The discussion revolves around determining the outer limits of integration for a triple integral related to the intersection of a cone and a spherical cap. Participants explore the geometric boundaries that define the volume of interest, specifically focusing on the limits of integration in the context of the given graph and equations.
Participants do not reach a consensus on the initial determination of the outer limits of integration, but there is agreement on the geometric interpretation of the boundary condition involving the cone and spherical cap. The discussion includes multiple perspectives and clarifications without resolving all uncertainties.
Participants rely on specific geometric interpretations and algebraic manipulations to explore the problem, with some steps and assumptions remaining implicit. The discussion does not resolve all mathematical details or the full implications of the derived equations.
egio said: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!
EDIT: I solved it! I had to substitute z = √(x2 + y2) into x2 + y2 + z2 = 8 then solve for the radius r. Thanks again!slider142 said:What is the boundary at which the cone and spherical cap intersect? As such, between which two x coordinates does the volume lie?
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:egio said: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..
Oh, interesting! Didn't see it that way. Awesome! Thank you for making it even clearer. :)slider142 said: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?