Roni1985 said:Homework Statement
[tex]\int[/tex][tex]\int[/tex]min(x,y)*dx*dy
both limits are from 0 to 1
what is the general solution of this question ?
Homework Equations
The Attempt at a Solution
I attempted to solve by saying that x<y, finding the integral
and then, y<x and finding its integral.
meaning, I divided my double integral into two double integrals that go from 0 to 1.
Actually, I have never done this before and this was just a guess.
Any help would be appreciated.
dickdick said:That's a really good guess. I can't think of a better way to handle it.
Roni1985 said:Hello,
Well, I think the answer is wrong because I need to get is 1/3 if I remember correctly... but I keep getting 1.
Thanks.
Roni1985 said:Homework Statement
[tex]\int[/tex][tex]\int[/tex]min(x,y)*dx*dy
both limits are from 0 to 1
what is the general solution of this question ?
Homework Equations
The Attempt at a Solution
I attempted to solve by saying that x<y, finding the integral
and then, y<x and finding its integral.
meaning, I divided my double integral into two double integrals that go from 0 to 1.
Actually, I have never done this before and this was just a guess.
Any help would be appreciated.
chiro said:Have you tried using a signum function?
You can use a signum function to calculate either a minimum or a maximum. However I'm not sure if an integral exists for it. If one did I would think that it would be some special function that's non-polynomial.
In case your interested here is how you calculate min(x,y)
min(x,y) = y * (signum(x-y)+1)/2 + x * (signum(y - x)+1)/2. The signum function in this case is defined as:
signum(x) = -1 if x < 0, +1 if x > 0, 0 if x = 0.
This kind of function has a Fourier series so you may want to ask someone who knows about these kind of functions.
dickdick said:Sure. Split it into two integrals. I'd be curious to see how you manage to get 1. Can you show your work? 1/3 is correct.
A double integral is a mathematical concept used in calculus to calculate the volume under a surface in a two-dimensional space. It is represented by a curved "S" symbol and consists of two integrals, an outer integral and an inner integral.
The "min" function in a double integral refers to the minimum value between two variables. In the context of min(x,y), it means that the value of the function at a certain point is equal to the smaller of the two variables x and y.
The "min" function is used in a double integral to determine the range of integration for the inner integral. Since the value of the function is equal to the smaller of the two variables, it helps to set the boundaries for the integration in order to accurately calculate the volume under the surface.
To solve a double integral of min(x,y), you must first set up the integral by determining the range of integration for both the inner and outer integrals. Then, you can evaluate the integral by using the standard techniques of integration, such as substitution or integration by parts.
A double integral of min(x,y) has various applications in real-world problems, such as calculating the volume of a solid with varying cross-sections, finding the center of mass of an object, and determining the probability of a certain event in statistics.