Question

Verify Green's theorem for , [(x - y)dx + 3xydy) where ç is the boundary of the region
bounded by the parabola x² = - 4y and y? = 4x.
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Answer 1

Answer:

Verify Green's theorem for , [(x - y)dx + 3xydy) where ç is the boundary of the region

Verify Green's theorem for , [(x - y)dx + 3xydy) where ç is the boundary of the region

Step-by-step explanation:

hope it's help you ✌✌

Answer 2

Answer:

Verification as follows:

Step-by-step explanation:

P= $x-y$ and Q =$3xy$ .

Point of intersection of the curve = (0,0) and (4,-4).

According to green theorem,

$\oint (Pdx+Qdy) = \int\int \left(\frac{\partial Q}{\partial x} -\frac{\partial P}{\partial y} \right)dA$

LHS:

Line Intregral along the curve $y^2=4x$ is $I_1$(say)

$\int\limits^4_0 {(x+2\sqrt{x} ) } \, dx + \frac{3}{4} \int\limits^{-4}_0 {y^3} \, dy=I_1$

Line Intregral along the curve $x^2=-4y$ is $I_2$(say)

$\int\limits^4_0 {(x+x^2/4) } \, dx + \frac{3}{8} \int\limits^{4}_0 {x^4} \, dx=I_2$

RHS :

$\frac{\partial Q}{\partial x} =3y \,\, and \,\,\,\frac{\partial P}{\partial y}=-1\\\implies \int^{4}_0\int\limits^{-\frac{x^2}{4} }_{-2\sqrt{x} } {(3y+1)} \, dy dx= I_3$

According to green's theorem:

$I_1 +I_2=I_3$ (you can check that)