Evaluating 2D Integrals: f(x,y)=min(x,y)

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SUMMARY

The discussion focuses on evaluating the two-dimensional integral of the function f(x,y) = min(x,y) over the region defined by 0 ≤ x ≤ 2 and 0 ≤ y ≤ 1. Participants clarify that min(x,y) represents the minimum value between x and y, leading to the need to divide the integration region into two parts: where f(x,y) = x (for x < y) and where f(x,y) = y (for x ≥ y). The solution involves performing double integration separately for these two regions to obtain the final result.

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  • Knowledge of the minimum function in mathematical analysis
  • Basic skills in setting up integration limits for two-dimensional regions
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  • Explore the concept of region division in multi-variable calculus
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Homework Statement


Evaluate the following definite two-dimensional integrals over the specified domains of integration.

f(x,y)=min(x,y), over the region {(x,y) : 0 \leq x \leq 2, 0 \leq y \leq 1}


Homework Equations





The Attempt at a Solution


I'm not even sure where to start because I'm not sure what the problem even means by f(x,y)=min(x,y). HELP!
 
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min(x,y) means minimum of x and y. For example, in the region x < y, f(x,y) = x.
 
So if I integrate first from 0 \leq y \leq 1 then the x bounds, my f(x,y)=y? I'm a little confused over this. I don't remember going over this in calculus and this homework is supposed to be a calculus review.
 
Divide the region {(x,y) : 0 < x < 2, 0 < y < 1} into two parts, one where f(x,y) = x, and one where f(x,y) = y. Then do the usual double integration for the two regions separately.
 

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