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

#### Solution:

###### Using Green’s Theorm, $\oint_c 4xy \ dx + 2xy \ dy = \iint\limits_R \frac{\delta }{\delta x} 2xy - \frac{\delta }{\delta y} 4xy \ dA$ $= \iint\limits_R 2y - 4x \ dA$ $= \int_{-2}^4\int_1^2 2y - 4x \ dy \ dx$ $= \int_{-2}^4 [y^2 - 4xy]_1^2 \ dx$ $= \int_{-2}^4 4 - 8x - 1 + 4x \ dx$ $= \int_{-2}^4 3 - 4x \ dx$ $= [3x - 2x^2]_{-2}^4$ $= 12 - 32 + 6 + 8$ $= -6$


#### Solution:

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, $\oint_c (x^2 - 3y) + 3x \ dy = \iint\limits_R \frac{\delta }{\delta x} 3x - \frac{\delta }{\delta y} (x^2 - 3y) \ dA$ $= \iint\limits_R 3 + 3 \ dA$ $= \int_0^{2\pi} \int_0^2 6 r \ dr \ d\theta$ $= 6 \int_0^{2\pi} [\frac{r^2}{2}]_0^2 \ dr \ d\theta$ $= 12 \int_0^{2\pi} \ d\theta$ $= 12 [\theta]_0^{2\pi}$ $= 12 \times 2\pi$ $= 24\pi$


#### Solution:

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, $\oint_c ln(1 + y) \ dx - \frac{xy}{1 + y} \ dy = \iint\limits_R \frac{\delta }{\delta x} (-\frac{xy}{1 + y}) - \frac{\delta }{\delta y} ln(1 + y) \ dA$ $= \iint\limits_R (-\frac{y}{1 + y} - \frac{1}{1 + y}) \ dA$ $= \int_0^2 \int_0^{-2x + 4} (-\frac{y}{1 + y} - \frac{1}{1 + y}) \ dy \ dx$ $= \int_0^2 \int_0^{-2x + 4} -\frac{1 + y}{1 + y} \ dy \ dx$ $= -1 \int_0^2 \int_0^{-2x + 4} \ dy \ dx$ $= -1 \int_0^2 [y]_0^{-2x + 4} \ dx$ $= -1 \int_0^2 -2x + 4 \ dx$ $= [x^2 - 4x]_0^2$ $= 4 - 8 - 0 + 0$ $= -4$