What is the solution to a multiple integral problem using polar coordinates?

In summary: Yes, that was a bit of a problem for me too... the easiest way I see would be something like this:First substitute x = r^2 \cos\theta \sin\theta and write the integral in the form\int_0^{\cdots} \frac{1}{x^2 + a x + b} \, dx = \int_0^{\cdots} \frac{1}{(x - x_1)(x - x_2)} \, dxthen note that this integral is equal to\frac{\log(x + a) - \log(x + b)}{b - a}.In principle it looks easy, but
  • #1
kidsmoker
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0

Homework Statement



By transforming to polar coordinates, show that

[tex]I = \int\int_{T}\frac{1}{(1+x^2)(1+y^2)}dxdy = \int^{\pi/4}_{0}\frac{log(\sqrt{2}cos(\theta))}{cos(2\theta)}d\theta[/tex]

where T is the triangle with successive vertices (0,0),(1,0),(1,1).


Homework Equations



[tex]I = \int\int_{K} f(x,y)dxdy = \int\int_{K'} g(u,v)*J*dudv[/tex]

where J is the Jacobian.

The Attempt at a Solution



The Jacobian is r, as always with a transformation to polar coordinates, so we get that

[tex]I = \int^{1}_{0}\int^{x}_{0}\frac{1}{(1+x^2)(1+y^2)}dydx = \int^{\pi/4}_{0}\int^{\sqrt2}_{1}\frac{r}{(1+r^2cos^2(\theta))(1+r^2sin^2(\theta))}drd\theta[/tex]

Firstly, is this correct? Secondly, if it is, could you give me a hint as to how to solve it to get the answer given? The obvious thing seems to be to split it into partial fractions, but I did try this once and didn't seem to get anywhere?

Thanks!
 
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  • #2
The transformation of coordinates looks correct, but you may want to check your integration boundaries. The triangle is "standing on it's tip" so for x = 0 you want to integrate y from 0 to 1, etc. In polar coordinates then, r doesn't run from 1 to [itex]\sqrt{2}[/itex]: it always starts at 0 but it only runs to [itex]\sqrt{2}[/itex] for [itex]\theta = \pi / 4[/itex].

It may help if you draw the triangle.
 
  • #3
Sorry I've just realized i got the coordinates of one of the vertices of the triangle muddled up. It should be (0,0),(1,0),(1,1) not (0,0),(0,1),(1,1) which i originally wrote.

The integration boundaries in polar coordinates do confuse me. With the correct vertices, surely r=1 at [itex]\theta = 0[/itex] and [itex]r=\sqrt{2}[/itex] at [itex]\theta = \pi / 4[/itex]. So these are my boundaries? Or am I misunderstanding it?

Thanks.
 
  • #4
kidsmoker said:
The integration boundaries in polar coordinates do confuse me. With the correct vertices, surely r=1 at [itex]\theta = 0[/itex] and [itex]r=\sqrt{2}[/itex] at [itex]\theta = \pi / 4[/itex]. So these are my boundaries?

Yes, you are correct about those values. But let's suppose that the theta integration is the outer one and the r-integration the inner one. That means that for every value of theta you will have to do the r-integral. So if [itex]\theta = 0[/itex] then r should run from 0 to 1, and if [itex]\theta = \pi / 4[/itex] then r should run from 0 to [itex]r=\sqrt{2}[/itex]. If [itex]\theta = \pi / 8[/itex] then r runs from 0 to ... ?

It is precisely analagous to the expression
[tex]I = \int^{1}_{0}\int^{x}_{0}\frac{1}{(1+x^2)(1+y^2)} dy dx[/tex]
you gave: when you do the inner integration over y, your x is fixed. You can pretend that you are walking along the x-axis, and for each point there doing the y-integral. However the upper boundary for y depends on what the "current" value of x is, so you have to integrate y from 0 to x and not from 0 to 1, despite the fact that y runs to 0 for x = 0 and y runs all the way to 1 for x = 1.

Do you understand?
 
  • #5
Ah okay... I think i see now. So I should have wrote

[tex]I = \int^{1}_{0}\int^{x}_{0}\frac{1}{(1+x^2)(1+y^2)}dy dx = \int^{\pi/4}_{0}\int^{\sqrt2}_{0}\frac{r}{(1+r^2cos^2(\theta ))(1+r^2sin^2(\theta))}drd\theta[/tex] ... ?
 
  • #6
No, the point I was trying to make is that instead of [itex]\sqrt{2}[/itex], you should have something which depends on [itex]\theta[/itex] - just like you have an x as integration boundary in the first form.
 
  • #7
Ahh, i see! Right so it should be

[tex]I = \int^{1}_{0}\int^{x}_{0}\frac{1}{(1+x^2)(1+y^2)}dy dx = \int^{\pi/4}_{0}\int^{1/cos\theta}_{0}\frac{r}{(1+r^2cos^2(\theta ))(1+r^2sin^2(\theta))}drd\theta[/tex]

I'm still left unable to solve the integral though :-( Any tips for the best method to use?

Thanks!
 
  • #8
kidsmoker said:
Ahh, i see! Right so it should be

[tex]I = \int^{1}_{0}\int^{x}_{0}\frac{1}{(1+x^2)(1+y^2)}dy dx = \int^{\pi/4}_{0}\int^{1/cos\theta}_{0}\frac{r}{(1+r^2cos^2(\theta ))(1+r^2sin^2(\theta))}drd\theta[/tex]
Very good, that's what I had in mind too :smile:

kidsmoker said:
I'm still left unable to solve the integral though :-( Any tips for the best method to use?

Thanks!

Yes, that was a bit of a problem for me too... the easiest way I see would be something like this:
First substitute [tex]x = r^2 \cos\theta \sin\theta[/tex] and write the integral in the form
[tex]\int_0^{\cdots} \frac{1}{x^2 + a x + b} \, dx = \int_0^{\cdots} \frac{1}{(x - x_1)(x - x_2)} \, dx[/tex]
then note that this integral is equal to
[tex]\frac{\log(x + a) - \log(x + b)}{b - a}[/tex].
In principle it looks easy, but prepare your trig identities... you will need them :sad:
 
  • #9
Ah okay cool. I managed to do it using the substitution t=r^2 and then splitting it into partial fractions. Was a bit long-winded but got the right answer.

Thanks!
 

FAQ: What is the solution to a multiple integral problem using polar coordinates?

What is a multiple integral problem?

A multiple integral problem is a mathematical concept that involves calculating the area or volume of a multidimensional object or region. It is an extension of single-variable integration and involves evaluating a function over multiple variables.

What are the different types of multiple integrals?

There are two main types of multiple integrals: double integrals and triple integrals. Double integrals are used to calculate the area under a surface in a two-dimensional space, while triple integrals are used to calculate the volume under a surface in a three-dimensional space.

How do you solve a multiple integral problem?

To solve a multiple integral problem, you first need to set up the integral by determining the limits of integration and the integrand (the function being integrated). Next, you can use various integration techniques such as substitution, integration by parts, or partial fractions to evaluate the integral. Finally, you can solve the resulting expression to find the area or volume of the given region.

What are the applications of multiple integrals?

Multiple integrals have various applications in physics, engineering, and other fields of science. They are used to calculate the mass, center of mass, and moments of inertia of three-dimensional objects. They are also used in probability and statistics to calculate the joint probability of multiple events.

Are there any limitations to multiple integrals?

Multiple integrals have some limitations, such as being difficult to evaluate for complex functions and regions. They also require a good understanding of single-variable integration and multivariable calculus concepts. Additionally, some regions may have non-standard boundaries that make it challenging to set up the integral properly.

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