instructed to write integrals over planar domain

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Example 1

We evaluate the iterated integral

$$ \int_0^1 \int_{1-x}^{1-x^2} \,xy\,dy\,dx. $$

To evaluate the integral symbolically, we can proceed in two stages.
syms x y
firstint =
(x^2*(x - 1)^2*(x + 2))/2
answer =

We can even perform the two integrations in a single step:

ans =

There is, of course, no need to evaluate such a simple integral numerically. However, if we change the integrand to, say, exp(x^2 - y^2), then MATLAB will be unable to evaluate the integral symbolically, although it can express the result of the first integration in terms of erf(x), which is the (renormalized) antiderivative of exp(-x^2).

Warning: Explicit integral could not be found. 
ans =
int((pi^(1/2)*exp(x^2)*(erf(x - 1) - erf(x^2 - 1)))/2, x = 0..1)

However, we can evaluate the integral numerically, using double.

Warning: Explicit integral could not be found. 

ans =


Problem 1

Evaluate the iterated integrals

$$\int_1^4 \int_{x}^{x^2} \,(x + y^3)\,dy\,dx.  $$


$$ \int_1^4 \int_{x}^{x^2} \,\cosh(x + y^3)\,dy\,dx. $$

Finding the Limits

We will now address the problem of determining limits for a double integral from a geometric description of the region of integration. While MATLAB cannot do that for us, it can provide some guidance through its graphics and can also confirm that the limits we have chosen define the region we intended. For a first example, we will consider the bounded region between the curves y = 2x+2 and y = exp(x). We begin by plotting the two curves on the same axes. You may need to experiment with the interval to get a useful plot; it should be large enough to show the region of interest, but small enough so that the region of interest occupies most of the plot.

ezplot(y1,[-1,2]); hold on;
ezplot(y2,[-1,2]); hold off;
title('region bounded by y = 2x+2 and y = e^x')

We can see from the plot that, to define the bounded region between the two graphs, exp(x) should be the lower limit for y, and 2x+2 should be the upper limit. In order to find the limits for x, we need the values for which the two functions coincide. We can try to use solve to find them.

ans =
- lambertw(0, -1/2*exp(-1)) - 1

Hmm..., not clear what this means, but we can convert to numerical values:

lims =


That seems to be the lower limit, but there should have been an upper limit around 1.7. We can search for its numerical value using fzero:

lims(2) = fzero (@(t) subs(y1-y2, x, t), 1.7)
lims =

  -0.768039047013466   1.678346990016661

Now that we have found the limits, we can use the following M-file verticalRegion to visualize the region they define so that we can check that it is the same as the region we had in mind. This program plots the two curves in two different colors: red for the function entered first, and blue for the function entered second, and shades the region in between in cyan.

type verticalRegion
function out = verticalRegion(a, b, f, g)
%VERTICALREGION is a version for Matlab of the routine on page 157
%  of "Multivariable Calculus and Mathematica" for viewing the region
%  bounded by two curves. f and g must be names of functions;
%  a and b are the limits.  The function f is plotted in red and the
%  function g in blue, and lines are drawn connecting the two curves
%  to illustrate the region they bound. The function g must have a variable
%  in it, so if it's supposed to be a constant, try a fudge like
%  '0 + eps*x'.
%This guarantees these are floating point
% numbers and not just symbolic expressions like sqrt(2)
hold on
for xxx=a:(b-a)/20:b,
 Y=[fin(xxx), gin(xxx)];
 plot([xxx, xxx], Y, 'c');
title(['region between ',char(f),' and ',char(g)])
hold off


Evidently our limits define the right region. We can now use them to integrate any function we like over the region in question. To find the area of the region, for example, we integrate the function 1. Since our limits for x are numerical, a symbolic calculation is not of much use directly, so we use double to convert to a numerical answer.

area =


Of course, the area between two curves could just as easily have been evaluated as the integral of the difference between the two functions of x whose graphs define the region; that is to say the first integration in this case is so simple that one can write down the result as easily as the integral. However, we can also integrate any function we like over the same region, changing only the integrand. For instance:

newint =


The numerical evaluation in this case was performed using numerical routines borrowed from MuPAD, which takes care of symbolic calculations. MATLAB has its own double integrator, called dblquad. There is a complication in using dblquad; it does not accept variable limits. Thus there are two ways to use dblquad:

  1. expand the region of integration to a rectangle, and make the integrand = 0 outside of the region where one wants to integrate, or
  2. make a change of variables in the inner integral. Fortunately, it is always the same one. We set y=y1+u(y2-y1) so that as u varies from 0 to 1, y varies from y1 to y2, and then integrate over x and u instead of over x and y. Note that dy=(y2-y1)du. This gives the identity

$$ \int_{x_1}^{x_2} \!\int_{y_1}^{y_2} \,f(x,y)\,dy\,dx = $$

$$ \phantom{x}\qquad \int_{x_1}^{x_2}\! \int_0^1\,f(x,y_1 + u(y_2 - y_1))(y_2 - y_1)\,du\,dx. $$

dblquad demands a vectorized function (not symbolic expression) as an input, depending on vector-valued x and scalar-valued y. We implement both methods, and use format long to compare the results.

format long
newint =


Next, the method using integration over a rectangle bigger than the true region of integration. We have to multiply the original integrand g by something that is 0 outside the region of integration. But y<=y2 is 0 (false) if y>y2 and 1 (true) otherwise. Similarly, y>=y1 is 0 (false) if y<y1 and 1 (true) otherwise.

g = @(x,y) x.^2+y^2;
h = @(x,y) g(x, y).*(y<=2*x+2).*(y>=exp(x));
newint1 =


Finally the method using a change of variables:

g = @(x,y) x.^2+y.^2;
h1 = @(x,u) g(x, exp(x)+u*(2*x+2-exp(x))).*(2*x+2-exp(x));
newint2 =


The answers provided by all three methods are clearly in the same ballpark, and those produced by int and the change of variables method agree to eight significant digits. The method using "truncation" is less accurate, because it results in integrating a discontinuous function, whereas the usual algorithms for numerical integration work best with an integrand that is smooth.

Because of the syntactical complications of using dblquad on a region that is not rectangular, we have created M-files symint2 and numint2, which have identical calling syntax and implement, respectively, symbolic double integration using int, and numerical double integration using dblquad. The advantage of the identical syntax is that if symint2 fails, one can try numint2 instead with a minimum of editing. We illustrate them by repeating the two previous computations.

type symint2
function out=symint2(integrand,var1,varlim1,varlim2,var2,lim1,lim2)
%SYMINT2 Symbolically evaluate double integral. 
%   NUMINT2 is basically a "wrapper" for iterated application of INT.
%   The INNER integral should be written first.
%   Example:
%       syms x y
%       symint2(x,y,0,x,x,1,2) integrates
%       the function x over the trapezoid 1<x<2, 0<y<x,
%       and gives the symbolic output 7/3.
%   See also NUMINT2.
type numint2
function out=numint2(integrand,var1,varlim1,varlim2,var2,lim1,lim2)
%NUMINT2 Numerically evaluate double integral. 
%   NUMINT2 is basically a "wrapper" for DBLQUAD, to
%   get around the fact that the latter is only designed for integration
%   over a rectangle, and requires inline vectorized input.  The
%   INNER integral should be written first.
%   Example:
%       syms x y
%       numint2(x,y,0,x,x,1,2) integrates
%       the function x over the trapezoid 1<x<2, 0<y<x,
%       and gives the numerical output 2.3333.
%   See also SYMINT2.
syms uU
Warning: Explicit integral could not be found. 

ans =


ans =


Note: As this example illustrates, in some problems, dblquad and numint2 do not work that well, and spit out long lists of warning messages. In next week's lesson, we will discuss another package called nit, or "numerical integration toolbox", that usually corrects the problem.

Problem 2

Let R be the region bounded by the curves y=3*x^2 and y=|2*x+3|.

a) Determine the limits of integration for a double integral over R; confirm your limits by using verticalRegion.

b) Evaluate

$$ \int\!\int_R x\exp(-y^2)\,dA, $$

both using double and int (or double and symint2, and using dblquad (or numint2).

Integrating over Implicitly Defined Regions

Let us now turn to the process of integrating over a region bounded by a level curve of a function of two variables. Once again, the real issue is the determination of the limits of integration. We will consider the region bounded by the ellipse 3*x^2+4*y^2=37. Although it is not strictly necessary in this case, we will begin by plotting the ellipse.

ezplot(3*x^2+4*y^2-37, [-4, 4, -4, 4]), axis equal

We can see from the plot that this region is vertically simple. We obtain our y limits by solving the equation of the ellipse for y in terms of x.

ylims =
 -(37 - 3*x^2)^(1/2)/2
  (37 - 3*x^2)^(1/2)/2

The two solutions give the upper and lower limits for y; we can find the limits for x by setting their difference to 0 and solving for x.

y1 =
(37 - 3*x^2)^(1/2)/2
y2 =
-(37 - 3*x^2)^(1/2)/2
xlims =
x1 =
x2 =

We can confirm our limits, using verticalRegion. The first two arguments to verticalRegion must be numerical, rather than symbolic. Therefore, we must apply double to x1 and x2. verticalRegion(double(x1),double(x2),y1,y2)

We can now integrate any function we desire over the region bounded by the ellipse.

Problem 3


$$ \sqrt{x^2 + y^2 + 1} $$

a) Over the region bounded by the ellipse 3*x^2+4*y^2=36.

b) Over the region bounded by the curve x^4+y^4=20.

Polar Coordinates

We conclude this lesson with a discussion of integration in polar coordinates. We consider integrating the function exp(-x^2-y^2) over the disk bounded by the circle (x-1)^2 + y^2 = 1. We begin by expressing the function whose vanishing defines the boundary curve in polar coordinates.

syms r th
polarfun=simplify(subs((x-1)^2+y^2-1, [x,y],[r*cos(th),r*sin(th)]))
polarfun =
r^2 - 2*r*cos(th)

Next we solve the equation of the boundary curve for r in terms of theta, to obtain the limits for r.

rlims =

Since the disk in which we are interested lies entirely to the right of the x-axis, the appropriate limits for theta are -pi/2 and pi/2. Bearing in mind that we must include a factor of |r |when integrating in polar coordinates, we can now set up and evaluate an iterated integral in any of the usual ways. For example:

Warning: Explicit integral could not be found. 

ans =


Problem 4

Reevaluate the integral above using numint2.

Additional Problems

1) Evaluate

$$ \int\!\int_R \,x^2 + y^2\, dA,$$

where R is the triangle with vertices at (0,0), (0,2) and (3,0). Hint: begin by determining the equation of the oblique side of the triangle. Use verticalRegion to confirm your limits.

2) Evaluate

$$ \int\!\int_R \,x\sin(y^2)\, dA,$$

where R is the region bounded by the closed curve

$$x^4 + y^2 = 8. $$

3) Evaluate

$$ \int\!\int_R \,\sqrt{x^2 + y^2 +r}\, dA,$$

where R is the region bounded by the cardioid

$$ r = 2 + 2\cos\theta.$$

4) Compute the area of the region bounded by the two curves x^2 - y^2 = 1 and y = x^2 - 3*x + 1.

instructed to write integrals over planar domain

instructed to write integrals over planar domain Posted on 26-04-2014  Evaluating a multiple integral involves expressing it as an iterated integral, which can then be evaluated either symbolically or numerically. We begin by discussing the evaluation of iterated integrals. 5/10 1924


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