# Thread: Double Integral over a general region

1. ## Double Integral over a general region

Hi All,

I have to find the volume of the given solid "enclosed by the cylinders z=x^2, y=x^2, and the planes z=0, y=4. But I'm not sure it makes sense. z=x^2 and y=x^2 aren't cylinders, though, they are parabolas. And they don't actually "enclose" any area, do they? Maybe I'm drawing it wrong, but I just can't make sense of it.

2. ## Re: Double Integral over a general region

Originally Posted by csuram3
z=x^2 and y=x^2 aren't cylinders, though, they are parabolas.
No. They aren't circular cylinders. But they are parabolic cylinders (cylinders with a parabolic generating curve).

Originally Posted by csuram3
And they don't actually "enclose" any area, do they? Maybe I'm drawing it wrong, but I just can't make sense of it.
They don't enclose an "area," they enclose a volume. These are three-dimensional surfaces, remember.

3. ## Re: Double Integral over a general region

Originally Posted by Reckoner
No. They aren't circular cylinders. But they are parabolic cylinders (cylinders with a parabolic generating curve).

They don't enclose an "area," they enclose a volume. These are three-dimensional surfaces, remember.
Thanks for the reply reckoner. I guess I misspoke. I understand that they are 3-dimensional. But the way I'm understanding it, z=x^2 opens "upward" to infinity, so without some upper bound on z, wouldn't the volume inside of this shape equal infinity?

Maybe an easier question for someone to answer would be, what should I set as my limits of integration? I know y should be between 0 and 4, and x should be between -2 and 2, but I guess I don't know what function I'm integrating.

The point of this section was also to integrate using functions as bounds (for 1 or both variables), as opposed to numbers.

4. ## Re: Double Integral over a general region

Originally Posted by csuram3
Thanks for the reply reckoner. I guess I misspoke. I understand that they are 3-dimensional. But the way I'm understanding it, z=x^2 opens "upward" to infinity, so without some upper bound on z, wouldn't the volume inside of this shape equal infinity?
It would, but we are also using the planes $y=4$ and $z=0$ as bounds. The bounded region is actually beneath $z=x^2$. Maybe later I'll make a graph to help you visualize it. I don't think I have time right now.

Originally Posted by csuram3
Maybe an easier question for someone to answer would be, what should I set as my limits of integration? I know y should be between 0 and 4, and x should be between -2 and 2, but I guess I don't know what function I'm integrating.
We can integrate $z=x^2$. For our limits, we could do either

$x^2\leq y\leq4$

$-2\leq x\leq2$

or

$-\sqrt y\leq x\leq\sqrt y$.

$0\leq y\leq4$

Both of these represent the same region. In some problems, changing the order of integration can make it easier to evaluate the integrals. In this case it doesn't matter too much.

5. ## Re: Double Integral over a general region

Thank did the trick! Thanks for the help! I was just picturing the area above z=x^2, not below it.