- #1

urnlint

- 1

- 0

## Homework Statement

Grr it got all got erased...This is a math problem that a student I tutor got wrong on his calculus test. I am having problems remembering what to do. I have looked up other problems, but the paraboloids seem simpler for some reason. Probably because someone is explaining it to me.

Find the volume between the two parabaloids x = [itex]y^{2}[/itex] + [itex]z^{2}[/itex] and y = [itex]x^{2}[/itex] + [itex]z^{2}[/itex]

Hint: x - y divides [itex]x^{2}[/itex] - [itex]y^{2}[/itex] (I do not know how this helps. Probably because I am doing it wrong)

## The Attempt at a Solution

Set z = 0 to find the "shadow" in the xy plane: y = [itex]x^{0.5}[/itex] and y = [itex]x^{2}[/itex]

They intersect at (0,0) and (1,1) because:

[itex]x^{2}[/itex] = [itex]x^{0.5}[/itex]

[itex]x^{2}[/itex] - [itex]x^{0.5}[/itex] = 0

[itex]x^{0.5}[/itex]([itex]x^{1.5}[/itex] - 1) = 0

[itex]x^{0.5}[/itex] = 0 gives x = 0 and y = 0 and

[itex]x^{1.5}[/itex] - 1 = 0

[itex]x^{1.5}[/itex] = 1 gives x = 1 and y =1

I think I understood up until now, but am sort of lost, and I do not have the answer given by the teacher to check myself. Do I do ∫∫∫ 1 dzdydx where 0≤x≤1, [itex]x^{2}[/itex]≤y≤[itex]x^{0.5}[/itex] and for the interval of z I just solve one of the given paraboloids for z, which will give a ±square root? Like z = ±√(x - [itex]y^{2}[/itex]) Then I would do trig substitution? Or would I subtract right from left so that I would just do a double integral of √(x - [itex]y^{2}[/itex]) - √(y - [itex]x^{2}[/itex]) with the above intervals for x and y?