Doubt in evaluating a double integral $\iint _R (y-1)dA$

Question

Using Green's Theorem, calculate $$ I =\int_r \vec{F}(r) \cdot d\vec{r} $$

$\vec{F}(x,y) := \langle y,xy \rangle$ and $C$ is a unit circle with center $(0,0),$ oriented counterclockwise.

Green's theorem is

$$ I =\int_r \vec{F}(r) \cdot d\vec{r} = \iint_R (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA $$

I have calculated $\dfrac{\partial Q}{\partial x}=y$ and $\dfrac{\partial P}{\partial y}=1$.

I am having trouble with the bounds for the double integral. What should I use for them? Should I be converting this into polar coordinates?

Answer 1

Simply note that your integral is,

$\displaystyle I =\int_r \vec{F}(r) \cdot d\vec{r} = \iint_R (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}) \ dA = \iint_R (y-1) \ dA$.

$y$ is an odd function so its integral over the circle centered at origin will be zero (due to symmetry above and below x-axis). So using the fact that area of unit circle is $\pi$, your answer can be simply written as,

$\displaystyle I = - \iint_R 1 \ dA = - \pi$