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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.