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Evaluate a definite integral with two variables?

  • Thread starter LilTaru
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  • #1
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Homework Statement



Evaluate.

[tex]\int[/tex] y2(1 - y3/a2)-2 dy

Upper limit = 0
Lower limit = -a


Homework Equations





The Attempt at a Solution



The a's make me really confused and I have no idea how to even start this question.
What I started with was:

Set u = 1 - y3/a2
du = (2y2 - 3y2a)/a3

Then I know you are supposed to make the integral contain only u's, but I am so confused! Am I on the right track? If so, where do I go from here? If not, can someone lead me on the right track?
 

Answers and Replies

  • #2
vela
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Just treat a as a constant.
 
  • #3
Char. Limit
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It might also make it easier to multiply top and bottom by a4.
 
  • #4
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I think that is the part that confuses me... If a is constant, how do we take the antiderivative?
 
  • #5
Char. Limit
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With respect to y. Just treat a as you would treat a 3, or a pi.
 
  • #6
vela
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Just for clarification, is the problem

[tex]\int_{-a}^0 y^2\left(\frac{1-y^3}{a^2}\right)^{-2}\,dy[/tex]

or

[tex]\int_{-a}^0 y^2\left(1-\frac{y^3}{a^2}\right)^{-2}\,dy[/tex]

? I'm guessing from what you wrote down for du, it is the former, but what you wrote in your original post, with the lack of parentheses, means the latter.
 
  • #7
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It's the latter option. I always forget to include parentheses!
 
  • #8
vela
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Oy, brain fart on my part. I think your du is consistent with the latter integral if you're making the mistake I think you're making. How you wrote the integral in your original post is fine. It's just that a lot of posters leave out parentheses so when I see problems written like this, it's a question that immediately pops into my mind.

To be a bit more explicit about what we're telling you...

Consider the function f(x)=9x2. When you differentiate it, you don't look at it as a product of 9 and x2 and apply the product rule, right? You just say the 9 is a constant so you can just pull it out front and then just differentiate the x2. Similarly, if you had g(x)=x2/32, you wouldn't use the quotient rule. You'd just pull the constant 1/32 out front and differentiate the x2. When we say treat a as a constant, we mean you can just pull it out front. You don't treat is as a function of x (or y in your case) even though it's written using a letter rather than numbers.
 
  • #9
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Ohhhhh! Wow... now it makes so much more sense... So, I can just pull 1/a2 out of the integral and than it becomes:

[tex]\int[/tex] y2(a2 - y3)-2 ???

Since we need to factor out 1/a2 from both 1 and y3/a2.
And then continue with substitution? This looks like it will work to me...
 
  • #10
Char. Limit
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Ohhhhh! Wow... now it makes so much more sense... So, I can just pull 1/a2 out of the integral and than it becomes:

[tex]\int[/tex] y2(a2 - y3)-2 ???

Since we need to factor out 1/a2 from both 1 and y3/a2.
And then continue with substitution? This looks like it will work to me...
Watch out. Because the 1/a^2 is inside another 2, you need to pull out 1/a^4.
 
  • #11
vela
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Well, you can't just erase it altogether, and you must still follow the rules of algebra. After you pull the factor out front, you should have

[tex]a^4\int_{-a}^0 y^2(a^2-y^3)^{-2}\,dy[/tex]
 
  • #12
HallsofIvy
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Yes, use the obvious substitution.

(You forgot the "dy" which is very important- for one thing, it was the "dy" that told you that, at least for the purposes of this integration, a is a constant.)
 
  • #13
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Yes. Sorry. I meant with the dy and with 1/a2 at the front. And thanks for catching it was 1/a4. Forgot about the squaring! Thanks a lot everyone! I think I can solve it now!
 
  • #14
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Solved it! I think it's correct... Got the answer as
(a2 - a6 + a7)/3

Since:

Set u = a^2 - y^3; du = -3y^2 dy
Let G(u) = 1/3(u)-1 (the antiderivative)
a^4 * G(0) - G(-a) (evaluation of the integral)
= a^4 * (1/3a^2) - ((a^2 + a^3)/3)
= (a2 - a6 + a7)/3

Is this correct?
 

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