Answer:
-48π
Step-by-step explanation:
Recall that Green's theorem states the following:
Let P, Q be [tex]\bf C^1[/tex]-functions on a region A, which is the interior of a closed piecewise [tex]\bf C^1[/tex]-path C, parametrized counterclockwise. Then
[tex]\bf \displaystyle\oint_CPdx+Qdy=\displaystyle\iint_A(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx[/tex]
We have
[tex]\bf P(x, y) = 2y^3\\\\Q(x,y)=-2x^3[/tex]
[tex]\bf \displaystyle\frac{\partial Q}{\partial x}=-6x^2\\\\\displaystyle\frac{\partial P}{\partial y}=6y^2[/tex]
Therefore
[tex]\bf \oint_C2y^3dx-2x^3Qdy=\displaystyle\iint_A(-6x^2-6y^2)dydx[/tex]
where A is the interior of the circle of radius 2 centered at the origin.
If we change to polar coordinates
x = rcos(t)
y = rsin(t)
Then A can be described with the inequalities
0 < r < 2
0 < t < 2π
and
[tex]\bf \displaystyle\iint_A(-6x^2-6y^2)dydx=-6\displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^{2} (r^2cos^2t+r^2sin^2t)rdrdt=\\\\-6\displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^{2} (r^2)rdrdt=-6\left(\displaystyle\int_{0}^{2\pi}dt\displaystyle\int_{0}^{2}r^3dr \right)=\\\\=\boxed{-48\pi}[/tex]
The result of the integral [tex]\int\limits^a_b {2y^3\ dx - 2x^3\ dy}[/tex] using the green theorem is [tex]-48\pi[/tex]
The integral is given as:
[tex]\int\limits^a_b {2y^3\ dx - 2x^3\ dy}[/tex]
By Green Theorem, we have:
[tex]\int\limits^a_b {Pdx + Qdy} = \int\int_A(\frac{dQ}{dx} - \frac{dP}{dy}) dy dx[/tex]
By comparison, we have:
[tex]P(x,y) = 2y^3[/tex]
[tex]Q(x,y) = -2x^3[/tex]
Differentiate both equations
[tex]\frac{dP}{dy} =6y^2[/tex]
[tex]\frac{dQ}{dx} = -6x^2[/tex]
So, the integral becomes:
[tex]\int\limits^a_b {2y^3dx - 2x^3dy} = \int\int_A(-6x^2 - 6y^2) dy dx[/tex]
Factor out -6
[tex]\int\limits^a_b {2y^3dx - 2x^3dy} = -6\int\int_A(x^2 +y^2) dy dx[/tex]
Express as polar coordinates
[tex]\int\limits^a_b {2y^3dx - 2x^3dy} = -6\int\limits^{2\pi}_0\int\limits^2_0 r^2 \ r dr d\theta[/tex]
This gives
[tex]\int\limits^a_b {2y^3dx - 2x^3dy} = -6\int\limits^{2\pi}_0\int\limits^2_0 r^3 dr d\theta[/tex]
Integrate
[tex]\int\limits^a_b {2y^3dx - 2x^3dy} = -6\int\limits^{2\pi}_0 \frac{r^4}{4}|\limits^2_0 d\theta[/tex]
Expand
[tex]\int\limits^a_b {2y^3dx - 2x^3dy} = -6\int\limits^{2\pi}_0 \frac{2^4}{4}d\theta[/tex]
[tex]\int\limits^a_b {2y^3dx - 2x^3dy} = -6\int\limits^{2\pi}_0 4d\theta[/tex]
Factor out 4
[tex]\int\limits^a_b {2y^3dx - 2x^3dy} = -24\int\limits^{2\pi}_0 d\theta[/tex]
Integrate
[tex]\int\limits^a_b {2y^3dx - 2x^3dy} = -24 [\theta]|\limits^{2\pi}_0[/tex]
Expand
[tex]\int\limits^a_b {2y^3dx - 2x^3dy} = -24 [2\pi - 0][/tex]
This gives
[tex]\int\limits^a_b {2y^3dx - 2x^3dy} = -24 [2\pi][/tex]
Expand
[tex]\int\limits^a_b {2y^3dx - 2x^3dy} = -48\pi[/tex]
Hence, the result of the integral is [tex]-48\pi[/tex]
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