Question #204889

Using Green's theorem evaluate the integral: ∫C (y³dx + 3x²dy) where C is the contour along the circle x² + y² = 1 taken counter clockwise.


1
Expert's answer
2021-06-09T17:54:23-0400

According to Green's theorem, CPdx+Qdy=D(QxPy)dA\int_C P dx + Q dy = \iint_D \left(\frac{ \partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA.

For given vector field, QxPy=6x3y2\frac{ \partial Q}{\partial x} - \frac{\partial P}{\partial y} = 6 x - 3 y^2, hence we need to evaluate surface integral(6x3y2)dxdy\iint (6 x - 3 y^2) dx dy over the circle of radius 1, centered at the origin.

It is convenient to do it in polar coordinates, where surface element is dS=rdrdφdS = r dr d\varphi, and 0r10\leq r \leq 1, 0φ2π0\leq \varphi \leq 2 \pi:

02πdφ01r(6rcosφ3r2sin2φ)=02π(63cosφ34sin2φ)dφ=\int_0^{2\pi} d\varphi \int_0^1 r (6 r \cos \varphi - 3 r^2 \sin^2 \varphi) = \int_0^{2\pi}(\frac{6}{3}\cos \varphi - \frac{3}{4} \sin^2 \varphi)d \varphi =

[2sinφ34(φ214sin2φ)]02π=34π\left[ 2 \sin \varphi - \frac{3}{4}(\frac{\varphi}{2} - \frac{1}{4} \sin 2 \varphi)\right]_0^{2 \pi} = -\frac{3}{4}\pi.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS