Find the volume of a given set

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SUMMARY

The discussion centers on evaluating the volume of a three-dimensional set defined by the inequalities 0≤x≤1, 0≤y≤1, and 1≤z≤e^{x+y}. A participant questioned the validity of modifying the upper limit for z to 0≤z≤e^{x+y}-1, while still obtaining the correct volume through the double integral ∫∫_{A}(e^{x+y}-1)dxdy. The consensus confirms that the original setup is correct, as the volume calculation involves the difference between the upper and lower bounds of z.

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Homework Statement


Evaluate the volume of the given set:


Homework Equations


[itex]0≤x≤1, 0≤y≤1, 1≤z≤e^{x+y}[/itex]


The Attempt at a Solution


I got the 'right' answer but I think the solution is not correct. I did chance the last term given above to [itex]0≤z≤e^{x+y}-1[/itex], and then as usual, I evaluated [itex]\int\int_{A}e^{x+y}-1dxdy[/itex] and the result was correct.

But can I make that change on z? If I can, why? (I sense this is a stupid question, nevertheless I couldn't find the answer hehe)

Many thanks
 
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carlosbgois said:

Homework Statement


Evaluate the volume of the given set:


Homework Equations


[itex]0≤x≤1, 0≤y≤1, 1≤z≤e^{x+y}[/itex]


The Attempt at a Solution


I got the 'right' answer but I think the solution is not correct. I did chance the last term given above to [itex]0≤z≤e^{x+y}-1[/itex], and then as usual, I evaluated [itex]\int\int_{A}e^{x+y}-1dxdy[/itex] and the result was correct.

But can I make that change on z? If I can, why? (I sense this is a stupid question, nevertheless I couldn't find the answer hehe)

Many thanks

That setup is correct. When you do a volume by a double integral like that, the integrand is always ##z_{upper}-z_{lower}##, which is what you have.
 

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