We can have a function, like this one:
Disk Graph 1.3.5 MAC OS X Disk Graph 1.3.5 – a tool that allows you to inspect your disk. Size: 5.28 MB Disk Graph is a tool that allows you to inspect your disk and easily find the files that take away most of your disk space. With its beautiful interface and its pie-like graph, locating big files has never been easier. Features of DiskGraph. An example of a graph that is not a unit disk graph is the star K 1,7 with one central node connected to seven leaves: if each of seven unit disks touches a common unit disk, some two of the seven disks must touch each other (as the kissing number in the plane is 6). Therefore, unit disk graphs cannot contain an induced K 1,7 subgraph.
And revolve it around the x-axis like this:
To find its volume we can add up a series of disks:
Each disk's face is a circle:
The area of a circle is π times radius squared:
A = π r2
And the radius r is the value of the function at that point f(x), so:
A = π f(x)2
And the volume is found by summing all those disks using Integration:
Volume = π f(x)2 dx
And that is our formula for Solids of Revolution by Disks
In other words, to find the volume of revolution of a function f(x): integrate pi times the square of the function.
Example: A Cone
Take the very simple function y=x between 0 and b
Rotate it around the x-axis .. and we have a cone!
The radius of any disk is the function f(x), which in our case is simply x
What is its volume? Integrate pi times the square of the function x :
Volume = π x2 dx
First, let's have our pi outside (yum).
Seriously, it is OK to bring a constant outside the integral:
Volume = πx2 dx
Using Integration Rules we find the integral of x2 is x3/3 + C
To calculate this definite integral, we calculate the value of that function for b and for 0 and subtract, like this:
Vicinity 1 1 2 – stay concentrated focused or relaxation. Volume = π (b3/3 − 03/3)
= π b3/3
Compare that result with the more general volume of a cone: Volume = 13 π r2 h
When both r=b and h=b we get:
Volume = 13 π b3
As an interesting exercise, why not try to work out the more general case of any value of r and h yourself?
We can also rotate about other lines, such as x = −1
Example: Our Cone, But About x = −1
So we have this:
Rotated about x = −1 it looks like this:
The cone is now bigger, with its sharp end cut off (a truncated cone)
Let's draw in a sample disk so we can work out what to do:
OK. Now what is the radius? It is our function y=x plus an extra 1:
y = x + 1
Then integrate pi times the square of that function:
Volume =π (x+1)2 dx
Pi outside, and expand (x+1)2 to x2+2x+1 :
Volume = π(x2 + 2x + 1) dx
Capture one 12 0 1 – raw workflow software manual. Using Integration Rules we find the integral of x2+2x+1 is x3/3 + x2 + x + C
And going between 0 and b we get:
Volume = π (b3/3+b2+b − (03/3+02+0))
= π (b3/3+b2+b)
Now for another type of function:
Example: The Square Function
Take y = x2 between x=0.6 and x=1.6
Rotate it around the x-axis:
What is its volume? Integrate pi times the square of x2:
Volume = π (x2)2 dx
Simplify by having pi outside, and also (x2)2 = x4 :
Volume = πx4 dx
The integral of x4 is x5/5 + C
And going between 0.6 and 1.6 we get:
Volume = π ( 1.65/5 − 0.65/5 )
≈ 6.54
Can you rotate y = x2 about x = −1 ?
In summary:
- Have pi outside
- Integrate the function squared
- Subtract the lower end from the higher end
About The Y Axis
We can also rotate about the Y axis:
Example: The Square Function
Take y=x2, but this time using the y-axis between y=0.4 and y=1.4
Rotate it around the y-axis:
And now we want to integrate in the y direction!
So we want something like x = g(y) instead of y = f(x). In this case it is:
x = √(y)
Now integrate pi times the square of √(y)2 (and dx is now dy):
Volume = π √(y)2 dy
Simplify with pi outside, and √(y)2 = y :
Volume = πy dy
The integral of y is y2/2
And lastly, going between 0.4 and 1.4 we get:
Volume = π ( 1.42/2 − 0.42/2 )
≈ 2.83..
Washer Method
Washers: Disks with Holes
What if we want the volume between two functions?
Example: Volume between the functions y=x and y=x3 from x=0 to 1
These are the functions:
Rotated around the x-axis:
The disks are now 'washers':
And they have the area of an annulus:
In our case R = x and r = x3
In effect this is the same as the disk method, except we subtract one disk from another.
And so our integration looks like:
Volume =π (x)2 − π (x3)2 dx
Have pi outside (on both functions) and simplify (x3)2 = x6:
Volume = π x2 − x6 dx
The integral of x2 is x3/3 and the integral of x6 is x7/7
And so, going between 0 and 1 we get:
Volume = π [ (13/3 − 17/7 ) − (0−0) ]
≈ 0.598..
So the Washer method is like the Disk method, but with the inner disk subtracted from the outer disk.
Disk Graph 2 1 3 0 5
Latest version Released:
Dijkstra/A*
![Disk Disk](https://media.cheggcdn.com/media/151/151ad616-e11d-4e47-a94d-072badd3e1cf/phpZGQSyt.png)
Project description
Dijkstar is an implementation of Dijkstra’s single-source shortest-pathsalgorithm. If a destination node is given, the algorithm halts when thatnode is reached; otherwise it continues until paths from the source nodeto all other nodes are found.
Accepts an optional cost (or “weight”) function that will be called onevery iteration.
Also accepts an optional heuristic function that is used to push thealgorithm toward a destination instead of fanning out in everydirection. Using such a heuristic function converts Dijkstra to A* (andthis is where the name “Dijkstar” comes from).
Performance is decent on a graph with 100,000+ nodes. Runs in around .5seconds on average.
Example:
In this example, the edges are just simple numeric values–110, 125,108–that could represent lengths, such as the length of a streetsegment between two intersections. find_path will use these valuesdirectly as costs.
Example with cost function:
The cost function is passed the current node (u), a neighbor (v) of thecurrent node, the edge that connects u to v, and the edge that wastraversed previously to get to the current node.
A cost function is most useful when computing costs dynamically. Ifcosts in your graph are fixed, a cost function will only add unnecessaryoverhead. In the example above, a penalty is added when the street namechanges.
When using a cost function, one recommendation is to compute a base cost whenpossible. For example, for a graph that represents a street network, the basecost for each street segment (edge) could be the length of the segmentmultiplied by the speed limit. There are two advantages to this: the size ofthe graph will be smaller and the cost function will be doing less work, whichmay improve performance.
Graph Export & Import
The graph can be saved to disk (pickled) like so:
And read back like this (load is a classmethod that returns apopulated Graph instance):
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