need help on path and vector field problem

**yzc717** · December 18th 2008, 01:27 AM

how do I find a scalar function on problem 4?

**JD-Styles** · December 18th 2008, 04:44 AM

Sorry the answer I gave you was wrong. Don't use it!

**yzc717** · December 18th 2008, 11:29 PM

do i have to draw the vector field first, then i can compute the volume?

how do i do it?

**mr fantastic** · December 19th 2008, 12:11 AM

Originally Posted by JD-Styles

$\nabla f=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})$

So, if we're working in the reverse order:

$f=\int \frac{\partial f}{\partial x}dx+ \int \frac{\partial f}{\partial y}dy+\int \frac{\partial f}{\partial z}dz+C$

So basically, integrate $2z^2$ wrt x, $(e^zcosy+x^3)$ wrt y, and $(e^zsiny-5z)$ wrt z.

I'm not sure the OP will get the right idea from this.

IF a scalar function f exists such that $F = \nabla f$ then a solution to the following simultaneous partial differential equations will exist:

$\frac{\partial f}{\partial x} = 2z^2$ .... (1)

$\frac{\partial f}{\partial y} = e^z \cos y + x^3$ .... (2)

$\frac{\partial f}{\partial z} = e^z \sin y - 5z$ .... (3)

**Mush** · December 19th 2008, 03:03 AM

Originally Posted by yzc717

how do I find a scalar function on problem 4?

$\nabla f = F$

Now... nabla f is given by:

$\frac{d f}{dx}\vec{i}+\frac{d f}{dy}\vec{j}+ \frac{d f}{dz}\vec{k}$

And you know that $F = 2z^2\vec{i} + (e^z cos(y) + x^3)\vec{j} + (e^z sin(y) -5z)\vec{k}$

So, in essence, you have to find a scalar function f that satisfies these equations:

$\frac{d f}{dx}=2z^2$

$\frac{d f}{dy}=(e^z cos(y) + x^3)$

$\frac{d f}{dz}=(e^z sin(y) -5z)$

A little integration is needed...

**JD-Styles** · December 19th 2008, 04:53 AM

I'll try to redeem myself here. I'd argue that no such function exists because:

Suppose such a function did exist. Look at the partial derivative wrt y:

$\frac{\partial f}{\partial y}=e^zcosy+x^3$

$f(x,y,z)=e^zsiny+x^3y+h(x,z)$ (2)

Here, h(x,z) is a function that depends only on x and z (which if derived wrt y becomes zero). Now if we take the partial derivative wrt x we should get $2z^2$ :

$\frac{\partial f}{\partial x}=3x^2y+\frac{\partial h(x,z)}{\partial x}=2z^2$

$\frac{\partial h(x,z)}{\partial x}=2z^2-3x^2y$

$h(x,z)=2xz^2-x^3y+g(z)$

But this is a contradiction because we have a term with y in a function that can only contain x and z. Thus no such function exists.

Another, maybe simpler, way to look at it is to argue that if such a function existed then:

$\frac{\partial ^2f}{\partial y \partial x}=\frac{\partial ^2f}{\partial x \partial y}$

$\frac{\partial }{\partial y}(2z^2)=\frac{\partial }{\partial x}(e^zcosy+x^3)$

$0=3x^2$ which is clearly a contradiction.

**yzc717** · December 19th 2008, 08:11 AM

how to find the volume,

and i got the first one, since curl does not equal to zero, f is not a gradient vector field, it not conservative.

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