Help setting up a triple integral

[tatiana_eggs]

Homework Statement

Hi guys, I need help setting up an integral.

Problem: Compute the integral f(x,y,z)=xyz over the solid region bounded below by plane z=-x, above by z=x, and otherwise b the parabolic cylinder x=2-y^2

This is not a surface integral, is it? Because the problems assigned to me before and after this question dealt with surface integrals.

Homework Equations

I'm going to assume it's not a surface integral. I can see clearly that my bounds for z will be between -x and x. I will hazard a guess that my y bounds might be sqrt(2-x) and -sqrt(2-x).

How can I figure out my x bounds algebraically. I have terrible 3-d intuition. I can draw a 3-d graph pretty well but it's hard for me to visualize it.

The Attempt at a Solution

I'm great at solving integrals once they're set up. Just need help with my bounds.
Thanks

 

[jackmell]

You're integrating over the footprint:

$$x=2-y^2$$

which I assume you want the volume in the first and fourth quadrant.

You can draw that simple sideways parabola easily. Now look at the figure and how would I have to integrate over all of it? You have the y direction right as it goes from $-\sqrt{2-x}$ up to $\sqrt{2-x}$. Now, it's not hard to see what x would have to go from right?

 

[tatiana_eggs]

Thank you for the reply, Jack, in my drawing I see that my bound for $$x$$ would start at $$x=2$$. I want to say my lower bound is $$x=-2$$ because of where $$z=-x$$ crosses $$x=2-y^2$$. Is that right?

 

[jackmell]

I don't see how you would think the lower bound on x would be -2 unless you're attempting to compute the volume of something other than what I think it is. The two sheets z=x and z=-x intersect along the y-axis right? And the paraboloid x=2-y^2 is just the parabola x=2-y^2 that extends without bounds along the z-axis but I assume we want that bounded by the sheets z=x and z=-x. I don't know how to better explain it other than just drawing it and showing you. It's that green part in there right? If so, can you not then say what the range on x is?

 

[tatiana_eggs]

Ah of course, 0 to 2 will be my x bounds. Thank you for the time you took to make the diagram. I see my mistake before. I redrew it with help from your picture and it is clear that my lower bound is x=0.

 

[jackmell]

Try to learn how to quickly draw them by hand and then learn how to draw them in Mathematica to confirm your drawing:

c1 = ContourPlot3D[{z == x, z == -x, 
    x == 2 - y^2}, {x, -5, 5}, 
   {y, -5, 5}, {z, -5, 5}, 
   AspectRatio -> 4/3]

c2 = ContourPlot3D[{x == 2 - y^2}, 
   {x, -5, 5}, {y, -5, 5}, {z, -5, 5}, 
   AspectRatio -> 4/3, RegionFunction -> 
    Function[{x, y, z}, 0 <= x <= 2 && 
      z <= x && z >= -x], 
   ContourStyle -> Green]

Show[{c1, c2}]