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Use Green's theorem to evaluate the line integral along the given positively oriented curve.

int_(C)7y^(3)dx-7x^(3)dy,quad C" is the circle "x^(2)+y^(2)=4

Use Green's theorem to evaluate the line integral along the given positively oriented curve. $\newline$ $\int_{C} 7 y^{3} d x-7 x^{3} d y, \quad C \text { is the circle } x^{2}+y^{2}=4$

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Find indefinite integrals using the substitution

Full solution

Q. Use Green's theorem to evaluate the line integral along the given positively oriented curve.

\newline

\int_{C} 7 y^{3} d x-7 x^{3} d y, \quad C \text { is the circle } x^{2}+y^{2}=4

Rephrase the problem: Step $1$ : Rephrase the problem. $\newline$ question_prompt: Use Green's theorem to evaluate the line integral along the given positively oriented curve for the function $7y^3 \, dx - 7x^3 \, dy$ , where $C$ is the circle $x^2 + y^2 = 4$ .
Apply Green's theorem: Step $2$ : Apply Green's theorem. $\newline$ Green's theorem relates a line integral around a simple, closed, positively oriented curve $C$ to a double integral over the plane region $D$ bounded by $C$ . Green's theorem states: $\newline$ $\int_C P \,dx + Q \,dy = \int\int_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA,$ $\newline$ where $P = 7y^3$ and $Q = -7x^3$ .
Calculate partial derivatives: Step $3$ : Calculate partial derivatives. $\newline$ $\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (-7x^3) = -21x^2$ , $\newline$ $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (7y^3) = 21y^2$ .
Set up the double integral: Step $4$ : Set up the double integral. $\int\int_D (-21x^2 - 21y^2) \, dA$ .
Convert to polar coordinates: Step $5$ : Convert to polar coordinates. $\newline$ Since $D$ is a circle of radius $2$ , convert the integral to polar coordinates where $x = r \cos(\theta)$ and $y = r \sin(\theta)$ . Then, $x^2 + y^2 = r^2$ and $dA = r \, dr \, d\theta$ . $\newline$ $\int\int_D (-21r^2) r \, dr \, d\theta = -21 \int_0^{2\pi} \int_0^2 r^3 \, dr \, d\theta$ .
Evaluate the integral: Step $6$ : Evaluate the integral. $\newline$ $-21 \int_0^{2\pi} [r^4/4]_0^2 d\theta = -21 \int_0^{2\pi} (16/4) d\theta = -21 \int_0^{2\pi} 4 d\theta = -21 [4\theta]_0^{2\pi} = -21 \times 8\pi = -168\pi$ .

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