Question #198609

Verify Green’s theorem in the plane for $ (xy + Sy?)dx + 3xdy,

where c is the closed curve of the region bounded by y = x and y = x?.


Expert's answer

Green’s theorem:

∮CPdx+Qdy=∬D(Qx−Py)dA\oint_CPdx+Qdy=\iint_D(Q_x-P_y)dA


We have:

P=xy+y,Q=3xP=xy+y,Q=3x

Then:

Qx=3,Py=x+1Q_x=3,P_y=x+1

∮C(xy+y)dx+3xdy=∫01(2−x)dx∫x1−xdy=∫01(2−x)(1−2x)dx=\oint_C(xy+y)dx+3xdy=\int^1_0(2-x)dx\int^{1-x}_xdy=\int^1_0(2-x)(1-2x)dx=

=(2x−5x2/2+2x3/3)∣01=2−5/2+2/3=1/6=(2x-5x^2/2+2x^3/3)|^1_0=2-5/2+2/3=1/6




∮C(xy+y)dx+3xdy=∫10.5[(x(1−x)+1−x)dx+3xd(1−x)]+\oint_C(xy+y)dx+3xdy=\int_1^{0.5}[(x(1-x)+1-x)dx+3xd(1-x)]+

+∫0.50[(x2+x)dx+3xdx]=∫10.5(1−x2−3x)dx+∫0.50(x2+4x)dx=+\int_{0.5}^{0}[(x^2+x)dx+3xdx]=\intop^{0.5}_1(1-x^2-3x)dx+\int^0_{0.5}(x^2+4x)dx=

=(x−x3/3−3x2/2)∣10.5+(x3/3+2x2)∣0.50==(x-x^3/3-3x^2/2)|^{0.5}_1+(x^3/3+2x^2)|^0_{0.5}=


=0.25−0.125/3−0.75/2−1+1/3+1.5−0.125/3−0.5=1/4−1/12−3/8+1/3=1/6=0.25-0.125/3-0.75/2-1+1/3+1.5-0.125/3-0.5=1/4-1/12-3/8+1/3=1/6


We have the same results, so Green’s theorem is verified.


LATEST TUTORIALS
APPROVED BY CLIENTS