- #1
egio
- 14
- 3
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!
Any help would be appreciated!
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:
[tex]\sqrt{x^2 + y^2} = \sqrt{8 - x^2 - y^2}[/tex]
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:
[tex]x^2 + y^2 = 8 - x^2 - y^2[/tex]
which can be simplified to the equation
[tex]x^2 + y^2 = 4[/tex]
This is a circle of radius 2. Do you see how that relates to the boundary for the outer integral?
The outer limit of integration is used to determine the boundaries of a region in three-dimensional space that is being integrated over. It helps to define the limits of the triple integral and ensures that the correct volume or surface area is calculated.
The outer limit of integration can be determined by looking at the boundaries of the region in three-dimensional space and identifying the maximum and minimum values for each variable. These values will then be used as the outer limits of integration for each variable in the integral.
Yes, the outer limit of integration can be negative, depending on the region being integrated over. It is important to carefully consider the boundaries of the region when determining the outer limits, as they may be positive, negative, or a combination of both.
Yes, there are some special cases where finding the outer limit of integration may be more complicated. For example, when the region being integrated over is not a simple shape, such as a sphere or cylinder, the outer limits may need to be determined using a combination of equations or geometric methods.
If the outer limit of integration is incorrect, the calculated volume or surface area will also be incorrect. It is important to carefully consider the boundaries of the region and ensure that the correct outer limits are used in order to obtain an accurate result from the triple integral.