I am trying to apply the Gaussian-Legendre Quadrature rule to a Double Integral, namely
$$
\int^1_0 \int^1_0 \text{sin}(x^2+y^2)dxdy
$$
I have done the following:
Define $\phi_n(x)$ is the Legendre polynomial of degree $n$. Using the fact that
\begin{align*}
\int^b_{a} f(x)dx &= \frac{b-a}{2}\int^1_{-1}f\left(\frac{b-a}{2}x+\frac{a+b}{2} \right)dx\\ &= \frac{b-a}{2}\sum^{n}_{i=1}w_if\left(\frac{b-a}{2}x_i+\frac{a+b}{2}\right)
\end{align*}
with
$$
w_i = \frac{2}{(1-x_i^2)[\phi_n '(x_i)]^2}
$$
where each $x_i$ is a root of the polynomial $\phi_n(x)$, I have applied this logic to the double integral case and obtained
$$
\int^1_0\int^1_0 F(x,y)dxdy = \frac{1}{4}\sum^n_{i=1}\sum^n_{j=1}w_i w_jF\left(\frac{1}{2}x_i+\frac{1}{2},\frac{1}{2}y_i+\frac{1}{2}\right)
$$
with
$$
w_i = \frac{2}{(1-x_i^2)[\phi_n '(x_i)]^2} \ \ \text{ and } \ \ w_j = \frac{2}{(1-y_j^2)[\phi_n '(y_j)]^2}
$$
However, this logic has not worked under implementation (using C++ code).
Is the logic here sound? Or have I oversimplied the issue?
Your help will be appreciated.
Note, I looked throughly online and on this website for an and couldn't find anything conclusive – if there is a link, I would appreciate it if you could direct me over.
Best Answer
In what way has it "not worked"?
Have you checked your quadrature method on $1$-variable integrals? How does it do on $\int_0^1 \int_0^1 \sin(x^2)\; dx\; dy$ and $\int_0^1 \int_0^1 \cos(y^2)\; dx\; dy$?