I am so confused on these integrals.
Here is the question.
Problem
$$G(x,y)=\ln(x^2+y^2)/2$$
Calculate the 2-D Laplacian $\Delta^2G$
For the interior $D$ of the circle $C$ of radius $a$ calculate $$\int\int\Delta^2GdA$$
My questions are, is the Laplacian a vector field or a scalar? How do I set the integral up?
I would really appreciate some help!
The laplacian operator defined on the scalar field gives a scalar. Here, it is defined as $$\Delta G\left(x,y\right):=\partial_x^2G\left(x,y\right)+\partial_y^2G\left(x,y\right).$$ We compute $$\partial_xG\left(x,y\right) =\frac{x}{x^2+y^2},$$ $$\partial_x^2G\left(x,y\right) =\frac{y^2-x^2}{\left(x^2+y^2\right)^2}$$ and by as $x$ and $y$ plays the same role in the expression of $G$, we deduce that $$\partial_y^2G\left(x,y\right) =\frac{x^2-y^2}{\left(x^2+y^2\right)^2}.$$ We then see that $\partial_x^2G\left(x,y\right)+\partial_y^2G\left(x,y\right)=0$ then the integral is zero. In fact, $G$ is harmonic, i.e. $\Delta G\equiv0$.
EDIT: I specify something following Ian's comment. The above calculus shows that $G$ is indeed harmonic where $G$ is well defined (i.e. when $x\neq0$ and $y\neq0$). The fact is there was an ambiguity in the OP's question to me. I answered to the question "what the scalar laplacian is?" and supposed $G$ well defined to do it. Now, if the question was computing the integral on the disk, what I said is not correct (see comments).