Changing order of integration limits

Question

$$\int_{1}^{3} \int_{0}^y x+y-1 \, dx \, dy = 9$$

How would I change the order of integration here? Wouldn't this require two integrals?

$$\int_{0}^{1} \int_{1}^3 x+y-1 \, dy \, dx + \int_{1}^{3} \int_{x}^3 x+y-1 \, dy \, dx = 9$$

Why does this integral below work?

$$\int_{0}^{3} \int_{x}^3 x+y-1 \, dy \, dx = 9$$

Is this just a coincidence? I'm not sure how to plot this in 3D. Here's a graph of the xy plane and the boundaries of the region on it. This is how I visualized the region. Perhaps I did something wrong?

I am finding the volume between the surface z = x + y and z = 1

Answer 1

$\displaystyle \int_1^3\!\!\int_0^y x+y+1 \operatorname{d}x\operatorname{d}y$, integrates over the interval $y\in[1,3], x\in[0,y]$.

As you noted, this is equivalent to the sum of the intervals: $x\in[0,1], y\in[1,3]$ and $x\in(1,3], y\in[x, 3]$

$\displaystyle \int_1^3\!\!\int_0^y x+y-1 \operatorname{d}x\operatorname{d}y = \displaystyle \int_0^1\!\!\int_1^3 x+y-1 \operatorname{d}y\operatorname{d}x+ \displaystyle \int_1^3\!\!\int_x^3 x+y-1 \operatorname{d}y\operatorname{d}x = 9$

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Note that the integral over the lower corner of the triangle (below y=1) is : $\displaystyle \int_0^1\!\!\int_x^1 x+y-1 \operatorname{d}y\operatorname{d}x = 0$, and hence by coincidence the integral over the entire triangle is equal to that of the truncated triangle (the required interval).

It will not hold in general.

Answer 2

The three equalities:

$$\int_{1}^{3} \int_{0}^y x+y-1 \, dx \, dy = 9$$

$$\int_{0}^{1} \int_{1}^3 x+y-1 \, dy \, dx + \int_{1}^{3} \int_{x}^3 x+y-1 \, dy \, dx = 9$$

$$\int_{0}^{3} \int_{x}^3 x+y-1 \, dy \, dx = 9$$

are correct. However,

$$\int_{0}^{3} \int_{x}^3 x+y-1 \, dy \, dx = \int_{1}^{3} \int_{0}^y x+y-1 \, dx \, dy = 9$$ is ONLY correct because

$$\int_{0}^1 \int_x^1 x+y-1 \,dy\,dx = 0$$

That is the integral evaluated over the little triangle in the bottom left, just below the green, in your picture. That is the difference between the second and third integrals at the start of this post. In general,

$$\int_{0}^{3} \int_{x}^3 f(x,y) \, dy \, dx \neq \int_{1}^{3} \int_{0}^y f(x,y) \, dx \, dy$$

Therefore, this is purely a coincidence and it should be:

$$\int_{1}^{3} \int_{0}^y f(x,y) \, dx \, dy = \int_{0}^{1} \int_{1}^3 f(x,y) \, dy \, dx + \int_{1}^{3} \int_{x}^3 f(x,y) \, dy \, dx$$