Vector Calculus Green’s Theorem Math Examples

Vector Calculus Green’s Theorem Math Examples:

These are from the book Calculus Early Transcendentals 10th Edition. Refer to 15.4 exercise maths.

Green’s Theorm:

Let, R be a simple connected plane region where boundary is a simple, closed, piecewise smooth curve oriented counter clockwise. If f(x,y) \text{and} g(x,y) are continuous and have continuous first partial derivatives on some open set containing R, then

\int f(x,y) \ dx + g(x,y) \ dy = \iint\limits_R (\frac{\delta g}{\delta x} - \frac{\delta f}{\delta y}) \ dA

Excercise 3:

Using Green’s Theorem evaluate the integral assuming the curve C is oriented Counter clockwise, \oint_c 4xy \ dx + 2xy \ dy, where C is the Rectangle bounded by x = -2, x = 4, y = 1, y = 2 .

Solution:

greens theorm exercise 3
greens theorm exercise 3
Using Green’s Theorm,
\oint_c 4xy \ dx + 2xy \ dy = \iint\limits_R \frac{\delta }{\delta x} 2xy - \frac{\delta }{\delta y} 4xy \ dA 

              = \iint\limits_R 2y - 4x \ dA 

              = \int_{-2}^4\int_1^2 2y - 4x \ dy \ dx 

              = \int_{-2}^4 [y^2 - 4xy]_1^2 \ dx 

              = \int_{-2}^4 4 - 8x - 1 + 4x \ dx 
 
              = \int_{-2}^4 3 - 4x \ dx 

              = [3x - 2x^2]_{-2}^4 

              = 12 - 32 + 6 + 8 

              = -6 

Excercise 7:

Using Green’s Theorem evaluate the integral assuming the curve C is oriented Counter clockwise, \oint_c (x^2 - 3y) \ dx + 3x \ dy, where C is the circle x^2 + y^2 = 4 .

Solution:

greens theorm exercise 7
greens theorm exercise 7

Since this is a full circle the interval of \int_{\theta} \text{is from} \ 0 \ to \ 2\pi
Also, x^2 + y^2 = 4 = 2^2 . So the interval of \int_{r} \text{is from} \ 0 \ to \ 2

Using Green’s Theorm,
\oint_c (x^2 - 3y) + 3x \ dy = \iint\limits_R \frac{\delta }{\delta x} 3x - \frac{\delta }{\delta y} (x^2 - 3y) \ dA 

               = \iint\limits_R 3 + 3 \ dA 

               = \int_0^{2\pi} \int_0^2 6 r \ dr \ d\theta 
 
               = 6 \int_0^{2\pi} [\frac{r^2}{2}]_0^2 \ dr \ d\theta 

               = 12 \int_0^{2\pi} \ d\theta 

               = 12 [\theta]_0^{2\pi} 

               = 12 \times 2\pi 

               = 24\pi 

Excercise 9:

Using Green’s Theorem evaluate the integral assuming the curve C is oriented Counter clockwise,
\oint_c ln(1 + y) \ dx - \frac{xy}{1 + y} \ dy, where C is the triangle with vertices (0,0), (2,0), (0,4) .

Solution:

greens theorm example 9
greens theorm example 9

The upper bounds for y can be found using points (2,0) and (0,4).

Here,
x_1 = 2, y_1 = 0, x_2 = 0, y_2 = 4,

\frac{y - y_1}{x - x_1} = \frac{y_1 - y_2}{x_1 - x_2} 
             
\frac{y - 0}{x - 2} = \frac{0 - 4}{2 - 0} 

y = -2x + 4 
Using Green’s Theorm,
\oint_c ln(1 + y) \ dx - \frac{xy}{1 + y} \ dy = \iint\limits_R \frac{\delta }{\delta x} (-\frac{xy}{1 + y}) - \frac{\delta }{\delta y} ln(1 + y) \ dA 

                  = \iint\limits_R (-\frac{y}{1 + y} - \frac{1}{1 + y}) \ dA 

                  = \int_0^2 \int_0^{-2x + 4} (-\frac{y}{1 + y} - \frac{1}{1 + y}) \ dy \ dx 

                  = \int_0^2 \int_0^{-2x + 4} -\frac{1 + y}{1 + y} \ dy \ dx 

                  = -1 \int_0^2 \int_0^{-2x + 4} \ dy \ dx 

                  = -1 \int_0^2 [y]_0^{-2x + 4} \ dx 

                  = -1 \int_0^2 -2x + 4 \ dx 

                  = [x^2 - 4x]_0^2 

                  = 4 - 8 - 0 + 0 

                  = -4 

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