How Can I Find the Volume of a Solid Bounded by Three Coordinate Planes?

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SUMMARY

The volume of the solid bounded by the three coordinate planes, the plane x + y + z = 2, and the plane z = x + y is calculated using the correct limits in a triple integral. The correct integral setup is \int_{0}^{1}\int_{0}^{1-x}\int_{x+y}^{2-y-x}dzdydx, which yields a volume of 1/3. The previous attempts resulted in incorrect limits, leading to erroneous volume calculations. The key to solving this problem lies in accurately identifying the intersection of the planes and the appropriate bounds for the integrals.

PREREQUISITES
  • Understanding of triple integrals in multivariable calculus
  • Familiarity with the equations of planes in three-dimensional space
  • Ability to visualize geometric regions defined by inequalities
  • Knowledge of integration techniques for calculating volumes
NEXT STEPS
  • Study the method of triple integrals for volume calculation in multivariable calculus
  • Learn how to find intersections of planes in three-dimensional space
  • Explore graphical software tools for visualizing 3D geometric regions
  • Practice solving similar volume problems with different bounding planes
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable calculus and volume calculations, as well as anyone looking to improve their understanding of geometric interpretations in three dimensions.

hqjb
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Can someone help me with this?

Homework Statement


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.

Homework Equations


x + y + z = 2
z = x + y

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)
 

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hqjb said:
Can someone help me with this?

Homework Statement


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.

Homework Equations


x + y + z = 2
z = x + y

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)
Hello hqjb. Welcome to PF !

Where do the planes, x+ y+ z = 2 and z = x+ y intersect ?
 
SammyS said:
Hello hqjb. Welcome to PF !

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

Hey thanks.
I suppose that's 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 I am not sure why
 
Last edited:
hqjb said:
Hey thanks.
I suppose that's 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}^{1}\int_{0}^{1-x}\int_{0}^{2-2y-2x}dzdydx
but I'm not sure why
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.
 
SammyS said:
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.

Yes, I am 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?
 
hqjb said:
Yes, I'm 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?
The region only touches the xy-plane (z = 0) at a single point, the origin.
 
SammyS said:
The region only touches the xy-plane (z = 0) at a single point, the origin.

Alright understood, so i shouldn't use the x-y trace in this case because there isn't one.
And the bounds are 0 < y < 1-x , 0 < x < 1, x+y < z < 2-x-y ?

Thanks for your patience and help.
 

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