How Well Are You Known?
- Thread starter Bellaabzug21
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[-lim h->(pi/2) tan x]/10
You wanna go, bro? Huh? You want a math war? Is that what you want? You think you stand a chance against me? HAVE YOU SEEN MY SIG?!?!
You wanna go, bro? Huh? You want a math war? Is that what you want? You think you stand a chance against me? HAVE YOU SEEN MY SIG?!?!
9/10
Integrals are more annoying than derivatives, but not as much so as MOAR INTEGRALS!!!11!!!!1
Integrals are more annoying than derivatives, but not as much so as MOAR INTEGRALS!!!11!!!!1
9/10
In order to find the mass of an arbitrary 3D shape with a given density function, you first need to first define the boundaries in a particular way, then integrate the density function multiplied by a "constant" expression that depends on the coordinate system you're using (for Cartesian coords, it's just 1; for spherical, it's rho^2 * sin(theta)) three times, keeping track of which variables you're treating as constants and which ones drop out because you're integrating them, and then you might find that one of the integrals is impossible to solve, so you have to go back to square one and define the boundaries a different way. There's six different ways to do that (for three dimensions), so if you've got a problem where only one works and you don't have a very clear idea of what you're doing, you might guess wrong five times before finding the right one.
I'm glad I'm through with Calc III.
The complexity of solving a nested integral problem goes up much faster than the number of integrals involved.
In order to find the mass of an arbitrary 3D shape with a given density function, you first need to first define the boundaries in a particular way, then integrate the density function multiplied by a "constant" expression that depends on the coordinate system you're using (for Cartesian coords, it's just 1; for spherical, it's rho^2 * sin(theta)) three times, keeping track of which variables you're treating as constants and which ones drop out because you're integrating them, and then you might find that one of the integrals is impossible to solve, so you have to go back to square one and define the boundaries a different way. There's six different ways to do that (for three dimensions), so if you've got a problem where only one works and you don't have a very clear idea of what you're doing, you might guess wrong five times before finding the right one.
I'm glad I'm through with Calc III.
-e^(i pi)
What if you want to compute the integral of some function on a surface? Surch as the surface area which is just constant 1 integrated over the surface.
You need to parametrize the surface, making sure that your parametrization is an embedding. You then compute the square root of the Gram determinant g and then you can proceed to compute the integral like Someone Else 37 described.
The difficulty in often lies in handling the integral boundaries correctly. It's the easiest if you an transform it so that they become a cuboid.
What if you want to compute the integral of some function on a surface? Surch as the surface area which is just constant 1 integrated over the surface.
You need to parametrize the surface, making sure that your parametrization is an embedding. You then compute the square root of the Gram determinant g and then you can proceed to compute the integral like Someone Else 37 described.
The difficulty in often lies in handling the integral boundaries correctly. It's the easiest if you an transform it so that they become a cuboid.
9/10
at least not how it is commonly defined.
and even if it was, if you divided it by twice infinity it would still be zero.
Anway, it is indeed undefined.
not really
at least not how it is commonly defined.and even if it was, if you divided it by twice infinity it would still be zero.
Anway, it is indeed undefined.