Evaluating Triple Integral in 1st Octant: Q & 3x+2y+z=6

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SUMMARY

The discussion focuses on evaluating the triple integral \(\int\int\int_{Q}(1-x) \, dz \, dy \, dx\) over the solid Q in the first octant, bounded by the plane defined by the equation \(3x + 2y + z = 6\). The correct limits for the integral are established as \(z\) from 0 to \(6 - 3x - 2y\), \(y\) from 0 to \(-\frac{3}{2}x + 3\), and \(x\) from 0 to 2. The final computed value of the integral is confirmed to be 3 after resolving the limits and performing the necessary algebraic calculations.

PREREQUISITES
  • Understanding of triple integrals in multivariable calculus
  • Familiarity with the concept of solid regions in the first octant
  • Knowledge of how to derive limits of integration from plane equations
  • Proficiency in algebraic manipulation and solving equations
NEXT STEPS
  • Study the method of setting up triple integrals in different coordinate systems
  • Learn how to visualize and interpret solid regions in three-dimensional space
  • Explore the application of the Jacobian in changing variables for multiple integrals
  • Practice solving similar problems involving planes and triple integrals
USEFUL FOR

Students in multivariable calculus, educators teaching integration techniques, and anyone looking to enhance their understanding of triple integrals and solid geometry in the first octant.

triden
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Homework Statement



Evaluate \underbrace{\int\int\int}_{Q}(1-x) dzdydx

Where Q is the solid that lies in the first octant and below the plane:
3x + 2y + z = 6


The Attempt at a Solution



I guess my main problem is finding the integral limits. For dz, I arranged the equation of the plane so that z = 6-3x-2y. My limits for the dz integral would then be 0 to 6-3x-2y.

Not really sure if this is right or how to get the limits for dy and dx. Its been a year since I was in my multivariable class.

Chris
 
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That is correct for z limits. For x and y look at the triangle in the xy plane (z = 0).
 
Ah...ok, I think I get it now. Just to confirm:

Since there is now a triangle in the x-y plane with a slope of (-3/2)x for the hypotenuse, my dy limits would be 0 to (-3/2)x and then my dx limits would be 0 to 2.

Chris
 
triden said:
Ah...ok, I think I get it now. Just to confirm:

Since there is now a triangle in the x-y plane with a slope of (-3/2)x for the hypotenuse, my dy limits would be 0 to (-3/2)x and then my dx limits would be 0 to 2.

Chris

Not quite. When you put z = 0, the equation of that slanted line is not y = -(3/2)x because it doesn't go through the origin. Just solve it for y...
 
oops, yeah should have caught that. y = -(3/2)x + 3

Got an answer of 3 after a bit of algebra...seems about right!

Thanks a lot for your help, I appreciate it.
 

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