Integral of (x^2+y^2)^(-3/2)dy

  • Thread starter FlyingButtress
  • Start date
  • Tags
    Integral
In summary, the conversation discusses a difficult integral problem in a Physics book that could not be solved using traditional methods. The problem involves an integral of (x^2+y^2)^(-3/2)dy and the attempt at a solution included trying u-sub and trig sub, but both were unsuccessful. The conversation also mentions using WolframAlpha to find a solution, but only the end solution is available and not the steps. Eventually, using trigonometric identities, the proper solution of y/[(x^2)*√(x^2+y^2)]+constant is obtained.
  • #1
FlyingButtress
6
0

Homework Statement



Hello, I came across an unusual integral problem in my Physics book that I could not solve. The book simply skips the work and says to check the integral tables, but I wasn't able to find one that satisfied this problem.

Homework Equations



This is the problem: integral of (x^2+y^2)^(-3/2)dy

The Attempt at a Solution



Tried u-sub, but the lack of y makes it impossible. Couldn't do trig sub because of the negative exponent.

When I plugged it into WolframAlpha, it gave me a tidy solution, but the solutions are unavailable even in their app. It loads for a bit and shows "solution unavailable."

The proper solution should be y/[(x^2)*√(x^2+y^2)]+constant.

Here is the actual URL:
http://www.wolframalpha.com/input/?i=integral+of+(x^2+y^2)^(-3/2)dy&incTime=true

This is the second time coming across this integral in this chapter and I cannot solve it for the life of me. Any insights would be greatly appreciated. Thanks!
 
Physics news on Phys.org
  • #2
The x can be regarded as a constant, right? Is it part of a double integral? Can you use IBP?
 
Last edited:
  • #3
Yes, x could be regarded as a constant, and no, not a part of a double integral. The physics parts of the problem were constants, so I just uploaded the core variables.

IBP didn't work for me because when I used tabular integration because the derivative and integral of each part would continue indefinitely if that makes sense.

Also if it helps, I worked through my problem using this guide I found:
http://faculty.wwu.edu/vawter/PhysicsNet/Topics/ElectricForce/LineChargeDer.html

I only switched the x and y because my problem was a horizontal version of this example.
 
Last edited by a moderator:
  • #4
I'm not sure what you mean by trig not working due to the negative exponent. I took a shot at it and got the same result as wolfram alpha. Try to take [itex] y = x tan \theta [/itex]. so the integral becomes:

[itex] \int \frac{1}{(x^{2} + y^{2})^{\frac{3}{2}}} dx = \int \frac{sec^{2} \theta d\theta}{(x^{2} + x^{2}tan^{2} \theta)^{\frac{3}{2}}} [/itex]

With trigonometric identities the solution should be easy.

Edit: wooah that LaTeX screw up, going to fix it.
 
  • #5
Yes trigoniometrics is the way to go here.
 
  • #6
Wow, that was easy. I don't know how that didn't spring to mind; I really need to brush up on my math haha.

Thank you!
 
  • #7
FlyingButtress said:

Homework Statement



Hello, I came across an unusual integral problem in my Physics book that I could not solve. The book simply skips the work and says to check the integral tables, but I wasn't able to find one that satisfied this problem.

Homework Equations



This is the problem: integral of (x^2+y^2)^(-3/2)dy

The Attempt at a Solution



Tried u-sub, but the lack of y makes it impossible. Couldn't do trig sub because of the negative exponent.

When I plugged it into WolframAlpha, it gave me a tidy solution, but the solutions are unavailable even in their app. It loads for a bit and shows "solution unavailable."

The proper solution should be y/[(x^2)*√(x^2+y^2)]+constant.

Here is the actual URL:
http://www.wolframalpha.com/input/?i=integral+of+(x^2+y^2)^(-3/2)dy&incTime=true

This is the second time coming across this integral in this chapter and I cannot solve it for the life of me. Any insights would be greatly appreciated. Thanks!

What do you mean that in Wolfram it gives a tidy solution, but the solutions are unavailable? When I click on your link it gives a solution.
 
  • #8
Ray Vickson said:
What do you mean that in Wolfram it gives a tidy solution, but the solutions are unavailable? When I click on your link it gives a solution.

Oops, I left out a word there. Meant to say "solution steps." Usually in their app it shows the procedure of obtaining the solution. By "tidy" I meant that the end solution looked elegant and nothing like what I had on my paper at the time...
 

Related to Integral of (x^2+y^2)^(-3/2)dy

1. What is the formula for finding the integral of (x^2+y^2)^(-3/2)dy?

The formula for finding the integral of (x^2+y^2)^(-3/2)dy is ∫ (x^2+y^2)^(-3/2)dy = -1/√(x^2+y^2) + C, where C is the constant of integration.

2. What is the domain of the function (x^2+y^2)^(-3/2)dy?

The domain of the function (x^2+y^2)^(-3/2)dy is all real numbers except for (0,0) since the function is undefined at that point.

3. How do you solve for the integral of (x^2+y^2)^(-3/2)dy?

To solve for the integral of (x^2+y^2)^(-3/2)dy, you can use the substitution method by letting u = x^2+y^2 and then using the power rule for integration, followed by substituting back in for u.

4. What is the significance of the constant of integration in the integral of (x^2+y^2)^(-3/2)dy?

The constant of integration represents the family of curves that can be obtained from the integral. It is added to account for any additional solutions that may have been lost during the integration process.

5. Can the integral of (x^2+y^2)^(-3/2)dy be evaluated using definite integration?

Yes, the integral of (x^2+y^2)^(-3/2)dy can be evaluated using definite integration by substituting in the limits of integration and then solving for the resulting expression.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
405
  • Calculus and Beyond Homework Help
Replies
4
Views
341
  • Calculus and Beyond Homework Help
Replies
7
Views
897
  • Calculus and Beyond Homework Help
Replies
2
Views
335
  • Calculus and Beyond Homework Help
Replies
20
Views
765
  • Calculus and Beyond Homework Help
Replies
5
Views
734
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
990
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
21
Views
2K
Back
Top