# Evaluate a definite integral with two variables?

## Homework Statement

Evaluate.

$$\int$$ y2(1 - y3/a2)-2 dy

Upper limit = 0
Lower limit = -a

## 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?

vela
Staff Emeritus
Homework Helper
Just treat a as a constant.

Char. Limit
Gold Member
It might also make it easier to multiply top and bottom by a4.

I think that is the part that confuses me... If a is constant, how do we take the antiderivative?

Char. Limit
Gold Member
With respect to y. Just treat a as you would treat a 3, or a pi.

vela
Staff Emeritus
Homework Helper
Just for clarification, is the problem

$$\int_{-a}^0 y^2\left(\frac{1-y^3}{a^2}\right)^{-2}\,dy$$

or

$$\int_{-a}^0 y^2\left(1-\frac{y^3}{a^2}\right)^{-2}\,dy$$

? 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.

It's the latter option. I always forget to include parentheses!

vela
Staff Emeritus
Homework Helper
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.

Ohhhhh! Wow... now it makes so much more sense... So, I can just pull 1/a2 out of the integral and than it becomes:

$$\int$$ 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...

Char. Limit
Gold Member
Ohhhhh! Wow... now it makes so much more sense... So, I can just pull 1/a2 out of the integral and than it becomes:

$$\int$$ 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.

vela
Staff Emeritus
Homework Helper
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

$$a^4\int_{-a}^0 y^2(a^2-y^3)^{-2}\,dy$$

HallsofIvy
Homework Helper
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.)

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!

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?