# Triple Integral and where to start

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• Dec 6th 2008, 10:30 PM
markmil2002
Triple Integral and where to start
I'm not looking for an answer, more or less where to start.

Here is the integral:

6/(1 + 48z - z^3) dzdydx

I can't tell where to start with this one besides pulling the 6 out of the numerator as a constant. I looked through all sorts of integration thechniques and I found no luck.
• Dec 6th 2008, 10:38 PM
robeuler
Quote:

Originally Posted by markmil2002
I'm not looking for an answer, more or less where to start.

Here is the integral:

6/(1 + 48z - z^3) dzdydx

I can't tell where to start with this one besides pulling the 6 out of the numerator as a constant. I looked through all sorts of integration thechniques and I found no luck.

Taking the 6 out is a good first start. One way of approaching this problem is partial fractions. You'll have 3 fractions left which you can integrate over all three variables one by one, then add the results back together.
• Dec 6th 2008, 10:43 PM
markmil2002
Look at the denominator. You can't just split the denominator into 3 parts. The whole denominator for that fraction is (1+48z-z^3) The only part of the numerator is the 6.
• Dec 6th 2008, 10:55 PM
robeuler
Quote:

Originally Posted by markmil2002
Look at the denominator. You can't just split the denominator into 3 parts. The whole denominator for that fraction is (1+48z-z^3) The only part of the numerator is the 6.

It isn't trivial to do, but by partial fractions decomposition you can. I am not saying you will get 1/1 +1/48z - 1/z^3. This method is usually covered in a calculus 2 course.

Partial fraction - Wikipedia, the free encyclopedia
• Dec 7th 2008, 12:54 AM
mr fantastic
Quote:

Originally Posted by markmil2002
I'm not looking for an answer, more or less where to start.

Here is the integral:

6/(1 + 48z - z^3) dzdydx

I can't tell where to start with this one besides pulling the 6 out of the numerator as a constant. I looked through all sorts of integration thechniques and I found no luck.

You haven't given any integral terminals. Depending on what they are things might be a lot simpler ....
• Dec 7th 2008, 05:10 AM
markmil2002
Well, I didn't give any because I wanted to know how to start. I don't see a way to split it up for partial fractions though. Especially since it's a 3rd degree, 1st degree, and zero degree. The quardratic equation doesn't work, and I don't see a way to make a perfect square either.
• Dec 7th 2008, 05:18 AM
mr fantastic
Quote:

Originally Posted by markmil2002
Well, I didn't give any because I wanted to know how to start. I don't see a way to split it up for partial fractions though. Especially since it's a 3rd degree, 1st degree, and zero degree. The quardratic equation doesn't work, and I don't see a way to make a perfect square either.

Let me blunter than my previous post:

Your question cannot be answered properly unless you also give the integral terminals. Perhaps you should review some examples from your textbook or class notes to see why that's the case.

Here's an example of a double integral that illustrates my point:

Find $\int \int \frac{\sin y}{y} \, dy \, dx$.

Find $\int_{x = 0}^{x=1} \int_{y = 0}^{y = x} \frac{\sin y}{y} \, dy \, dx$.
• Dec 7th 2008, 05:45 AM
markmil2002
Well I don't think the limits will necessarily change things, but the limits for the first integral are 0 and y. 0 being the lower limit and y being the upper limit. The next integral has limits of x and 4, followed by the third integral having limits of 0 and 4.
• Dec 7th 2008, 02:03 PM
markmil2002
Quote:

Originally Posted by mr fantastic
Let me blunter than my previous post:

Your question cannot be answered properly unless you also give the integral terminals. Perhaps you should review some examples from your textbook or class notes to see why that's the case.

Here's an example of a double integral that illustrates my point:

Find $\int \int \frac{\sin y}{y} \, dy \, dx$.

Find $\int_{x = 0}^{x=1} \int_{y = 0}^{y = x} \frac{\sin y}{y} \, dy \, dx$.

I also looked at changing it to cylindrical and spherical. Niether one seems to make things easier though.
• Dec 7th 2008, 02:26 PM
mr fantastic
Quote:

Originally Posted by markmil2002
I also looked at changing it to cylindrical and spherical. Niether one seems to make things easier though.

Is this triple integral the specific question you've been asked to solve or is it the result of a calculation you've done in order to answer a different question?
• Dec 7th 2008, 08:12 PM
markmil2002
It's a triple integral I am trying to solve.
• Dec 7th 2008, 08:17 PM
mr fantastic
Quote:

Originally Posted by markmil2002
It's a triple integral I am trying to solve.

Yes I know that. What I'm asking is were you given it to solve or is it the result of trying to solve another question.
• Dec 7th 2008, 08:47 PM
markmil2002
Just trying to solve it. It was given to me just to solve. It has nothing to do with anything else.
• Dec 8th 2008, 02:31 PM
markmil2002
I guess nobody can help?(Punch)
• Dec 8th 2008, 02:55 PM
mr fantastic
Quote:

Originally Posted by markmil2002
I guess nobody can help?(Punch)

I'd suggest reversing the order of the integration but the difficulties remain.

Perhaps you could ask the person who gave you the question exactly how they were expecting you to solve it.

Personally, I doubt there's a closed form solution.
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