# Homework Help: Triple Integral to find vol.

1. Mar 14, 2012

### hqjb

Can someone help me with this?

1. The problem statement, all variables and given/known data
Find the volume V of the solid S bounded by the three coordinate planes, bounded above
by the plane x+ y+ z = 2, and bounded below by the plane z = x+ y.

2. Relevant equations
x + y + z = 2
z = x + y

3. The attempt at a solution
$\int_{0}^{2}\int_{-x}^{2-x}\int_{x+y}^{2-y-x}dzdydx$
So I used the above triple integral and got -4(did it twice), wolfram-alpha's calculator gives me 0 and the textbook answer is 1/3

So obviously I did something wrong in the triple integral and in identifying the limits.
But I just want to know the right limits for the above question as I have problems identifying them (I drew traces(attached) on the planes but not sure if the region's right)

#### Attached Files:

• ###### region.jpg
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2. Mar 14, 2012

### SammyS

Staff Emeritus
Hello hqjb. Welcome to PF !

Where do the planes, x+ y+ z = 2 and z = x+ y intersect ?

3. Mar 14, 2012

### hqjb

Hey thanks.
I suppose thats more of a hint than an answer right?

I think you'll get y=1-x,
tried and got the right answer with $\int_{0}^{2}\int_{0}^{1-x}\int_{0}^{2-y-x}dzdydx$
but Im not sure why

Last edited: Mar 14, 2012
4. Mar 14, 2012

### SammyS

Staff Emeritus
The lower bound for the z integral is wrong --- should be x+y .

You will often get hints and other guidance, so that you will then understand how to solve a problem.

5. Mar 14, 2012

### hqjb

Yes, Im hoping to understand this too. That wasn't an attempt to answer but a random integral that got the answer lol.

I finally got it after plotting the thing in 3d. my x-y trace was wrong it should be the line of intersection between the two planes.

Why can't it be if I let z=0 for both equations and draw the region?

6. Mar 14, 2012

### SammyS

Staff Emeritus
The region only touches the xy-plane (z = 0) at a single point, the origin.

7. Mar 14, 2012

### hqjb

Alright understood, so i shouldnt use the x-y trace in this case because there isnt one.
And the bounds are 0 < y < 1-x , 0 < x < 1, x+y < z < 2-x-y ?

Thanks for your patience and help.