[EXAMPLES] Regions and order of integration

Let

f(x,y)=xy^2

and let $\Omega$ be the triangle with vertices

(0,0),\qquad(1,1),\qquad(1,-1).

Vertical slices give

0\le x\le1,\qquad -x\le y\le x.

Therefore

\begin{aligned} \iint_\Omega xy^2\,dA &=\int_0^1\int_{-x}^{x}xy^2\,dy\,dx\\ &=\int_0^1x\left[\frac{y^3}{3}\right]_{-x}^{x}dx\\ &=\int_0^1\frac{2x^4}{3}\,dx =\frac{2}{15}. \end{aligned}

The same triangle can be described horizontally by

-1\le y\le1,\qquad |y|\le x\le1.

The reversed-order calculation is

\begin{aligned} \iint_\Omega xy^2\,dA &=\int_{-1}^1\int_{|y|}^{1}xy^2\,dx\,dy\\ &=\int_{-1}^1\frac{y^2}{2}(1-y^2)\,dy =\frac{2}{15}. \end{aligned}

[routine]

Consider

\int_0^1\int_{2x}^{2}\sin(y^2)\,dy\,dx.

The inner integral has no elementary antiderivative. The region is

0\le x\le1,\qquad 2x\le y\le2.

Its $y$ -values range from $0$ to $2$ . For fixed $y$ , the inequality $2x\le y$ gives $0\le x\le y/2$ . Hence

\begin{aligned} \int_0^1\int_{2x}^{2}\sin(y^2)\,dy\,dx &=\int_0^2\int_0^{y/2}\sin(y^2)\,dx\,dy\\ &=\int_0^2\frac y2\sin(y^2)\,dy\\ &=\frac14\int_0^4\sin u\,du \qquad(u=y^2)\\ &=\frac{1-\cos4}{4}. \end{aligned}

Reversing the order turns the difficult inner integral into a direct substitution.

[exam]

Consider

\int_{-1}^{1}\int_{x^2}^{2-x^2}f(x,y)\,dy\,dx.

The region lies between

y=x^2,\qquad y=2-x^2.

Its $y$ -range is $0\le y\le2$ . For $0\le y\le1$ , the binding condition is $x^2\le y$ :

-\sqrt y\le x\le\sqrt y.

For $1\le y\le2$ , the binding condition is $y\le2-x^2$ :

-\sqrt{2-y}\le x\le\sqrt{2-y}.

The reversed order is therefore

\begin{aligned} \int_{-1}^{1}\int_{x^2}^{2-x^2}f(x,y)\,dy\,dx &=\int_0^1\int_{-\sqrt y}^{\sqrt y}f(x,y)\,dx\,dy\\ &\quad+\int_1^2\int_{-\sqrt{2-y}}^{\sqrt{2-y}} f(x,y)\,dx\,dy. \end{aligned}

A single pair of horizontal bounds would either omit part of the region or include points outside it.

[challenge]

Examples from Regions and order of integration