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

[theBEAST]

Homework Statement

Here is the question along with the solution and sketch.

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...?

 

[HallsofIvy]

Homework Statement

Here is the question along with the solution and sketch.

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.