# Choosing the limits of integration

1. Jun 23, 2006

### Maxwell

If I'm asked to find the volume of a solid that lies below the surface z = f(x,y), and above to region in the xy-plane bounded by a certain curve...and I'm only given 3 limits of integration, what do I do?

For example:

z = 9 - x - y

Given y = 0, x = 3, y = (2x)/3

At first I thought I could get another limit for my x integral by solving y = (2x)/3 for x -- however, I then end up with an answer that has an x or a y in it. So I figure this must be wrong.

I then chose the limits for my x integral to be 0 and 3, (I choose 0 randomly) and I got an answer of 18. The books answer was 19. So I believe I am close.

Is that how I find the last limit if I am only given 3?

Thank you!

2. Jun 23, 2006

### 0rthodontist

You should draw a picture of your region. The outer integral you set up has to have constant limits, and the inside integral will go between two functions that you can see from the picture.

3. Jun 23, 2006

### nrqed

I am not following here. Can you double check that there is no typo here. Did you mean *z=0* on the second line?

EDIT: I see that the question states that the volume is between the xy plane and the z function so the lower limit of z is 0. Fine. But the limits on x and y are still not clear. There should be two limits on x, it does not make sense to simply give one value of x. So there must be a typo.

Pat

Last edited: Jun 23, 2006
4. Jun 23, 2006

### shmoe

The given curves (lines if you prefer) in the xy plane y = 0, x = 3, y = (2x)/3, define the bounded region (the wording of the problem is confusing as stated).

Maxwell, plot these three lines as suggested above if you haven't already. What do your limits of integration look like? For practice you should set up the integral as dxdy and also dydx (that is change the order of integration and set up the limits).

5. Jun 23, 2006

### nrqed

Oh! yes, of course (slapping myself hard!)
That was obvious. Sorry for my dumb mistake. Thanks shmoe.

To Maxwell: Do the z integral first, from 0 (since it starts from the xy plane) up to the function given for z.

Now, do the x and y integral. As Shmoe suggested, it would be a good practice to do it two ways: x and then y or y and then x.

let's say you do y and then x. The way to think about it is the following. Since you integrate y first and then x, you must imagine having x fixed to some value as you vary y. Then the question is: for an arbitrary value of x within your region, the value of y varies from what to what? (hint: the lower limit is a number but the upper limit is a function of x)

Patrick

6. Jun 24, 2006

### michael3

(S_0_3)(S_0_2x/3)(S_0_9-x-y)1 dzdydx

Performing first integration

(S_0_3)(S_0_2x/3) 9-x-y dydx

Performing Second Integration

(S_0_3) 6x-(2x^2)/3 - (2x^2)/9 dx

Performing Third Integration

3x^2-(2x^3)/9-(4x^3)/54 | plugging 3 in for all x values gives you a final value of 19.

7. Jun 24, 2006

### nrqed

Michael, we ask that people don't post full solutions. The goal is to help the original poster to work out the problem by himself/herself. The goal is to help him/her *learn*.

regards

Patrick

8. Jun 25, 2006

### michael3

waaa, stop whining at me for u being lazy