Question 1 of 7: Gauss quadrature of a field variable over a rectangular domain

Question 1: Gauss quadrature of a field variable over a rectangular domain (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A single four-node bilinear isoparametric element covering the whole rectangle, with the integrand and limits printed in the question.

Find. The value of $g$ from Gauss–Legendre quadrature with the smallest sufficient number of points, presented in vector-matrix form with the Jacobian, and an explanation of how it compares with the printed exact value.

Approach. Map the rectangle onto the parent square with the given bilinear shape functions, form the Jacobian, choose the quadrature order from the polynomial degree of the mapped integrand, and sum the weighted station values.

1. Map the physical rectangle onto the parent square. The isoparametric statement is $x=\sum_{i=1}^{4}N_i x_i$ and $y=\sum_{i=1}^{4}N_i y_i$. With the nodal coordinate matrix $\mathbf{X}=\begin{bmatrix}1&2\\7&2\\7&6\\1&6\end{bmatrix}$ the sums collapse to the affine map $$x = 4 + 3\xi,\qquad y = 4 + 2\eta$$ because opposite sides of the rectangle are parallel and equal.
2. Form the Jacobian matrix in the standard array format. The Jacobian of the map is the derivative array of the shape functions post-multiplied by the nodal coordinates, $$[\mathbf{J}]=\frac{1}{4}\begin{bmatrix}-(1-\eta)&(1-\eta)&(1+\eta)&-(1+\eta)\\-(1-\xi)&-(1+\xi)&(1+\xi)&(1-\xi)\end{bmatrix}\begin{bmatrix}x_1&y_1\\x_2&y_2\\x_3&y_3\\x_4&y_4\end{bmatrix}=\begin{bmatrix}3&0\\0&2\end{bmatrix}$$ Every $\xi$ and $\eta$ has cancelled, so the Jacobian is constant over the element and $$\boxed{\,|\mathbf{J}| = (3)(2)-(0)(0) = 6\,}$$ The determinant is positive everywhere, confirming a valid orientation-preserving mapping.
3. Choose the number of Gauss points from the polynomial degree. Substituting the map, the mapped integrand $f\bigl(x(\xi),y(\eta)\bigr)=x+x^{3}y$ is cubic in $\xi$ (the $x^{3}$ term) and linear in $\eta$. An $n$-point Gauss–Legendre rule is exact for polynomials of degree $\le 2n-1$, so the requirement $2n-1 \ge 3$ gives $n=2$ in the $\xi$ direction, and $n=1$ would already suffice in $\eta$. Taking the conventional square rule, $$n_\xi \times n_\eta = 2\times 2 = 4 \text{ points},\qquad \xi_i,\eta_j = \pm\tfrac{1}{\sqrt{3}} = \pm 0.5773503,\qquad w_i = w_j = 1$$
4. Evaluate the integrand at the four stations. The stations map to $x = 4 \pm 3/\sqrt{3} = 2.2679492$ or $5.7320508$ and $y = 4 \pm 2/\sqrt{3} = 2.8452995$ or $5.1547005$. Collecting the integrand values in the array $\mathbf{F}$, with rows indexed by $\xi$ and columns by $\eta$:

5. Assemble the quadrature in vector-matrix form. With the weight vector $\mathbf{w}=\{1\ \ 1\}^{T}$ and the station array $\mathbf{F}$, the whole evaluation is a single quadratic form: $$g \simeq |\mathbf{J}|\;\mathbf{w}^{T}\mathbf{F}\,\mathbf{w}= 6\begin{Bmatrix}1&1\end{Bmatrix}\begin{bmatrix}35.45954&62.39963\\541.60037&976.54047\end{bmatrix}\begin{Bmatrix}1\\1\end{Bmatrix}$$ Carrying out the products, $\mathbf{w}^{T}\mathbf{F}\mathbf{w} = 1616.00000$ and $$\boxed{\,g \simeq 6 \times 1616.00000 = 9696.000\,}$$
6. Part (b) — compare with the exact value. The quadrature result is identical to the printed exact value $g_{\text{exact}} = 9696$, to every digit carried; the discrepancy is $0.000\%$ and is pure floating-point round-off. This is not a coincidence and it is the point the question is testing: the mapping is affine, so $|\mathbf{J}|$ is a constant that comes outside the integral and does not raise the polynomial degree; the mapped integrand is a polynomial of degree 3 in $\xi$ and 1 in $\eta$; and the two-point Gauss–Legendre rule integrates any polynomial up to degree $2(2)-1 = 3$ exactly. Whenever the integrand is polynomial and the element is a parallelogram, Gauss quadrature of sufficient order is not an approximation at all.
7. Show the contrast with an insufficient rule. Had a single point been used, the rule would sample only the centroid $\xi=\eta=0$, i.e. $x=4$, $y=4$, with weight $w=2$ in each direction: $$g_{1\text{-pt}} = (2)(2)\,f(4,4)\,|\mathbf{J}| = 4(260)(6) = 6240$$ an error of $(6240-9696)/9696 = -35.64\%$. The one-point rule is exact only through degree $2(1)-1 = 1$, and the integrand is cubic, so the cubic and quadratic content is simply lost. The comparison shows that the phrase “appropriate number of gauss points” means enough for the polynomial degree, not “one per node”.