When can you factor out terms from a double integral?

Question

Consider the following double integral:

$$ \int \int_{S} (x+z) dS$$

Over the region in first octant of plane $ x+y+z=1$

I know the bounds are :

$$ 0 \leq x \leq 1$$ $$ 0 \leq y \leq 1$$ $$ 0 \leq z \leq 1$$

To start, I first rewrite this integral as:

$$ \int \int_{S} (1-y) \sqrt{3} dx dy$$

And,

$$ \sqrt{3} \int \int_S (1-y) dx dy$$

Now here is the part I get confused:

I asked some people and they told I have to do integral with 'y' first (outer integral) and then do the inner integral. This is due to the fact that limits of integral on 'y' depend on 'x'. But, I had this alternative scheme in mind:

Suppose we take the imiplict expression of $ x+y+z=1$ and rewrite it as $ x+z=1-y$ So, suppose we fixed $ 1-y = C$ then we could just vary 'x' and 'z' whilst we integrated and hence factor out the expression from integral.

This gives me:

$$ \sqrt{3} \int (1-y) x dy$$

Similar arguments to keep 'x' constant while I integrate with 'y',

$$ \sqrt{3} x \frac{(1-y)^2}{2}$$

Plugging in the bounds:

$$ 0$$

But, I'm not sure where exactly I went wrong...

Answer 1

Continue with

$$I= \iint_S (1-y) dS =\sqrt{3} \iint_{x>0,y>0,x+y<1}(1-y) dx dy\\ = \sqrt3 \int_0^1\int_0^{1-y} (1-y)dxdy = \sqrt3\int_0^1 (1-y)^2dy = \frac1{\sqrt3}$$

Answer 2

In order to start talking about breaking out a factor we must make sure that the double integral over a $2$-dimensional region $D$: $$\iint_{D} f(x,y) \, dA,$$ can be represented as a double iterated integral (so we can consider the integrals seperatley): $$\int_{X}\left(\int_{Y}f(x,y) \, dx \right)dy \ \text{or} \int_{Y}\left(\int_{X}f(x,y) \, dy \right)dx. $$ This can be guranteed (and both integrals are equal) if you can show it to satisfy Fubini's theorem (altough this is not always easy as you might have to compute both integrals).

Suppose for the sake of argument that Fubini's theorem is satisfied in the following example;

Consider $D = \{(x,y): a \leq y \leq b, \alpha(y)\leq x \leq \beta(y)\}$ to be the area we are going to integrate your function $h(x,y)=f(x)g(y)$ over. Assuming this, we can by Fubini's theorem write $$\int_{a}^{b}\left(\int_{\alpha(y)}^{\beta(y)}h(x,y) \,dx\right)dy.$$ Now fix any $y=s\in (a,b)$, then the inner integral is $$\int_{\alpha(s)}^{\beta(s)}h(x,s)\,dx = \int_{\alpha(s)}^{\beta(s)}f(x)g(s)\,dx =g(s)\int_{\alpha(s)}^{\beta(s)}f(x)\,dx. $$ which can intuitively be seen by noting that we are integrating over a horizontal strip in the $x,y$-plane. By definition, a horizontal line in the $x,y$-plane is parallel to the $x$-axis and lines parallel to the $x$-axis can be described by $y=k\in\mathbb{R}$. So naturally $g(y)$ is constant along any such strip as it dependes only on $y$. So we can say in general that it holds for all of the interval $(a,b)$.

We can therefore write \begin{align*}\int_{a}^{b} \left( \int_{ \alpha(y)} ^ { \beta(y)} h(x,y) \, dx \right) dy &= \int_{a}^{b} \left( \int_{ \alpha(y)}^{ \beta(y)} f(x)g(y) \, dx \right)dy\\ &= \int_{a}^{b} g(y) \left(\int_{ \alpha(y)}^{ \beta(y)} f(x) \, dx \right)dy.\end{align*} If you want to break out $f(x)$ then you must find a way to change $D$ into something like $D'=\{(x,y):a'\leq x\leq b', \alpha'(x)\leq y \leq \beta'(x)\}$ which still represents the same area. Then by Fubini's theorem $$\int_{a}^{b} \left( \int_{ \alpha(y)}^{ \beta(y)} f(x)g(y) \, dx \right)dy = \int_{a'}^{b'} \left( \int_{ \alpha'(y)}^{ \beta'(y)} f(x)g(y) \, dy \right)dx$$ but now the inner integral can be considered to be an integration over a vertical strip so $x$ is instead constant along such a strip.