pi}\biggl[2\cos t\bigl(4\sin^{2}t\ln(4\sin^{2}t+1) +48\sin t\cos^{2}t-2\cos t\bigr) -2\sin t\;\frac{1+(12\cos t+1)\,16\sin^{2}t\cos^{2}t}{2\cos t+3}\biggr]dt.

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