1 Functions as limits of integration

In Section 27.1 double integrals of the form

$I={\int \; <!--nolimits\; -->}_{x=a}^{x=b}{\int \; <!--nolimits\; -->}_{y=c}^{y=d}f\left(x,y\right)dydx$

were considered. They represent an integral over a rectangular region in the $xy$ plane. If the limits of integration of the inner integral are replaced with functions $G_1,G_2$ ,

$I={\int \; <!--nolimits\; -->}_{x=a}^{x=b}{\int \; <!--nolimits\; -->}_{G_1\left(x\right)}^{G_2\left(x\right)}f\left(x,y\right)dydx$

then the region described will not, in general, be a rectangle. The region will be a shape bounded by the curves (or lines) which these functions $G_1$ and $G_2$ describe.

As was indicated in 27.1

$I={\int \; <!--nolimits\; -->}_{x=a}^{x=b}{\int \; <!--nolimits\; -->}_{G_1\left(x\right)}^{G_2\left(x\right)}f\left(x,y\right)dydx$

can be interpreted as the volume lying above the region in the $xy$ plane defined by $G_1\left(x\right)$ and $G_2\left(x\right)$ , bounded above by the surface $z=f\left(x,y\right)$ . Not all double integrals are interpreted as volumes but this is often the case. If $z=f\left(x,y\right)<0$ anywhere in the relevant region, then the double integral no longer represents a volume.

Example 8

Evaluate the integral $I={\int \; <!--nolimits\; -->}_{x=0}^1{\int \; <!--nolimits\; -->}_{y=0}^{1-x}2xydydx$

Solution

Figure 10

Figure 11

Projecting the relevant part of the surface (Figure 10) down to the $xy$ plane produces the triangle shown in Figure 11. The extremes that $x$ takes are $x=0$ and $x=1$ and so these are the limits on the outer integral. For any value of $x$ , the variable $y$ varies between $y=0$ (at the bottom) and $y=1-x$ (at the top). Thus if the volume, shown in the diagram, under the function $f\left(x,y\right)$ , bounded by this triangle is required then the following integral is to be calculated.

${\int \; <!--nolimits\; -->}_{x=0}^1{\int \; <!--nolimits\; -->}_{y=0}^{1-x}f\left(x,y\right)dydx$

Once the correct limits have been determined, the integration is carried out in exactly the same manner as in Section 27.1

First consider the inner integral $g\left(x\right)={\int \; <!--nolimits\; -->}_{y=0}^{1-x}2xydy$

Integrating $2xy$ with respect to $y$ gives $x{y}^2+C$ so $g\left(x\right)={\left[x{y}^2\right]}_{y=0}^{1-x}=x{\left(1-x\right)}^2$

Note that, as is required, this is a function of $x$ , the variable of the outer integral. Now the outer integral is

Regions do not have to be bounded only by straight lines. Also the integrals may involve other tools of integration, such as substitution or integration by parts. Drawing a sketch of the limit functions in the plane and shading the region is a valuable tools when evaluating such integrals.

Example 9

Evaluate the volume under the surface given by $z=f\left(x,y\right)=2xsin\left(y\right)$ , over the region bounded above by the curve $y=x^2$ and below by the line $y=0$ , for $0\le x\le 1$ .

Figure 12

Solution

First sketch the curve $y=x^2$ and identify the region. This is the shaded region in Figure 12. The required integral is

Making the substitution $u=x^2$ so $du=2xdx$ and noting that the limits $x=0,1$ map to $u=0,1$ , gives

Example 10

Evaluate the volume under the surface given by $z=f\left(x,y\right)=x^2+\frac{1}{2}y$ , over the region bounded by the curves $y=2x$ and $y=x^2$ .

Figure 13

Solution

The sketch of the region is shown in Figure 13. The required integral is

$I={\int \; <!--nolimits\; -->}_{x=a}^b{\int \; <!--nolimits\; -->}_{y=x^2}^{2x}\left(x^2+\frac{1}{2}y\right)dydx$

To determine the limits for the integration with respect to $x$ , the points where the curves intersect are required. These points are the solutions of the equation $2x=x^2$ , so the required limits are $x=0$ and $x=2$ . Then the volume is given by

Example 11

1. Evaluate the volume under $z=f\left(x,y\right)=5x^{2}y$ , over the half of the unit circle that lies above the $x$ -axis. (Figure 14).

   Figure 14

   

2. Repeat 1. for $z=f\left(x,y\right)=1.$

Solution

1. This region is bounded by the circle $y^2+x^2=1$ and the line $y=0$ . Since only positive values of $y$ are required, the equation of the circle can be written $y=\sqrt{1-x^2}$ . Then the required volume is given by $$\begin{array}{rcll}I& =& {\int \; <!--nolimits\; -->}_{x=-1}^1{\int \; <!--nolimits\; -->}_{y=0}^{\sqrt{1-x^2}}\left(5x^{2}y\right)dydx={\int \; <!--nolimits\; -->}_{x=-1}^1{\left[\frac{5}{2}{x}^2{y}^2\right]}_{y=0}^{\sqrt{1-x^2}}dx& \\ & & & \\ & =& {\int \; <!--nolimits\; -->}_{x=-1}^1\frac{5}{2}{x}^2\left(1-x^2\right)dx=\frac{5}{2}{\left[\frac{x^3}{3}-\frac{x^5}{5}\right]}_{-1}^1=\frac{2}{3}& \end{array}$$
2. $$\begin{array}{rcll}I& =& {\int \; <!--nolimits\; -->}_{x=-1}^1{\int \; <!--nolimits\; -->}_{y=0}^{\sqrt{1-x^2}}1dydx={\int \; <!--nolimits\; -->}_{x=-1}^1{\left[y\right]}_{y=0}^{\sqrt{1-x^2}}dx& \\ & =& {\int \; <!--nolimits\; -->}_{x=-1}^1\sqrt{1-x^2}dx\left(\mathrm{\text{which by substituting}}x=sin\theta \right)=\frac{\pi }{2}& \end{array}$$

Note that by putting $f\left(x,y\right)=1$ we have found the volume of a semi-circular lamina of uniform height 1. This result is numerically the same as the area of the region in Figure 14. (This is a general result.)

Task!

Evaluate the following double integral over a non-rectangular region.

${\int \; <!--nolimits\; -->}_{x=0}^1{\int \; <!--nolimits\; -->}_{y=x}^1\left(x^2+y^2\right)dydx$

1. First sketch the region of the $xy$ -plane determined by the limits:

   Answer

   

2. Now evaluate the inner triangle:

   Answer

   In the triangle, $x$ varies between $x=0$ and $x=1$ . For every value of $x$ , $y$ varies between $y=x$ and $y=1$ .
   
   The inner integral is given by

   $$\begin{array}{rcll}\text{Inner}\text{Integral}={\int \; <!--nolimits\; -->}_{y=x}^1\left(x^2+y^2\right)dy& =& {\left[x^{2}y+\frac{1}{3}{y}^3\right]}_x^1& \\ & =& x^2\times 1+\frac{1}{3}\times 1^3-\left(x^2\times x+\frac{1}{3}{x}^3\right)& \\ & =& x^2+\frac{1}{3}-\frac{4}{3}{x}^3& \end{array}$$
3. Finally evaluate the outer integral:

   Answer

   The inner integral is placed in the outer integral to give

   $$\begin{array}{rcll}\text{Outer}\text{Integral}={\int \; <!--nolimits\; -->}_0^1\left(x^2-\frac{4}{3}{x}^3+\frac{1}{3}\right)dx& =& {\left[\frac{1}{3}{x}^3-\frac{1}{3}{x}^4+\frac{1}{3}x\right]}_0^1& \\ & =& \left(\frac{1}{3}-\frac{1}{3}+\frac{1}{3}\right)-0& \\ & =& \frac{1}{3}& \end{array}$$

   Note that the above Task is simply one of integrating a function over a region - there is no reference to a volume here. Another like this now follows.

Task!

Integrate the function $z=x^{2}y$ over the trapezium with vertices at $\left(0,0\right)$ , $\left(1,1\right)$ , $\left(1,2\right)$ and $\left(0,4\right)$ .

Answer

Evaluate the following integrals

1. ${\int \; <!--nolimits\; -->}_{x=0}^1{\int \; <!--nolimits\; -->}_{y=3x}^{x^2+2}xydydx$
2. ${\int \; <!--nolimits\; -->}_{x=1}^2{\int \; <!--nolimits\; -->}_{y=x^2+2}^{3x}xydydx$ [Hint: Note how the same curves can define different regions.]
3. ${\int \; <!--nolimits\; -->}_{x=1}^2{\int \; <!--nolimits\; -->}_{y=1}^{x^2}\frac{x}{y}dydx$ , [Hint: use integration by parts for the outer integral.]

Answer

1.1 Splitting the region of integration

Sometimes it is difficult or impossible to represent the region of integration by means of consistent limits on $x$ and $y$ . Instead, it is possible to divide the region of integration into two (or more) sub-regions, carry out a multiple integral on each region and add the integrals together. For example, suppose it is necessary to integrate the function $g\left(x,y\right)$ over the triangle defined by the three points $\left(0,0\right)$ , $\left(1,4\right)$ and $\left(2,-2\right)$ .

Figure 15

It is not possible to represent the triangle $ABC$ by means of limits on an inner integral and an outer integral. However, it can be split into the triangle $ABD$ and the triangle $BCD$ . $D$ is chosen to be the point on $AC$ directly beneath $B$ , that is, line $BD$ is parallel to the $y$ -axis so that $x$ is constant along it. Note that the sides of triangle $ABC$ are defined by sections of the lines $y=4x$ , $y=-x$ and $y=-6x+10$ .

In triangle $ABD$ , the variable $x$ takes values between $x=0$ and $x=1$ . For each value of $x$ , $y$ can take values between $y=-x$ (bottom) and $y=4x$ . Hence, the integral of the function $g\left(x,y\right)$ over triangle $ABD$ is

$I_1={\int \; <!--nolimits\; -->}_{x=0}^{x=1}{\int \; <!--nolimits\; -->}_{y=-x}^{y=4x}g\left(x,y\right)dydx$

Similarly, the integral of $g\left(x,y\right)$ over triangle $BCD$ is

$I_2={\int \; <!--nolimits\; -->}_{x=1}^{x=2}{\int \; <!--nolimits\; -->}_{y=-x}^{y=-6x+10}g\left(x,y\right)dydx$

and the integral over the full triangle is

$I=I_1+I_2={\int \; <!--nolimits\; -->}_{x=0}^{x=1}{\int \; <!--nolimits\; -->}_{y=-x}^{y=4x}g\left(x,y\right)dydx+{\int \; <!--nolimits\; -->}_{x=1}^{x=2}{\int \; <!--nolimits\; -->}_{y=-x}^{y=-6x+10}g\left(x,y\right)dydx$

Example 12

Integrate the function $g\left(x,y\right)=xy$ over the triangle $ABC$ .

Solution

Over triangle $ABD$ , the integral is

Over triangle $BCD$ , the integral is

So the total integral is $I_1+I_2=\frac{15}{8}+\frac{5}{8}=\frac{5}{2}$