Multi-variable integration with a e^u

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SUMMARY

The discussion focuses on calculating the mass of a rectangular box B defined by the bounds 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, and 0 ≤ z ≤ 1, with a density function ρ(x, y, z) = z e^(x+y). The integral setup involves a double integral after integrating with respect to z, leading to the expression ∫ from 0 to 1 ∫ from 0 to 2 (1/2)e^(x+y) dy dx. The participant seeks clarification on applying "u" substitution, specifically how to adjust the limits of integration for y when u = x + y.

PREREQUISITES
  • Understanding of triple integrals in multivariable calculus
  • Familiarity with density functions and mass calculations
  • Knowledge of "u" substitution in integration
  • Experience with exponential functions and their properties
NEXT STEPS
  • Study the application of "u" substitution in multiple integrals
  • Learn how to compute mass using density functions in three dimensions
  • Explore the properties of exponential functions in integration
  • Practice solving triple integrals with varying limits of integration
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Students and educators in multivariable calculus, mathematicians focusing on integration techniques, and anyone involved in physics or engineering applications requiring mass calculations of three-dimensional objects.

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Homework Statement



Find the mass of the rectangular box B where B is the box determined by
0 [tex]\leq[/tex] x [tex]\leq[/tex] 1, 0 [tex]\leq[/tex] y [tex]\leq[/tex] 2, and 0 [tex]\leq[/tex] z [tex]\leq[/tex] 1, and with density function [tex]\rho[/tex] ( x, y, z ) = z e^{x+y}.

Homework Equations



"u" substitution

The Attempt at a Solution



I believe I've taken the first integral with respect to dz correctly which led me to this integral

[tex]\int[/tex] from 0 to 1 [tex]\int[/tex] from 0 to 2 (1/2)e^(x+y) dy dx

I know i need to use a "u" substitution and have u=x+y but I'm unsure of how that changes the range of the integral with respect to y. if my equation is unclear please let me know. thank you!
 
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write out the triple integral and use
[tex]e^{x+y} = e^x e^y[/tex]
 
ah ok i understand how that works out! thank you! greatly appreciated!
 

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