Line integral along the edge of an isoparametrically mapped quadrilateral

I need to integrate a function along the edge of a quadrilateral (boundary integral). For example, the function is $$f(x,y)=x^3+y^3$$, the quadrilateral coordinates are $$(0,0),(2,-1),(3,2),(1,3)$$ and the concerned edge is the line joining $$(2,-1)$$ and $$(3,2)$$. So, the required integral is $$I=\int_{(2,-1)}^{(3,2)}x^3+y^3\ dl\ \ ;\ \ dl=\sqrt{dx^2+dy^2}$$ In another procedure I mapped the quadrilateral into a standard square in $$s,t$$ axes.
By using Lagrange interpolation, $$X(s,t)=\ \frac{3(s + 1)(t + 1)}{4} - \frac{(s + 1)(t - 1)}{2} - \frac{(s - 1)(t + 1)}{4}$$ $$Y(s,t)=\ \frac{(s + 1)(t - 1)}{4} - \frac{3(s - 1)(t + 1)}{4}+ \frac{(s + 1)(t + 1)}{2}$$ Since integration is from $$t=-1$$ to $$t=1$$, $$dl=J dt$$ where $$J=\sqrt{\left(\frac{\partial X}{\partial t}\right)^2+\left(\frac{\partial Y}{\partial t}\right)^2}$$. So the required integral is, $$I=\int_{-1}^1X^3+Y^3\ Jdt$$ Unfortunately, the second method gives the wrong result. I got 66.4078 while the actual result is 55.3399.
Where did I make mistake here?

• Did you compute $(X^3+Y^3)J$? – chris Dec 21 '16 at 15:32
• @chris, Yes. I have integrated it. Is there any conceptual error in procedure? – user294664 Dec 21 '16 at 16:54
• Your first integral goes from (2,-1) to (3,2). The exact solution is $35\sqrt{10}/2$. If that edge corresponds to $s=1$ and $-1 \leq t \leq 1$, your mapping would go from (2,-1) to (3,-2), so it's probably not correct. Besides, I would expect the determinant of the mapping's Jacobian, not the 2-norm. – chris Dec 22 '16 at 13:02
• @chris How would I get a better result numerically from mapping? – user294664 Dec 22 '16 at 15:02
• @chris, Could you explain how does my mapping go from (2,-1) to (3,-2) – user294664 Dec 22 '16 at 15:23

Denote the rectangular vertices, as follows: $$\mathbf{v}_{00}=\left(\matrix{0\\0}\right),\quad \mathbf{v}_{10}=\left(\matrix{2\\-1}\right),\quad \mathbf{v}_{11}=\left(\matrix{3\\2}\right),\quad \mathbf{v}_{01}=\left(\matrix{1\\3}\right)$$ Now, the position vector $$\mathbf{r}$$ inside (or on the boundary) of the quadrilateral can be expressed in terms of barycentric coordinates $$(\xi,\eta)$$: $$\mathbf{r}(\xi,\eta) = (1-\xi)(1-\eta)\mathbf{v}_{00} + \xi(1-\eta)\mathbf{v}_{10} + (1-\xi)\eta\mathbf{v}_{01} + \xi\eta\mathbf{v}_{11}, \quad \xi,\eta\in[0,1]$$ Now, the integral for the edge between $$\mathbf{v}_{10}$$ and $$\mathbf{v}_{11}$$, will correspond to $$\xi=1$$ and $$\eta \in[0,1]$$. Now, in a slighlty sloppy mathematical notation: $$\int\limits_{\mathbf{v}_{10}}^{\mathbf{v}_{11}}f(\mathbf{v})d\mathbf{v} = \int\limits_0^1d\eta f\big(\mathbf{r}(\xi=1,\eta)\big)J$$ where $$f(\mathbf{v}) = (\mathbf{v}\cdot\hat{x})^3 + (\mathbf{v}\cdot\hat{y})^3$$, which is just a fancy way to write $$f(x,y) = x^3+y^3$$ using position-vectors. For this integral, the Jacobian $$J$$ would be the length of the line between $$\mathbf{v}_{10}$$ and $$\mathbf{v}_{11}$$. So, $$J=\sqrt{10}$$
$$J\int\limits_0^1d\eta f\big(\mathbf{r}(\xi=1,\eta)\big)=\sqrt{10}\frac{35}{2}\approx 55.3399$$ which is the analytical result.