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27.12.2012
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Advanced Caluclus Unit XIV
integration of functions of two or more independent variables - Part iV
Objectives
From this session a learner is expected to achieve the following
- Familiarize with evaluating double integrals in polar form.
- Learn to find the area of a closed and bounded region in the polar coordinate plane.
- Study the method of changing Cartesian integrals into polar integrals.
- Familiarize with evaluating triple integrals in cylindrical coordinates
- Familiarize with evaluating triple integrals in spherical coordinates
Contents
1. Introduction
2. Double integrals in Polar Form
3. The area of a closed and bounded region R in the polar coordinate plane
4. Changing Cartesian Integrals into Polar Integrals
5. Triple Integrals in Cylindrical Coordinates
6. Triple Integrals in Spherical Coordinates
1. Introduction
Double integrals are sometimes easier to evaluate if we change from Cartesian coordinates to polar coordinates. Similarly, triple integrals can sometimes be easily evaluated if we change to cylindrical or spherical coordinates. In this session we describe double integrals in polar form, triple integrals in cylindrical coordinates and triple Integrals in spherical coordinates.
2. Double integrals in Polar Form
In the coming example we represent a given region by means of inequalities involving polar variables.
Example 1 By means of system of inequalities, represent the region R that lies inside the cardioid and outside the circle .
Solution
The given regionR isthe shaded portion in Fig. 1. We observe that a typical ray from the origin enters the region R where and leaves where. Hence The rays from the origin that intersect the region R run from Hence Thus the region is
Now for stating the formula for double integral in polar coordinates, suppose that a function is defined over a region R that is bounded by the rays and and by the continuous curves and Then the double integral of over the region R is given by
Example 2Find where the region of integration R is given by the system of inequalities and
Solution
Example3 Evaluate over the upper half of the circle .
Solution
Fig. 2
Referring Fig. 2, the polar equation of theupper half of the given circle is
and.
Hence the required integral is
, using reduction formula.
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3. The area of a closed and bounded region R in the polar coordinate plane
The area of a closed and bounded region R in the polar coordinate plane is given by
Remark If is the constant function whose value is 1, then the integral of over R is the area of R. Hence in view of Examples 1 and 2, the area of the region that lies inside the cardioid and outside the circle is
Example 4 Find the area enclosed by the lemniscate.
Solution
Fig 3
A rough sketch of the given lemniscate is given in Fig. 3. We note that the required area is four times the shaded region in the first-quadrant.
In the shaded portion, a typical ray from the origin (with ) leaves the shaded region where . Hence Also, in the shaded region varies from 0 to . Hence the the system of inequalities that represent the shaded portion is
Hence, the required area is
= 4.
4. Changing Cartesian Integrals into Polar Integrals
For changing a Cartesian integral into a polar integral, we proceed as follows:
- In the Cartesian integral substitute and and replace by .
- Supply polar limits of integration for the boundary of R.
The Cartesian integral then becomes
whereG denotes the region of integration in polar coordinates.
Example 5 Evaluate
whereR is the semicircular region bounded by the x-axis and the curve.
Solution
The shaded portion in Fig. 4 is the given region R. In Cartesian coordinates, the integral in question is a nonelementary integral and there is no direct way to integrate with respect to either x or y. Polar coordinates help us. Substituting and replacing by and using the system of inequalities in polar variables for the given region, the Cartesianintegral becomes
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5. Triple Integrals in Cylindrical Coordinates
For the evaluation of triple integral when a calculation involves a cylinder, we can often simplify our work by using cylindrical coordinates.
The volume element for subdividing a region in space with cylindrical coordinates is
Triple integrals in cylindrical coordinates are then evaluated as iterated integrals, as in the following example.
Example 6 Find the limits of integration in cylindrical coordinates for integrating a function over the region D bounded below by the plane, laterally by the circular cylinder, and above by the paraboloid.
Fig 5
Solution
A rough sketch is given in Fig. 5. The base of D is also the region’s projection R on the xy-plane. The boundary of R is the circle. Its polar coordinate equation is obtained as follows:
.
The z-limits of integration: A line M through a typical point in R parallel to the z-axis enters D at and leaves at . (Here we have used the fact, with and)
The r-limits of integration: A ray L through from the origin enters R at and leaves at .
The limits of integration: As L sweeps across R, the angle it makes with the positive x-axis runs from to.
Hence the region of integration is
Also the integral is
.
Example7Find the centroid of the solid (with density given by enclosed by the cylinder, bounded above by the paraboloid and below by the xy-plane.
Solution
First we note that with and
A rough sketch of the solid, bounded above by the paraboloid and below by the plane is given in Fig.6. Its base R is the disk in the xy-plane , that isR is the disk in the xy-plane.
The solid’s centroid is with
Fig6
where is the moment about the xy-plane, is the moment about the yz-plane, and so on.
But in the given problem, the solid’s centroidlies on its axis of symmetry, here the z-axis. This makes. Hence it remains to find. For this purpose we find the first moment using the formula
and mass using the formula
where is the density of the solid.
Now to find the limits of integration for the mass and moment integrals, we proceed as follows:
The z-limits: A line M through a typical point in the base parallel to the z-axis enters the solid at and leaves at .
The r- limits: A ray L through from the origin enters R at and leaves at .
The limits: As L sweeps over the base, the angle it makes with the positive x-axis runs from to. Also, since the value of is
.
Also, the mass is given by
Therefore,
and the centroid is.
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6. Triple Integrals in Spherical Coordinates
For the evaluation of triple integral when a calculation involves a sphere, we can often simplify our work by using spherical coordinates.
The volume element in spherical coordinates is the volume of a spherical wedge defined by the differentials . The wedge is approximately a rectangular box with one side a circular arc of length another side a circular arc of length, and thickness. Therefore the volume element in spherical coordinates is
,
and triple integrals take the form
In particular, the volume of a region D in space is given by (with )
To evaluate these integrals, we usually integrate first with respect to . The procedure for finding the limits of integration is illustrated in the following example.
Example 8 Find the volume of the upper region D cut from the solid sphere by the cone.
Solution
A rough of sketch of the region D and its projection R on the xy-plane are given in Fig. 7.
The volume is determined using the formula
To find the limits of integration for evaluating the integral, we proceed as follows:
(limits of integration): We consider a ray M from the origin making an angle with the positive z-axis. We also considerL, the projection of M on the xy-plane, along with the angle that L makes with the positive x-axis. Ray M enters D at and leaves at .
(limits of integration): The cone makes an angle of with the positive -axis. For any given, the angle can run from to.
(The limits of integration): The ray L sweeps over R as runs from 0 to.
Hence the region of integration is
and the volume is
.
Summary
In this session we have seen that double integrals are sometimes easier to evaluate if we change from Cartesian coordinates to polar coordinates. Also we have seen that, triple integrals can sometimes be easily evaluated if we change to cylindrical or spherical coordinates.
Assignments
1.Change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral.
2. Evaluate .
3. Find the area of the region cut from the first quadrant by the curve .
4. Evaluate the cylinder coordinate integral .
5. Evaluate the spherical coordinate integral .
References
- Gorakh Prasad, Integral Calculus, Pothishala Pvt. Ltd., Allahabad.
- N. Piskunov, Differential and Integral Calculus, Peace Publishers, Moscow.
- Shanti Narayan, A course of Mathematical Analysis, S. Chand and Company, New Delhi.
Quiz
1. The double integral of over theregion R that is bounded by the rays and and by the continuous curves and is given by ______
(a)
(b)
(c)
(d) none of the above
Ans. (c)
2. The area A of a closed and bounded region R in the polar coordinate plane is given by ______
(a)
(b)
(c)
(d)
Ans. (b)
3. By changing to polar coordinates, the Cartesian integral becomes ______
(a)
(b)
(c)
(d)
Ans. (d)
FAQ
1. What is the procedure for changing Cartesian integral to polar integral?
Answer: For changing a Cartesian integral into a polar integral, we proceed as follows:
- In the Cartesian integral substitute and and replace by .
- Supply polar limits of integration for the boundary of R.
The Cartesian integral then becomes
whereG denotes the region of integration in polar coordinates.
2. What is the procedure for evaluating triple integral in cylindrical Coordinates.
Answer: For the evaluation of triple integral when a calculation involves a cylinder, we can often simplify our work by using cylindrical coordinates. The volume element for subdividing a region in space with cylindrical coordinates is
Triple integrals in cylindrical coordinates are then evaluated as iterated integrals.
Glossary
integral sum of a function in a region: Let be a continuous function of two variables and defined on a region R bounded by a closed curve C . Let the region R be divided, in an arbitrary manner, into n sub-regions . So as not to inroduce new symbols we will denote by both the subregions and their areas. In each subregion take a point (Fig. 2); we will then have n points:
.
The sum
is an integral sum of the function in the region R.
Double integral of a function in a region: If a function is continuous in a closed region R, then the sequence of integral sums has a limit if the maximum diameter of the subregions approaches zero as This limit is the same for any sequence of integral sums, that is, it is independent either of the way R is partitioned into subregions or of the choice of the point inside a subregion. This limit is called the double integral of over the region R, and is denoted by
Region of integration: The region R under consideration of the double integral is called the region of integration.
Regular three-dimensional domain: A subset V of a three dimensional XYZ space is said to be a regular three-dimensional domain (or regular three-dimensional region) if
(i)every straight line parallel to z axis and drawn through the interior (i.e., not lying on the boundary S) point of V cuts the surface S at two points
(ii)entire V is projected on the xy-plane into a regular two-dimensional region R.