I How to find outer limit of integration for this triple integ

[egio]

Here's a graph and its triple integral. How are the limits of integration for the outer integral [-2,2]? I have no idea how this was found.

Any help would be appreciated!

 

[slider142]

Here's a graph and its triple integral. How are the limits of integration for the outer integral [-2,2]? I have no idea how this was found.

Any help would be appreciated!

What is the boundary at which the cone and spherical cap intersect? As such, between which two x coordinates does the volume lie?

 

[egio]

What is the boundary at which the cone and spherical cap intersect? As such, between which two x coordinates does the volume lie?

EDIT: I solved it! I had to substitute z = √(x² + y²) into x² + y² + z² = 8 then solve for the radius r. Thanks again!

The boundary has a radius of y = √(4 - x²). I'm not sure what to do with this equation. Hmm..

 

[slider142]

EDIT: I solved it! I had to substitute z = √(x² + y²) into x² + y² + z² = 8 then solve for the radius r. Thanks again!

The boundary has a radius of y = √(4 - x²). I'm not sure what to do with this equation. Hmm..

Not quite! The boundary occurs when both surfaces have the same z coordinate, as we can see from the diagram. Equating the z values of the variables in both equations, we see that the x and y coordinates of points on the boundary must satisfy the equation:
\sqrt{x^2 + y^2} = \sqrt{8 - x^2 - y^2}
Restricting ourselves to the domain of each square root function, we can simplify this requirement to all points in the domain that satisfy the equation:
x^2 + y^2 = 8 - x^2 - y^2
which can be simplified to the equation
x^2 + y^2 = 4
This is a circle of radius 2. Do you see how that relates to the boundary for the outer integral?

 

[egio]

Not quite! The boundary occurs when both surfaces have the same z coordinate, as we can see from the diagram. Equating the z values of the variables in both equations, we see that the x and y coordinates of points on the boundary must satisfy the equation:
\sqrt{x^2 + y^2} = \sqrt{8 - x^2 - y^2}
Restricting ourselves to the domain of each square root function, we can simplify this requirement to all points in the domain that satisfy the equation:
x^2 + y^2 = 8 - x^2 - y^2
which can be simplified to the equation
x^2 + y^2 = 4
This is a circle of radius 2. Do you see how that relates to the boundary for the outer integral?

Oh, interesting! Didn't see it that way. Awesome! Thank you for making it even clearer. :)