Double Integrals: Will this solution always give the correct answer?

[theBEAST]

Homework Statement

Here is the problem:
http://dl.dropbox.com/u/64325990/Photobook/question.PNG

Here is the answer:
http://dl.dropbox.com/u/64325990/Photobook/solution.PNG

So what the answer says is that you can find the volume under the surface minus the volume of the rectangle with height z=1. However I don't see how this will work for every question. Here I will illustrate why I think their solution will not always be correct:
http://dl.dropbox.com/u/64325990/Photobook/Photo%202012-05-31%205%2004%2056%20PM.jpg

As you can see, if it so happens that the max height of the surface is less than 1 then subtracting the rectangle will give you an answer too small. Does anyone agree with me?
For those who are interested, I decided to subtract 1 from the z function and find the surface under z = 1+x^2+(y-2)^2 which also gave the correct answer.
Thank you!

 

[HallsofIvy]

Actually, what I see is that you have the paraboloid upside down! With $z= 2+ x^2+ (y- 1)^2$, and squares never being negative, the smallest value of z will be 2 when x= 0 and y= 1.

 

[theBEAST]

Actually, what I see is that you have the paraboloid upside down! With $z= 2+ x^2+ (y- 1)^2$, and squares never being negative, the smallest value of z will be 2 when x= 0 and y= 1.

Ah yes, the illustration I made was just an example that would not work right? But the question given here, because the height is always greater than 2, then subtracting a rectangle would work. Thanks for clearing that up.