Integrating triangle in a 2D plane

Question

I am interested in integrating $(x^2y+y^2x)$ on the following loop: $(x=1,y=2)\rightarrow(x=2,y=1)\rightarrow(x=3,y=3)\rightarrow(x=1,y=2)$.

I know this loop forms a triangle with all three sides changing with respect to the $x$ and $y$ axis. I understand how to do the integration part, but unfortunately I am really stuck on how to set up the integration itself. Should I do a double integral for each side of the triangle and then add them?

If anyone can help me set up the integration for this problem it would be greatly appreciated. Thank you!

Answer 1

Yes, it can be done using two double integrals. Here you are dividing the triangle into two parts by the vertical line $x=2$.

The equation of the line joining $(1,2)$ and $(2,1)$ is $x+y=3$, and that of the line joining $(1,2)$ to $(3,3)$ is $x-2y+3=0$.

You can set $x=1$ to $x=2$, and then you will have the range of $y$ as $y=3-x$ to $y=\frac{x+3}{2}$.

The equation of the line joining $(2,1)$ to $(3,3)$ is $2x-y=3$.

Then you can set $x=2$ to $x=3$, and you will get the range of $y$ as $y=2x-3$ to $y=\frac{x+3}{2}$.

EDIT:

The question meant to ask integrating along the loop, not over the region bounded by it.

For each line, the curve can be parameterized. Example procedure for $y=3-x$.

Let $x=t$. Then, $y=3-t$. $t$ varies from $1$ to $2$.

The function $x^2y+y^2x$ then becomes $t^2(1-t)+t(1-t)^2$.

$$\oint {(x^2y+y^2x).dr}= \int_{1}^{2} {(t^2(1-t)+t(1-t)^2)|v(t)|dt}+otherterms$$.

where $$|v(t)|=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}=\sqrt{2}$$

Answer 2

You can always parameterize a triangle with $u,v$, where $u,v>0$ and $u+v<1$. This always leads to an integral $$\int_0^1 \int_0^u\ldots dv\,du$$

Convex interpolation inside the triangle with vertices A,B,C is then $$T=Au+Bv+C(1-u-v)$$ Imagine that "u" is the variable that tells you how much you move from C to A, and "v" the same from C to B. (u=0,v=0) is now C, (u=1,v=0) is now A, and (u=0,v=1) is B. That defines your variable change for both coordinates. For illustration, see sketches for Barycentric coordinates (we are integrating over two of barycentric coordinates, and the third one is expressed with the other two $1-u-v$).

From this you must also compute the Jacobian (which is the ratio how much your original triangle is bigger than the unit triangle (0,0)→(1,0)→(0,1) in (u,v) space where you are doing the integral).

In your case:

$$(x,y)=(u+2v+3(1-u-v),2u+v+3(1-u-v))=(3-2u-v,3-u-2v)$$ and the jacobian:

$$J=\det\begin{bmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} =(-2)(-2)-(-1)(-1)=3$$

This leaves you with

$$\int_0^1 \int_0^u \left((3-2u-v)^2(3-u-2v)+(3-u-2v)^2(3-2u-v)\right) 3 dv\,du$$