Is the Projection of a Triple Integral's Base Always a Rectangle?

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SUMMARY

The discussion centers on the projection of a triple integral's base in the xy-plane, specifically questioning whether it is always a rectangle. The participant argues against the initial sketch, asserting that the equations y = 1 - z and x = 1 - z² yield a non-rectangular relationship, specifically x = -(y - 1)² + 1. However, another participant clarifies that the projection of the boundaries defined by these equations does indeed form a rectangle, validating the original sketch.

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theBEAST
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Homework Statement


Here is the question along with the solution and sketch.
WCyAe4Y.png


I think the sketch is wrong because the projection in the xy plane shows a rectangular box. I don't think it is a rectangular box because you can solve for an equation relating x and y.

You know that y = 1-z and x = 1-z^2, so you can solve for some equation with x and y by eliminated z. Which gets you x = -(y-1)^2 + 1 for the relationship in the xy plane and it is clearly not a rectangle...?
 
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theBEAST said:

Homework Statement


Here is the question along with the solution and sketch.
WCyAe4Y.png


I think the sketch is wrong because the projection in the xy plane shows a rectangular box. I don't think it is a rectangular box because you can solve for an equation relating x and y.

You know that y = 1-z and x = 1-z^2, so you can solve for some equation with x and y by eliminated z. Which gets you x = -(y-1)^2 + 1 for the relationship in the xy plane and it is clearly not a rectangle...?
That last equation is, of course, x= y(2- y) but it is the projection of the boundary of the y= 1- z and x= 1- z^2, NOT the base itself. The sketch is correct.
 

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