# Line integrals

christopnz

## Homework Statement

find $$\int$$ 3x^2 dx +2yz dy + y^2 dz between the points A=(0,1,2) and b=(1,-1,7) by finding a suitable f

## Homework Equations

3. attempt
Isnt f just the partial intergral of the above equation?
f =x^3 + zy^2 + zy^2

but solution for f is f =x^3 + zy^2why is the y^2 term left out

Last edited:

eok20
they want you to find a scalar function f such that grad f = (3x^2, 2yz, y^2) and then use the fundamental theorem of line integrals

EngageEngage
It wouldn't be the partial integral (it sometimes is but not usually) of all the terms at once. To find this function f, you first integrate your the coefficient of the dx term with respect to x. Then you have
f = int 3x^2 dx + K(y,z). The constant is of y and z. Then you diffferentiate this equation that you just got with respect to y. And you set it equal to the coefficient of the dy term to find K(y,z). When you're done with this part you will have f = 3x^2 + something + K(z). Then you do the same thing again. Just to make sure you get it right, take its gradient when you're done to see if you get the right vector field.

Homework Helper

## Homework Statement

find $$\int$$ 3x^2 dx +2yz dy + y^2 dz between the points A=(0,1,2) and b=(1,-1,7) by finding a suitable f

## Homework Equations

3. attempt
Isnt f just the partial intergral of the above equation?
f =x^3 + zy^2 + zy^2

but solution for f is f =x^3 + zy^2why is the y^2 term left out
You have zy2 twice: f(x,y,z)= x3+ 2zy2. You should be able to see by differenting that df= 3x2dx+ 4zy dy+ 2y2dz which is not what you want!

No, you do NOT just integrate the dx, dy, and dz parts separately- in this particular example, you get both the dy and dz functions from the same expression.

You know that $\partial f/\partial x= 3x^2$ so you know that f= x3 "plus a constant". But since you are using "partial" integration, that "constant" could be any function of y and z: if f(x,y,z)= x3+ g(y,z), for g any function of y and z, the fx= 3x2.

You also know that, for the same f, fy= gy(y,z)= 2yz. From that, g(y,z)= y2z + some function of z only! If g(y,z)= y2z+ h(z), for h any function of z only, then gy= 2yz.

Again, for that same g, gz= y2+ h'(z)= y2 so h'(z)= 0. That is, h(z) really is a constant: C. Then g(y,z)= y2z+ C and so f(x,y,z)= x3+ y2z.