Volume with triple integral

In summary, the author attempted to solve a problem involving volume between a given plane and the xy-plane, but was having difficulties due to a mistake he made. After breaking the problem down into smaller parts, he was able to find the answer through visualization.
  • #1
178
3

Homework Statement



Find the volume between the plane x+y+z = 1 and the xy-plane, for x+y[itex]\leq[/itex]2, x[itex]\geq[/itex]0, y[itex]\geq[/itex]0.

The Attempt at a Solution



First, the plane is above the xy-plane for y < 1-x and below the xy-plane for y > 1-x, so we'll need two integrals. This is how I set them up.

[itex]\int[/itex][itex]^{1}_{0}[/itex][itex]\int[/itex][itex]^{1-x}_{0}[/itex][itex]\int[/itex][itex]^{1-x-y}_{0}[/itex]1dzdydx - [itex]\int[/itex][itex]^{2}_{0}[/itex][itex]\int[/itex][itex]^{2-x}_{1-x}[/itex][itex]\int[/itex][itex]^{1-x-y}_{0}[/itex]1dzdydx

This gives me an answer of 7/6, but my book has 1 as the answer. I'm assuming my setup is wrong, since I checked the evaluation of the integrals in wolfram alpha. But I can't see what I did wrong. The innermost integrals give z-value of the given plane. The middle integrals sum those stacks from 0 to 1-x above the xy-plane and from 1-x to 2-x below the xy-plane. And then the outermost integrals sum those slices from the y-axis to the boundary at 0=1-x and 0=2-x respectively.

Help would be much appreciated.
 
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  • #2
Weird. I tried evaluating your integrals, and I got (1/6)-(-3)=(19/6).

See what you get by evaluating the second integral again. I don't know if I made a mistake or not.
 
  • #3
Try drawing a picture of the volume that you're trying to find. When I took multi-V, I noticed that helped me visualize the question better, which is essential to setting up your integral equation.
 
  • #4
Opus_723 said:

Homework Statement



Find the volume between the plane x+y+z = 1 and the xy-plane, for x+y[itex]\leq[/itex]2, x[itex]\geq[/itex]0, y[itex]\geq[/itex]0.

The Attempt at a Solution



First, the plane is above the xy-plane for y < 1-x and below the xy-plane for y > 1-x, so we'll need two integrals. This is how I set them up.

[itex]\int[/itex][itex]^{1}_{0}[/itex][itex]\int[/itex][itex]^{1-x}_{0}[/itex][itex]\int[/itex][itex]^{1-x-y}_{0}[/itex]1dzdydx - [itex]\int[/itex][itex]^{2}_{0}[/itex][itex]\int[/itex][itex]^{2-x}_{1-x}[/itex][itex]\int[/itex][itex]^{1-x-y}_{0}[/itex]1dzdydx

This gives me an answer of 7/6, but my book has 1 as the answer. I'm assuming my setup is wrong, since I checked the evaluation of the integrals in wolfram alpha. But I can't see what I did wrong. The innermost integrals give z-value of the given plane. The middle integrals sum those stacks from 0 to 1-x above the xy-plane and from 1-x to 2-x below the xy-plane. And then the outermost integrals sum those slices from the y-axis to the boundary at 0=1-x and 0=2-x respectively.

Help would be much appreciated.

The dydx portion of your second integral needs to be broken into two sections. y goes from 1-x to 2-x only when x is between 0 and 1. When x is between 1 and 2 y goes from 0 to 2-x. Draw a picture of y = 1-x and y = 2-x in the xy plane and you will see it.
 
  • #6
LCKurtz said:
The dydx portion of your second integral needs to be broken into two sections. y goes from 1-x to 2-x only when x is between 0 and 1. When x is between 1 and 2 y goes from 0 to 2-x. Draw a picture of y = 1-x and y = 2-x in the xy plane and you will see it.

AHA! Thank you! I actually had already drawn the picture, but I've been staring at it forever without noticing that! Now I feel dumb. I'm kind of in a studying binge and I think I'm starting to burn out a bit. Thank you though.
 

1. What is a triple integral?

A triple integral is an extension of a single or double integral to three dimensions. It is calculated by integrating a three-dimensional function over a three-dimensional region.

2. How do you set up a triple integral?

To set up a triple integral, you need to define the limits of integration for each variable (x, y, z) and determine the order of integration. This is usually done by visualizing the 3D region and breaking it into smaller, simpler shapes such as spheres, cylinders, or rectangular prisms.

3. What is the difference between a triple integral and a regular integral?

The main difference between a triple integral and a regular integral is the number of dimensions involved. A regular integral is calculated over a one-dimensional interval, while a triple integral is calculated over a three-dimensional region.

4. How do you calculate volume using a triple integral?

To calculate volume using a triple integral, you need to integrate the function representing the volume over the region of interest. The limits of integration will depend on the shape of the region, and the order of integration can be changed depending on the complexity of the function.

5. What are some real-world applications of triple integrals?

Triple integrals have various applications in physics, engineering, and other fields. They are commonly used to calculate the volume of irregularly shaped objects, the mass and center of mass of a solid, and the moment of inertia of 3D objects. They also have applications in fluid mechanics, electromagnetism, and probability theory.

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