Finding a volume of a solid in polar coordinates using double integrals

 We continue with solving problems of volume of solids in polar coordinates using double integrals. Here is another example.

Example 2

Solution

Let's first express the equation of the region which is a disk in polar coordinates. We need to find θ and r.

The equation of the disk can be expressed as::

This is a set of circles with radiuses less than or equal to1. These circles are centered in (1,0). The center can be found by rewriting the left part of the above inequality as (x-1)² + (y-0)².  The angle θ varies from 0 to 2ℼ..

Expanding the square term in the equation (x-1)² + y² = 1, we have: 

By simplification, we get : 

By substituting x = cosθ and y = sinθ, we have:

Solving this equation, we find: 

The disk on the xy plane can be expressed on the following region as:

Let's express z in polar coordinates by substituting x and y in the equation of z. We obtain z = 4-r².

Let's calculate the volume. We have:

Let's calculate the inner integral:

=  

= 8cos²θ−4cos⁴θ

Let's integrate with respect to the outer integral:

Calculating this expression we have:

V = 5/2.2π +5/2sin4π-1/8sin8π- 0 = 5𝛑/2

Volume of a solid in polar coordinates using double integrals

 As in rectangular coordinates, if a solid S is bounded by the surfaces z = f(r, θ) and the surfaces r = a, r = b, θ = ɑ, θ = β, then the volume V of S can be found by integration using the formula:

If the base of the solid can be described as D = {(r, θ), ɑ ≤ θ ≤ β h₁(θ) ≤ θ ≤ h₂(θ)}, the volume V becomes:

Example

Solution

By the method of double integration, the volume is the iterated integral of the form:

Evaluating a Double Integral over General Polar Regions of Integration

 Goal: Evaluate a double integral over a general polar of integration

To evaluate the double integral of a continuous function over polar general regions using iterated integrals, we consider the types I and II regions used previously in the calculation of double integrals over general regions in rectangular coordinates. We write polar equations as r = f(θ) rather than θ = f(r). The general polar region is defined by:

 

 

The figure above represents the general polar region between ɑ ≤θ ≤ β and h₁ (θ) ≤θ ≤ h₂(θ)

Theorem

Example

Solution

Practice

Evaluating a double integral by converting from rectangular coordinates

 Goal: Evaluate a double integral by converting from rectangular coordinates to polar coordinates 

In the last post, we showed how to convert a double integral from rectangular coordinates to polar coordinates. The first step consists in sketching the region.

Example

Solution

The region is the set of circles comprised between the circle of radius 1 and the circle of radius 2. Let's start by sketching the region. The inferior limit of the set of circles is represented in red. It has 1 as radius. The superior limit is represented in blue. Its radius is 2. Note that x ≤ 0, therefore the region is located in the negative part of the plane of coordinates.

In polar coordinates, the radius of the region represented by the set of circles comprised between the red one and the blue one varies from 1 to 2. The θ angle varies from ℼ/2 to 3ℼ/2. Therefore, R is an annular region that can be represented by:

Practice

Double Integrals over a polar rectangular region

 Goal: Evaluate a double integral over a polar rectangular region

Definition

The double integral of a function f(r, θ) over a polar rectangular plane in the θ plane is defined as 

As in double integrals over a rectangular region, the double integral over a polar rectangular region can be expressed as in iterated integral in polar coordinates.

In polar coordinates dA is replaced by rdrdθ. The double integral of f(x,y) in rectangular coordinates can be expressed in polar coordinates by substitution. This is done by substituting x by rcosθ, y by rsinθ and dA by rdrdθ.

The properties for double integrals over rectangular coordinates apply also to the double integrals over polar coordinates.

Example 1

Solution

As we can see in the figure below, r =1 and r = 3 represent circles of radius r =1 and r =3 and 0≤ θ ≤ℼ covers the entire top half of the plane.

Example 2

Solution

The figure is similar to that in example 1 but with outer radius 2. Do it by yourself.

Practice