Question

Verify Green’s theorem in the plane for ∫_c{(xy + y²)dx + x²dy} where C is the closed curve of the region bounded by y = x and y = x².

Answer 1

Given:

\(\int \limits_C (xy + y^2)dx + x^2 dy \)   .....(1)

\(\int Pdx + Qdy\)    ......(2)

Comparing equation (1) and equation (2) we get

P = xy + y² and Q = x²

^{\(\therefore \frac{\partial P}{\partial y} = x + 2y\) and \(\frac {\partial Q}{\partial x} = 2x\)} 

Given : y = x and y = x²

Solving both equations we get,

x = x²

0 = x² − x

∴ x = 0 or x = 1

∴ Point of intersection is (0, 0) and (1, 1)

∴ By Green theorem,

\(\int Pdx + Qdy = \int \int _R \left(\frac {\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dx \, dy\)

1) Outer limit x: The vertical strip will slide from x = 0 to x = 1

2) Inner limit y:

a) Upper limit is equation of line : y = x

b) Lower limit is equation of parabola : y = x²

Answer 2

We shall find the points of intersection of y = x and y = x².

Equating the R.H.S.

∴ x = x² ⇒ x – x² = 0

x (1 – x) = 0

x = 0, 1

∴ y = 0, 1 and hence (0, 0), (1, 1) are the points of intersection.

We have Green’s theorem in a plane,