Evaluate a definite integral with two variables?

In summary, the problem is to find the integral of y2(1-y3/a2)-2 dy. For this, you treat a as a constant and use the substitution dy=1/a2. You then use the antiderivative, G(u), to solve for u.
  • #1
LilTaru
81
0

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?
 
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  • #2
Just treat a as a constant.
 
  • #3
It might also make it easier to multiply top and bottom by a4.
 
  • #4
I think that is the part that confuses me... If a is constant, how do we take the antiderivative?
 
  • #5
With respect to y. Just treat a as you would treat a 3, or a pi.
 
  • #6
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
It's the latter option. I always forget to include parentheses!
 
  • #8
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
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
LilTaru said:
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
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
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
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
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?
 

1. What is a definite integral with two variables?

A definite integral with two variables is an integral that involves two variables, usually denoted by x and y. It represents the area under a curve in a two-dimensional coordinate system.

2. How do you evaluate a definite integral with two variables?

To evaluate a definite integral with two variables, you first need to set up the integral by identifying the limits of integration for both variables. Then, you can use various integration techniques, such as substitution or integration by parts, to solve the integral and find the numerical value of the area under the curve.

3. What are the applications of evaluating a definite integral with two variables?

Evaluating a definite integral with two variables has many applications in science and engineering. It can be used to calculate areas, volumes, and other physical quantities. It is also used in probability and statistics to calculate probabilities and expected values.

4. What are some common mistakes when evaluating a definite integral with two variables?

Some common mistakes when evaluating a definite integral with two variables include forgetting to include the differential of one variable when integrating with respect to the other, mixing up the limits of integration, and using the wrong integration technique.

5. Can definite integrals with two variables have multiple solutions?

Yes, definite integrals with two variables can have multiple solutions. This usually occurs when the curve being integrated has multiple intersecting regions or when different integration techniques are used to solve the integral, resulting in different solutions.

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