work for closed curve x²+y²=1 in the plane z=0

Hi I am not sure if i am doing this right.The question im having issues with is in green text.

The text in blue are the other parts of the question(I included it because it may be necessary...but was easy to calculate.)..

F is a force field . F=2xyzi+ (x²z+2y)j+(x²y-2z)k

Calculate ∇xF,and show its conservative.

find potential function Φ for F,

Hence find the work done by F.around the closed curve x²+y²=1 in the plane z=0, transversed in a counterclockwise direction ,starting and ending at the point (1,0,0)

(Clearly this is the base of a cylinder.)

Here is my method :

letting x=costi (α)

y=sintj

since x²+y²=1,y=√1-x² (β)

then subbing (α) into (B)

I got r=sin²ti+cos²tj=1 t ranges from 0 to 2π

then dr=(2sintcost)idt-(2costsint)jdt

Re: work for closed curve x²+y²=1 in the plane z=0

Quote:

Originally Posted by

**n22** Hi I am not sure if i am doing this right.The question im having issues with is in green text.

The text in blue are the other parts of the question(I included it because it may be necessary...but was easy to calculate.)..

F is a force field . F=2xyzi+ (x²z+2y)j+(x²y-2z)k

Calculate ∇xF,and show its conservative.

find potential function Φ for F,

Hence find the work done by F.around the closed curve x²+y²=1 in the plane z=0, transversed in a counterclockwise direction ,starting and ending at the point (1,0,0)

(Clearly this is the base of a cylinder.)

Here is my method :

letting x=costi (α)

y=sintj

since x²+y²=1,y=√1-x² (β)

then subbing (α) into (B)

I got r=sin²ti+cos²tj=1 t ranges from 0 to 2π

then dr=(2sintcost)idt-(2costsint)jdt

Isn't the work done 0 just because you have a closed curve and a conservative field?

.

Re: work for closed curve x²+y²=1 in the plane z=0

[QUOTE=zzephod;791883]Isn't the work done 0 just because you have a closed curve and a conservative field?

Hi, yes you are correct.... but what if it were not a closed curve...............then how would one approach this question?

Re: work for closed curve x²+y²=1 in the plane z=0

[QUOTE=n22;791951] Quote:

Originally Posted by

**zzephod** Isn't the work done 0 just because you have a closed curve and a conservative field?

Hi, yes you are correct.... but what if it were not a closed curve...............then how would one approach this question?

You use $\displaystyle x=\cos(\theta)$, $\displaystyle y=\sin(\theta)$, $\displaystyle z=0$ and integrate $\displaystyle F(\cos(\theta).\sin(\theta),0)$ with respect to $\displaystyle \theta$ from $\displaystyle \theta=0$ to $\displaystyle 2\pi$.

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