# Green's theorem with unit normal and del operator

By appropriately choosing the functions P and Q in Green's theorem, show that

$$\iint_R\nabla^2 \phi\,\mathrm{d}A =\int \frac{\partial \phi}{\partial n} \, \mathrm{d}s,$$ where $$\frac{\partial}{\partial n}$$ denotes differentiation w.r.t. outward normal to $$C$$. Using the above result, evaluate $$\int_C \frac {\partial \phi}{\partial n}\,\mathrm{d}s$$ over the boundary curve $$C$$ of the region $$R$$, where

$$\phi=\ln(x^2+y^2)+x^2y^3 \qquad R: \text{1st quadrant of x^2+y^2=a^2}$$

For the $$1$$st part, I need to choose appropriate P and Q.

\begin{align} \iint_R\nabla^2 \phi \, \mathrm{d}A &=\iint_R \left(\frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}\right)\, \mathrm{d}A\\ \\ &=\iint_R \frac{\partial}{\partial x}\left(\frac{\partial \phi}{\partial x}\right)+\frac{\partial}{\partial y}\left(\frac{\partial \phi}{\partial y}\right)\, \mathrm{d}A \tag 1 \end{align}

By Green's theorem,

$$\iint=\iint \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\, \mathrm{d}A \tag 2$$

Comparing (1) and (2)

$$P=-\frac{\partial \phi}{\partial y}\\ Q=\frac{\partial \phi}{\partial x}$$

\begin{align} \iint_R & =\int -\frac{\partial \phi}{\partial y}\;dx+\frac{\partial \phi}{\partial x}\, \mathrm{d}y\\ \\ & =\int (-\frac{\partial \phi}{\partial y}\;\frac{dx}{ds}+\frac{\partial \phi}{\partial x}\;\frac{dy}{ds})\, \mathrm{d}s\\ \\ &=\int(\frac{\partial \phi}{\partial x}\hat{i}+\frac{\partial \phi}{\partial y}\hat{j}).(\frac{dy}{ds}\hat{i}-\frac{dx}{ds}\hat{j})\\ \end{align}

$$\therefore \nabla \phi=\frac{\partial \phi}{\partial x}\hat{i}+\frac{\partial \phi}{\partial y}\hat{j} \\ \hat{n}=\frac{dy}{ds}\hat{i}-\frac{dx}{ds}\hat{j}$$

Then for the 2nd part I calculate

$$\nabla \phi= (\frac{2x}{x^2+y^2}+2xy^3)\hat{i}+(\frac{xy}{x^2+y^2}+3x^2y^3)\hat{j}$$

And then after parametrising, I have made a big mess. I need some help.

I reached

$$\int_0^{2\pi}\left(-\frac{2}{a}-5a^5\sin^2 t\cos^3 t\right)\, \mathrm{d}t.$$

I am not sure whether this is good. I am thinking of using Walli's formula for this, but I am not sure where to start.