How can we set up a triple integral to solve this problem?

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Homework Help Overview

The discussion revolves around setting up triple integrals for evaluating integrals over specified bounded regions in three-dimensional space. The subject area includes calculus, specifically multivariable calculus and triple integrals.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of triple integrals, questioning the bounds for integration and the relationships between the variables x, y, and z. There are attempts to clarify the geometric interpretation of the bounded region and the limits of integration.

Discussion Status

Some participants are exploring the correct bounds for the variables and the implications of the surfaces defining the region. There is an ongoing dialogue about the interpretation of the equations and the constraints they impose on the integration limits. Guidance has been offered regarding visualizing the bounded region and the relationships between the variables.

Contextual Notes

Participants note uncertainties regarding the maximum values for x and y, as well as the need to clarify the intersection of surfaces in three-dimensional space. There is mention of specific functions and their roles in defining the region for integration.

stunner5000pt
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Triple Integral setup...

[tex]\int \int \int_{G} 6x (z+y^3) dx dy dz[/tex] G bounded by [tex]x = 0, \ x = y, \ z = y-y^2, \mbox{and} \ z=y^2 - y^3[/tex]
x from 0 to1
y from 0 to x
z from z=y-y^2 to y^2 - y^3
and the integration order becomes dz dy dx
would this give the right answer?

what aboiut this one
[tex]\int \int \int_{G} xy + xz dx dy dz[/tex]

G bounded by z = x, z=2-x, z = y^2
z goes from 2-y^2 to y^2
y goes from 2-x to x
x goes from 0 to 2
and the integration order to dz dy dx

I think the second one is wrong. Please do help!
 
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Can you explain your reasoning for 1? Particularly the x and y. What exactly are you trying to do in these problems?
 
the question is evaluate the integral over the given bounded region G

for the first one z goes like the y function, that s fine
for the y takes on a min value of 0 and max of x and
for the x the min value is zero but I am not sure about the max value ..
 
Last edited:
Think of x=0 and x=y as planes, not as lines.

Also, x takes on a min value of 0, and a max value of y. You don't know much about y.

Try imagining the region G in your head, its bounded by the yz plane, and the plane y=x and the z function.

You want to find how far y and x go given your constraints. Solving the functions would not be a bad idea.
 
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if i were to solve what am i solving for?
which functions would i use?

x=0, x= y, z = y - y^2 , z = y^2 - y^3
the intesection of which surfaces??
 
there's only two equations that could be solved here. You have two functions 'z' in R^3. For a triple integral, you want to find the domain of these curves on the xy plane, where z=0. Solve the z functions to find the range of the y function.


http://tutorial.math.lamar.edu/AllBrowsers/2415/TripleIntegrals.asp Example 2
 
Last edited by a moderator:
ok i got the y ranges from 0 to 1
now for the x part since x goes from 0 to y, x goes from 0 to 1
but y has to be a function of x so y goes from 0 to x?? and then x from 0 to 1?
 
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[tex]\int\int \int_{G} 6x (z+y^3) dx dy dz[/tex] = [tex]\int\int_{D} \int_{y-y^2}^{y^2-y^3} 6x(z+y^3)dz dA[/tex]

http://www.public.asu.edu/~hyousif/maple.JPG
http://www.public.asu.edu/~hyousif/xzplane.JPG

edit: Its the yz plane, not the xz plane.

Heres the x-y plane, which is your domain:
Notice its a triangle with vertices at (0,0) (0,1) (1,1) You can express this as a domain pretty easily.

http://www.public.asu.edu/~hyousif/xyplane.JPG
 
Last edited by a moderator:

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