Double integrals over a bounded, but non-rectangular region

Multiple integrals: Double integrals

Double integrals over a bounded, but non-rectangular region

The region of integration in a double integral need not necessarily be a rectangle; this already follows from the interpretation that it is the volume under the graph of a positive continuous function $f(x,y)$ over a closed region in the $xy$ plane. We start with two particular regions, which are one-sidedly simple.

A region of integration is $\boldsymbol{y}$ -simple if it is bounded by two vertical lines $x=a$ and $x=b$ together with two continuous functions $y=c(x)$ and $y=d(x)$ between these two lines. The figure below illustrates the situation.

y-eenvoudig integratiegebied

The double integral $\displaystyle \iint_R f(x,y)\,\dd(x,y)$ is then approximated by adding the volumne under the graph of $f$ of 'very thin' slices to the left starting in $x$ and with width $\dd x$ between $x=a$ and $x=b$ , approximated by $\displaystyle \left(\int_{y=c(x)}^{y=d(x)}f(x,y)\,\dd y\right)\!\cdot \dd x$ . Thus, in the infinitesimal case, we get the following iterated integral: $\iint_R f(x,y)\,\dd(x,y)=\int_{x=a}^{x=b}\left(\int_{y=c(x)}^{y=d(x)}f(x,y)\,\dd y\right)\dd x$

A region of integration is $\boldsymbol{x}$ -simple if it is bounded by two horizontal lines $y=c$ and $y=d$ together with two continuous functions $x=a(y)$ and $x=d(y)$ between these two lines. The figure below illustrates the situation.

x-eenvoudig integratiegebied

The double integral $\displaystyle \iint_R f(x,y)\,\dd(x,y)$ is then approximated by adding the volumes under the graph of $f$ of 'very thin' starting on the left in $y$ and with height $\dd y$ between $y=c$ and $y=d$ , approximated by $\displaystyle \left(\int_{x=a(y)}^{x=b(y)}f(x,y)\,\dd x\right)\!\cdot \dd y$ . Thus, in the infinitesimal case, we get the following iterated integral: $\iint_R f(x,y)\,\dd(x,y)=\int_{y=c}^{y=d}\left(\int_{x=a(y)}^{x=b(y)}f(x,y)\,\dd x\right)\dd y$

Many of the regions of integration over which we want to compute a double integral are $y$ -easy, $x$ -easy, or both. In the latter case, the following strong version of Fubini's theorem holds.

Suppose $f(x,y)$ is a continuous function on a region of integration $R$

If $R$ is $y$ -simple, i.e. $R=\{(x,y)\mid a\le x\le b, c(x)\le y\le d(x)\}$ , with continuous function $y=c(x)$ and $y=d(x)$ , then $\iint_R f(x,y)\,\dd(x,y)=\int_{x=a}^{x=b}\left(\int_{y=c(x)}^{y=d(x)}f(x,y)\,\dd y\right)\dd x$
If $R$ is $x$ -simple, i.e. $R=\{(x,y)\mid a(x)\le x\le b(x), c\le y\le d\}$ , with continuous function $x=a(y)$ and $x=b(y)$ , then $\iint_R f(x,y)\,\dd(x,y)=\int_{y=c}^{y=d}\left(\int_{x=a(y)}^{x=b(y)}f(x,y)\,\dd x\right)\dd y$
If $R$ is $x$ - and $y$ -simple, that is $R=\{(x,y)\mid a(x)\le x\le b(x), c\le y\le d\}\quad\text{and}\quad R=\{(x,y)\mid a\le x\le b, c(x)\le y\le d(x)\}$ for some continuous functions $x=a(y)$ , $x=b(y)$ , $y=c(x)$ and $y=d(x)$ , then $\iint_R f(x,y)\,\dd(x,y)=\int_{x=a}^{x=b}\left(\int_{y=c(x)}^{y=d(x)}f(x,y)\,\dd y\right)\dd x=\int_{y=a}^{y=b}\left(\int_{x=a(y)}^{x=b(y)}f(x,y)\,\dd x\right)\dd y$

In the above formulas we have explicitly indicated in the limits of integration which variables are substituted in a primitive. This is useful and prevents errors, but it is not necessary: instead of $\iint_R f(x,y)\,\dd(x,y)=\int_{y=c}^{y=d}\left(\int_{x=a(y)}^{x=b(y)}f(x,y)\,\dd x\right)\dd y$ you can also use or encounter them in mathematical texts the notations $\iint_R f(x,y)\,\dd(x,y)=\int_{c}^{d}\int_{a(y)}^{b(y)}f(x,y)\,\dd x\,\dd y$ and $\iint_R f(x,y)\,\dd(x,y)=\int_{c}^{d}\dd y\int_{a(y)}^{b(y)}f(x,y)\,\dd x$ The order in which $\dd x$ and $\dd y$ appear in the integral formulas already determines the limits of integration and the order of integration.

Calculate $\iint_R (6x+6y)\,\dd(x,y)$ with $R$ the region enclosed between the two parabolas $y=x^2$ and $x=y^2$ .

$\iint_R (6x+6y)\,\dd(x,y)={}$ ${{9}\over{5}}$

The parabolas $y=x^2$ and $x=y^2$ intersect at the origin and at the point $(1,1)$ . If we consider the region of integration as a $y$ -simple region, then we first keep $x$ fixed (with $0\le x\le 1$ and integrate for this fixed $x$ to $y$ . The integration limits are then $a(y)=x^2$ and $b(y)=\sqrt{x}$ ; see the figure below.

integratiegebied is y-eenvoudig

The iterated integral becomes: $\begin{aligned}\iint_R (6 x+6 y)\,\dd(x,y) &= \int_{x=0}^{x=1}\left(\int_{y=x^2}^{y=\sqrt{x}}(6 x+6 y)\,\dd y\right)\dd x\\[0.25cm] &= \int_{x=0}^{x=1}\biggl[(6 xy+3 y^2)\biggr]_{y=x^2}^{y=\sqrt{x}}\,\dd x\\[0.25cm] &= \int_{x=0}^{x=1}\bigl((6 x^{\frac{3}{2}}+3 x)-(6 x^3+3 x^4)\bigr)\,\dd x\\[0.25cm] &=\biggl[{{12}\over{5}} x^{\frac{5}{2}}+{{3}\over{2}}x^2-{{3}\over{2}} x^4-{{3}\over{5}} x^5\biggr]_{x=0}^{x=1}\\[0.25cm] &= {{12}\over{5}}+{{3}\over{2}}-{{3}\over{2}}-{{3}\over{5}}\\[0.25cm]&={{9}\over{5}} \end{aligned}$

We can also consider the region of integration as $x$ -simple. Then we first keep $y$ fixed (with $0\le y\le 1$ and integrate for this fixed $y$ to $x$ . The integration limits are then $c(x)=y^2$ and $d(x)=\sqrt{y}$ ; see the figure below.

x-eenvoudig gebied

The iterated integral becomes: $\begin{aligned}\iint_R (6 x+6 y)\,\dd(x,y) &= \int_{y=0}^{y=1}\left(\int_{x=y^2}^{x=\sqrt{y}}(6x+6y)\,\dd x\right)\dd y\\[0.25cm] &= \int_{y=0}^{y=1}\biggl[(3 x^2+6 xy)\biggr]_{x=y^2}^{x=\sqrt{y}}\,\dd x\\[0.25cm] &= \int_{y=0}^{y=1}\bigl((3 y+6 y^{\frac{3}{2}})-(3 y^4+6 y^3)\bigr)\,\dd x\\[0.25cm] &=\biggl[{{3}\over{2}} y^2+{{12}\over{5}} y^{\frac{5}{2}}-{{3}\over{5}} y^5-{{3}\over{2}} y^4\biggr]_{y=0}^{y=1}\\[0.25cm] &= {{3}\over{2}}+{{12}\over{5}}-{{3}\over{5}}-{{3}\over{2}}\\[0.25cm]&={{9}\over{5}} \end{aligned}$

The route is different, but the result is the same.

New example

In the exact calculation of a double integral $\iint_R f(x,y)\dd(x,y)$ over a non-rectangular region $R$ you can proceed as follows.

Sketch the region $R$ in the $xy$ plane.
Check whether the region of integration is $x$ -simple, $y$ -simple, or maybe both.
Decide on the basis of the integrand and the nature of the region of integration whether to integrate to $x$ or $y$ first.

When integrating first to $y$ (i.e. $R$ is $y$ -simple):

Determine the left bound $a$ and right bound $b$ for $x$ as extremes of $R$ in the horizontal direction.
Draw a vertical auxiliary line at position $x$ between $a$ and $b$ in the sketch of $R$
Determine the lower limit $c(x)$ and the upper limit $d(x)$ for the segment that remains within $R$ at horizontal position $x$
Calculate $\displaystyle \int_{y=c(x)}^{y=d(x)}\,f(x,y)\,\dd y$ . This is a function of $x$ .
Integrate the latter function with limits of integration $a$ and $b$ . So: $\iint_R f(x,y)\,\dd(x,y)=\int_{x=a}^{x=b}\left(\int_{y=c(x)}^{y=d(x)}f(x,y)\,\dd y\right)\dd x$

When integrating first to $x$ (i.e. $R$ is $x$ -simple):

Determine the $c$ lower bound and $d$ upper bound for $y$ as extremes of $R$ in the vertical direction.
Draw a horizontal auxiliary line at position $y$ between $c$ and $d$ in the sketch of $R$
Determine the left limit $a(y)$ and the right limit $b(y)$ for the segment that remains within $R$ at vertical position $y$
Calculate $\displaystyle \int_{x=a(y)}^{x=d(y)} f(x,y)\,\dd x$ . This is a function of $y$ .
Integrate the latter function with limits of integration $c$ and $d$ . So: $\iint_R f(x,y)\, \dd(x,y)=\int_{y=c}^{y=d}\left(\int_{x=a(y)}^{x=b(y)} f(x,y)\,\dd x\right)\dd y$