Integral with different variables

In summary, the conversation discusses doing an integral with respect to r2 multiplied by an integral with respect to theta2 and phi2. The term under the square root confuses one person, who asks if it can be integrated with either the r2 or theta2 part constant. The other person suggests using a change of variable and references Fubini's Theorem for reversing the order of integration. The first person asks how to check if the function is continuous and the other person suggests trying integration by theta first. The order of integration does not matter for this exercise.
  • #1
Viona
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Homework Statement
Integral with different variables
Relevant Equations
Integral with variables
I want to do this integral in the picture:
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where r1 and a are constants. I know I can integrate each part separately. There will be an integral with respect to r2 multiplied by integral with respect to theta2 and the last one with respect to phi2. But the term under square root confuses me. Can I integrate it with r2 part considering theta2 constant or integate it with theta2 part considering r2 constant?
 
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  • #3
Change of variable ##t=\cos\theta## would make
[tex]\int_{-1}^1 \frac{dt}{\sqrt{A-Bt}}[/tex]
Does it make sense ?
 
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  • #4
anuttarasammyak said:
Change of variable ##t=\cos\theta## would make
[tex]\int_{-1}^1 \frac{dt}{\sqrt{A-Bt}}[/tex]
Does it make sense ?
Yes. it seems good. But I want to ask: for this type of integral does the order matter? I should start by integrating w.r.t. r2 first or it is optional?
 
  • #5
scottdave said:
Yes, see this for more clarification - https://tutorial.math.lamar.edu/Classes/CalcIII/IteratedIntegrals.aspx

Check out the example problems (with solutions)
That was helpful. Thanks. I learned that this type of integrals are called iterated integrals. For this type of integral the order is important particularly when the integrand is not continuous on the domain of integration. Then I found a theorem called Fubini’s Theorem. I understood that we can reverse the order if the integrand is continuous on the domain of integration. Now I am wondering how to check if the function is continous or not?
 
  • #6
Viona said:
But I want to ask: for this type of integral does the order matter? I should start by integrating w.r.t. r2 first or it is optional?
Why don' you try integration by ##\theta## at first. The order should not matter for this exercise at least.
 
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FAQ: Integral with different variables

What is an integral with different variables?

An integral with different variables refers to an integral where the variable of integration differs from the variable used in the function being integrated. This often happens in the context of substitution, where a new variable is introduced to simplify the integral.

How do you perform a variable substitution in an integral?

To perform a variable substitution in an integral, you replace the original variable with a new one, usually to simplify the integration process. You also need to change the differential accordingly. For example, if you let \( u = g(x) \), then \( du = g'(x) dx \). The integral is then rewritten in terms of \( u \) and \( du \).

What is the purpose of using different variables in integration?

The purpose of using different variables in integration is to simplify the integral or to make it possible to integrate a function that is otherwise difficult to handle. Substitution can transform a complex integral into a simpler one, making it easier to evaluate.

Can you change the variable of integration in a definite integral?

Yes, you can change the variable of integration in a definite integral. When doing so, you must also change the limits of integration to correspond to the new variable. This ensures that the integral evaluates to the correct value. For example, if you substitute \( u = g(x) \), the limits of integration change to \( u(a) \) and \( u(b) \) if the original limits were \( a \) and \( b \).

What is an example of an integral with a variable substitution?

An example of an integral with a variable substitution is the integral of \( \int e^{x^2} 2x \, dx \). By letting \( u = x^2 \), we get \( du = 2x \, dx \). The integral then becomes \( \int e^u \, du \), which is much simpler to evaluate. The result is \( e^u + C \), or \( e^{x^2} + C \) when reverting back to the original variable.

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