Using Green's Theorem, calculate the closed integral over $C$ $$\oint_C(3x+4y)dx+(2x-3y)dy$$
where $C$ is the circle of radius $4$ units,with its centre at the origin of the $x,y$-plane.
My approach: From Green's Theorem, $$\oint_C(3x+4y)dx+(2x-3y)dy = \iint_A (\frac{\partial}{\partial y}(3x+4y) + \frac{\partial}{\partial x}(2x-3y)) dx \,dy$$
$\frac{\partial}{\partial y}(3x+4y)= 4$ and $\frac{\partial}{\partial x}(2x-3y)=2$, which combines to form $\iint 6\,dx\,dy$. Now my question is that from the question what is the limit of the two integrals I am going to take, is it one integral from $0$ to $2 \pi$ and another $0$ to $4$ or something else…
Please see my approach and tell me whether I am going or wrong or right??
Best Answer
Your approach is wrong because you applied the equation in Green's Theorem wrong. Also, once you see that the integrand in the integral over the area is a constant, the integral itself is just the area enclosed by the circle times that constant.
Let $P(x,y) = 3 x+4 y$ and $Q(x,y) = 2 x-3 y$. Then Green's Theorem states that the line integral over $C$ is
$$\iint_A dx dy\, \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) = \iint_A dx dy\,(2-4) = -32 \pi$$