limits of changing the order of a double integral

Question

I've solved this integral $\int_{1}^3 \int_{1-y}^{2-y}x^2y\,dxdy$ which is 3 1/3. Now I'm trying it with different integral order but can't get the limits right. My try was, $\int_{0}^1 \int_{1-x}^{2-x}x^2y\,dydx$.

Answer 1

A good drawing always helps for these sorts of questions.

As you can see, the region can be broken up into a triangle between $x=-2$ and $-1$, a pair of lines between $x=-1$ and $0$, and another triangle between $x=0$ and $1$.

So hopefully you can see that

$$\int_1^3 \int_{1-y}^{2-y} x^2y \ \text{d}x \ \text{d}y = \int_{-2}^{-1}\int_{1-x}^3 x^2y \ \text{d}y \ \text{d}x + \int_{-1}^{0} \int_{1-x}^{2-x}x^2y \ \text{d}y \ \text{d}x + \int_{0}^{1} \int_{1}^{2-x}x^2y \ \text{d}y \ \text{d}x$$

Answer 2

Hint: Try drawing the figure, bounded by \begin{align} x=2-y\\ x=1-y\\ y=1\\ y=3 \end{align}

And then set the limits with y as a function of x first.

Answer 3

First of all you need to draw a graph of the region of integration which is a parallelogram.

You notice that you need three integrals to cover the region if you changed the order of integration.

The x runs from -2 to 1 and you upper and lower limits for y changes on each sub-intervals of length 1.