## Vector Calculus Question about Surface Integrals

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
 PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks
 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?
 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.

Recognitions:
Gold Member
Science Advisor
Staff Emeritus

## 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.

 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?
 Thread Tools

 Similar Threads for: Vector Calculus Question about Surface Integrals Thread Forum Replies Calculus & Beyond Homework 4 Calculus & Beyond Homework 1 Calculus & Beyond Homework 0 Calculus & Beyond Homework 12 Calculus & Beyond Homework 9