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Double Integral Substitution

  • Thread starter killpoppop
  • Start date
  • #1
13
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Homework Statement



Evaluate the integral.

1|0 s|0 ( t . sqrt ( t2 + s2 ) dt ds


I hope the way I've written it makes some sort of sense.

The Attempt at a Solution



After getting my head around changing the order of integration I get hit with this question and for some reason am totally stumped.

First idea was to switch the order to leave you differentiating with respect to s first.
Which means your just integrating a fairly simple function?

Then instead of following this through my brain just kept saying substitution, substitution.

Using the original integral given.
Setting: u = t2+s2
du/2 = t.dt

Then substitute in accordingly. But I was confused by how you change the interval values.
With single integration your simply left with something like u = x + 1
I looked around and found to make a substitution you need to create two variables say and u and a v. This got me interested but also slightly more confused.

Writing this out has cleared my head and lead me to believe that the first method could work and I will try it out now.

What I would appreciate is a method on how to solve the integral (if both of mine are wrong), but now mainly an explanation on substitution with two integrals.

Thanks!
 
Last edited:

Answers and Replies

  • #2
13
0
Using my first method gets very messy. And doesn't work. After the first integration your left with an integration by parts which just gets messier the more you do it. Very stuck.
 
  • #3
33,507
5,192

Homework Statement



Evaluate the integral.

1|0 s|0 ( t . sqrt ( t2 + s2 ) dt ds


I hope the way I've written it makes some sort of sense.

The Attempt at a Solution



After getting my head around changing the order of integration I get hit with this question and for some reason am totally stumped.

First idea was to switch the order to leave you differentiating with respect to s first.
Which means your just integrating a fairly simple function?

Then instead of following this through my brain just kept saying substitution, substitution.

Using the original integral given.
Setting: u = t2+y
du/2 = t.dt

Then substitute in accordingly. But I was confused by how you change the interval values.
With single integration your simply left with something like u = x + 1
I looked around and found to make a substitution you need to create two variables say and u and a v. This got me interested but also slightly more confused.

Writing this out has cleared my head and lead me to believe that the first method could work and I will try it out now.

What I would appreciate is a method on how to solve the integral (if both of mine are wrong), but now mainly an explanation on substitution with two integrals.

Thanks!
Here's your integral. You can double-click it to see my LaTeX script.
[tex]\int_{s = 0}^1 \int_{t = 0}^s t \sqrt{t^2 + s^2} dt~ds[/tex]

Try this substitution first:
u = t2 + s2
du = 2tdt
s is a constant as far as t is concerned.
 
Last edited:
  • #4
13
0
Im just a little confused about what the interval would change to? Would this be right for the inner intergral?

Intervals are being changed using u = t2 + s2
Where t = s and t = 0


[tex]
1/2 \int_{u = s^2}^{u=2s^2} \sqrt{u} du
[/tex]
 
  • #5
33,507
5,192
Im just a little confused about what the interval would change to? Would this be right for the inner intergral?
That's integral - no such word as "intergral."
Intervals are being changed using u = t2 + s2
Where t = s and t = 0


[tex]
1/2 \int_{u = s^2}^{u=2s^2} \sqrt{u} du
[/tex]
You don't need to change the limits of integration, as long as you undo your substitution before evaluating the antiderivative at the limits of integration. I find it easier to keep track of things in multiple integrals if I include the variable in the lower limit of integration.

If you need to change them, the substitution is u = t2 + s2, so when t = 0, u = s2, and when t = s, u = 2s2, which is what you have.

Edit: I didn't notice until now, but you are missing the outside integral. Don't forget it.
 
  • #6
13
0
Ok cheers! Yeah couldn't be bothered to put up the outside INTEGRAL =]
Need to get the first one sorted first. Back to the pen and paper.
 
  • #7
13
0
1/12?
 

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