Question

use green's theorem to evaluate the line integral along the given positively oriented curve. c 9y3 dx − 9x3 dy, c is the circle x2 y2

Answer 1

The line integral along the given positively oriented curve is -216π. Using green's theorem, the required value is calculated.

What is green's theorem?

The theorem states that,

$\int_CPdx+Qdy = \int\int_D(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dx dy$

Where C is the curve.

Calculation:

The given line integral is

$\int_C9y^3dx-9x^3dy$

Where curve C is a circle x² + y² = 4;

Applying green's theorem,

P = 9y³; Q = -9x³

Then,

$\frac{\partial P}{\partial y} = \frac{\partial 9y^3}{\partial y} = 27y^2$

$\frac{\partial Q}{\partial x} = \frac{\partial -9x^3}{\partial x} = 27x^2$

$\int_C9y^3dx-9x^3dy = \int\int_D(-27x^2 - 27y^2)dx dy$

⇒ $-27\int\int_D(x^2 + y^2)dx dy$

Since it is given that the curve is a circle i.e., x² + y² = 2², then changing the limits as

0 ≤ r ≤ 2; and 0 ≤ θ ≤ 2π

Then the integral becomes

$-27\int\limits^{2\pi}_0\int\limits^2_0r^2. r dr d\theta$

⇒ $-27\int\limits^{2\pi}_0\int\limits^2_0 r^3dr d\theta$

⇒ $-27\int\limits^{2\pi}_0 (r^4/4)|_0^2 d\theta$

⇒ $-27\int\limits^{2\pi}_0 (16/4) d\theta$

⇒ $-108\int\limits^{2\pi}_0 d\theta$

⇒ $-108[2\pi - 0]$

⇒ -216π

Therefore, the required value is -216π.

Learn more about green's theorem here:

brainly.com/question/23265902

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