# Vector Calculus Question about Surface Integrals

by Conaissance99
Tags: calculus, integrals, surface, vector
 P: 2 Why is it that when the force field is z^2 and you take the surface integral over a sphere of radius a using spherical coordinates, that yields the flux to be (4pi a^3 )/ 3 BUT in a calculus book, the force field is z instead of z^2 evaluated using polar coordinates and it yields the same amount of flux, (4pi a^3 )/ 3. How can this be when the force is different (z^2 instead of z?) Isn't it when you for example, get five times the force, like 5z you would get the answer multiplied by a factor of 5. When you square z it should come out to be different shouldn't it? Any help greatly appreciated. Thanks in advance
 P: 2 Please note that this is not a homework question. Simply a question that if you change the value of the force in your surface integral in this case, shouldn't the answer be different?
 P: 350 Sorry I don't understand the question. What exactly is the integral being calculated? Alternatively, what exactly is the physical quantity being calculated? If it is the flux of a force field across the sphere, then what is the force field? You need to say what direction it is pointing.
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Vector Calculus Question about Surface Integrals

 Quote by Conaissance99 Why is it that when the force field is z^2 and you take the surface integral over a sphere of radius a using spherical coordinates, that yields the flux to be (4pi a^3 )/ 3 BUT in a calculus book, the force field is z instead of z^2 evaluated using polar coordinates and it yields the same amount of flux, (4pi a^3 )/ 3. How can this be when the force is different (z^2 instead of z?) Isn't it when you for example, get five times the force, like 5z you would get the answer multiplied by a factor of 5. When you square z it should come out to be different shouldn't it? Any help greatly appreciated. Thanks in advance
First, this doesn't make sense. Force is a vector quantity and the force field must be a vector function, not scalar. I will assume you mean something like <0, 0, z>. In that case, "the force field is z instead of z^2 evaluated using polar coordinates and it yields the same amount of flux, (4pi a^3 )/ 3" is incorrect. The integral over the top part of the sphere, z> 0, will cancel the integral over the bottom part, t< 0, and the integral is 0.
P: 428
 Quote by Conaissance99 Why is it that when the force field is z^2 and you take the surface integral over a sphere of radius a using spherical coordinates, that yields the flux to be (4pi a^3 )/ 3 BUT in a calculus book, the force field is z instead of z^2 evaluated using polar coordinates and it yields the same amount of flux, (4pi a^3 )/ 3. How can this be when the force is different (z^2 instead of z?) Isn't it when you for example, get five times the force, like 5z you would get the answer multiplied by a factor of 5. When you square z it should come out to be different shouldn't it? Any help greatly appreciated. Thanks in advance
What do you mean by polar coordinates? Did you mean spherical, cylindrical?

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