Volume Integral

1. Aug 28, 2014

Zondrina

1. The problem statement, all variables and given/known data

I ran into this tough integral. I am asked to compute the volume of the following region using a double integral with $a, b, c > 0$.

The first octant region bounded by the co-ordinate planes $x=0$, $y=0$, $z=0$ and the cylinders $a^2y = b(a^2-x^2)$ and $a^2z = c(a^2-x^2)$.

2. Relevant equations

$$V = \int \int_R f(x,y) dA$$

3. The attempt at a solution

They want this solved as a Cartesian integral, so the first thing I did was find $z$:

$$a^2z = c(a^2-x^2) \Rightarrow z = c(1 - \frac{x^2}{a^2})$$

Hence we want the volume integral:

$$V = \int \int_R f(x,y) dA = \int \int_R c(1 - \frac{x^2}{a^2}) dA$$

Now I'm not entirely certain what the proper limits are in terms of $x$ and $y$. If we consider the other cylinder $a^2y = b(a^2-x^2)$ with $y=0$, we get:

$$0= b(a^2-x^2) \Rightarrow x = \pm a$$

If $x = 0$, then:

$$a^2y = ba^2 \Rightarrow y = b$$

So I get the limits $0 ≤ x ≤ a$ and $0 ≤ y ≤ b$. Are these limits reasonable?

Last edited: Aug 28, 2014
2. Aug 28, 2014

LCKurtz

No, those limits are not reasonable. They would describe a rectangle in the xy plane. Have you drawn a picture of your surfaces and your domain in the xy plane?

3. Aug 28, 2014

Zondrina

Something just occurred to me... those are not cylinders. Even re-arranging the givens does not give you a cylinder.

$$\frac{y}{b} + \frac{x^2}{a^2} = 1$$
$$\frac{z}{c} + \frac{x^2}{a^2} = 1$$

4. Aug 28, 2014

LCKurtz

It's true they aren't cylinders, but they are cylindrical surfaces. Like I asked before, have you drawn any pictures?

5. Aug 28, 2014

Zondrina

Should look like this right:

That dotted line is poorly drawn.

Last edited: Aug 28, 2014
6. Aug 28, 2014

LCKurtz

The whole thing is poorly drawn. But what you need is a picture of the domain in the xy plane with equations and intercepts labelled. That's where you will find the proper xy limits. And you might also identify the type of curves you are drawing. They aren't circles...

7. Aug 28, 2014

Zondrina

They are parabolas.

I have $z = c(1 - \frac{x^2}{a^2})$, under it I have the x-y surface given by:
$$a^2y = b(a^2 - x^2) \Rightarrow y = b(1 - \frac{x^2}{a^2})$$

These are equal when $b = c$.

8. Aug 28, 2014

LCKurtz

But you don't have b = c. At the risk of repeating myself, read the red above again. You have a portion of a parabola as a boundary of your region. How do you get the limits? It's a simple calc II problem.

9. Aug 28, 2014

Zondrina

You read them right off the graph. I must clearly be misunderstanding something, because even reading them off a graph gives me $0 ≤ x ≤ a$ and $0 ≤ y ≤ b(1 - \frac{x^2}{a^2})$.

Last edited: Aug 28, 2014
10. Aug 28, 2014

Zondrina

I believe those are the correct limits in my post above now. Sorry for the double.

11. Aug 28, 2014

LCKurtz

Those are the correct limits, assuming you use the dydx order in your double integral. In your original post you had $0\le x \le a$ and $0\le y \le b$ which described a rectangle. What are you still misunderstanding?

12. Aug 28, 2014

Zondrina

I misunderstand why the question is so poorly designed. I was under the impression i was projecting cylinders.

Turns out that wasn't the case.