How to solve this double integration?

In summary, the conversation was about solving a double integration involving r and theta, with the rest being constants. The original equation had some mistakes in the denominator, which were pointed out and corrected by other members. The conversation also included discussions on how to properly format the equation using LaTeX and potential approaches for solving the integration.
  • #1
Red
19
0

Homework Statement


Hi Guys, I need help to solve this double integration. This integration is over r and theta. The rest are constant.

Homework Equations


∫[itex]^{R_{2}}_{r=0}[/itex] ∫[itex]^{\pi}_{\theta=0}[/itex] r[itex]^{2}[/itex] sin([itex]\theta[/itex]) dr d[itex]\theta[/itex] / ((D[itex]^{2}[/itex]+r[itex]^{2}[/itex] - 2rd cos([itex]\theta[/itex]))[itex]^{2}[/itex] - R[itex]_{1}[/itex]^2)[itex]^{3}[/itex], r from 0 to R[itex]_{2}[/itex], theta from 0 to pi.

The slash stand for division, I am not sure how to create a proper fraction here so I used slash.

The Attempt at a Solution


I have attempted to solve this problem by reducing it to s hollow shell, ie r is a constant, but the integratino seem to be the same for theta.

Thank you for your help!
 
Last edited:
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  • #2
I assume D is the same as d.
Are you sure about the integrand? It looks dimensionally wrong. R12 is a quadratic term being subtracted from a quartic term.
To get fractions in LaTex use \frac{}{}.
 
  • #3
Red said:
∫[itex]^{R_{2}}_{r=0}[/itex] ∫[itex]^{\pi}_{\theta=0}[/itex] r[itex]^{2}[/itex] sin([itex]\theta[/itex]) dr d[itex]\theta[/itex] / ((D[itex]^{2}[/itex]+r[itex]^{2}[/itex] - 2rd cos([itex]\theta[/itex]))[itex]^{2}[/itex] - R[itex]_{1}[/itex]^2)[itex]^{3}[/itex], r from 0 to R[itex]_{2}[/itex], theta from 0 to pi.
Oh messy - you used the equation editor right?
Here, let me help...

$$\int_{r=0}^{R_2} \int_{\theta=0}^{\pi}
\frac{ r^2 \sin\theta \;\text{d}r\text{d}\theta }{ \big( (D^2+r^2-2rD\cos\theta)^2-R_1^2\big)^3 }$$ ... this what you meant?

To see how I did that, use the "quote" button at the bottom of this post.
I have the same reservations about the denominator as haruspex... suspect the square about the cosine rule term is misplaced.

It would probably help to know what problem you are actually trying to solve here.
 
  • #4
Simon Bridge said:
Oh messy - you used the equation editor right?
Here, let me help...

$$\int_{r=0}^{R_2} \int_{\theta=0}^{\pi}
\frac{ r^2 \sin\theta \;\text{d}r\text{d}\theta }{ \big( (D^2+r^2-2rD\cos\theta)^2-R_1^2\big)^3 }$$ ... this what you meant?

To see how I did that, use the "quote" button at the bottom of this post.
I have the same reservations about the denominator as haruspex... suspect the square about the cosine rule term is misplaced.

It would probably help to know what problem you are actually trying to solve here.

Thank you Simon Bridge! It look much nicer now. I realized I made a mistake that you and haruspex pointed out, I have corrected the integration and it looks like this:

$$\int_{r=0}^{R_2} \int_{\theta=0}^{\pi}
\frac{ r^2 \sin\theta \;\text{d}r\text{d}\theta }{ \big( D^2+r^2-2rD\cos\theta-R_1^2\big)^3 }$$

I realized I can solve the theta part of this integration, however I do not know how to proceed on with the r part. This is what I get after solving the theta part:

$$\frac{ 1 }{ 4 D }\int_{r=0}^{R_2} \frac{ r }{ \big( (D-r)^2-R_1^2\big)^2 } - \frac{ r }{ \big( (D+r)^2-R_1^2\big)^2 } \text{d}r $$

Can you help me on this integration? Thank you!

I am using the Hamaker approach to derive the van der Waal potential between two sphere of radius R[itex]_{1}[/itex] and R[itex]_{2}[/itex], the distance between the center of the two sphere is D.
 
  • #5
I think you want a substitution like ##R_1\sec u = D-r## for the first one and ##R_1\sec u = D+r## for the second one.
But you may be able to just do it by partial fractions.
 
  • #6
Thank you for your help!
 
  • #7
Simon Bridge said:
To see how I did that, use the "quote" button at the bottom of this post.
You can also right click > Show math as > TeX commands.
 
  • #8
Thanks adjacent! I have just learn TeXstudio so I am kind of more familiar with this kind of text input now.
 

1. What is double integration?

Double integration is a mathematical technique used to find the area under a curve or the volume of a three-dimensional shape. It involves integrating a function over two variables, usually denoted as x and y, and is commonly used in physics, engineering, and economics.

2. How do I set up a double integration?

The first step in setting up a double integration is to identify the limits of integration, which are the boundaries within which you want to find the area or volume. These limits are usually given in the problem or can be determined by graphing the function. Then, you must determine the order of integration, which specifies which variable will be integrated first. This can be determined by looking at the function and the limits of integration.

3. What is the process for solving a double integration?

The process for solving a double integration involves performing two separate integrations, one after the other. The first integration is done with respect to one variable while holding the other variable constant. Then, the result of this integration is integrated again with respect to the second variable. This results in a numerical value, which represents the area or volume you are trying to find.

4. What are some common techniques for solving double integrations?

Some common techniques for solving double integrations include using the fundamental theorem of calculus, substitution, and integration by parts. It is important to choose the most appropriate technique based on the form of the function and the limits of integration to ensure an accurate solution.

5. How can I check if my double integration is correct?

One way to check if your double integration is correct is by using a graphing calculator or software to plot the function and the boundaries of the integration. The area or volume calculated through integration should match the shaded region on the graph. Another way is to use a known formula for the area or volume of a specific shape and compare it to your calculated value.

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