# Vector calculus problem

• formulajoe
In summary, the particle travels from (0,0,0) to (2,1,3) along the segments (0,0,0) ->(0,1,0)->(2,1,0)->(2,1,3). The integral is F dot product with dl. I can't figure this out. I tried doing it the way the book says to but I get a different answer than what the book gives.f

#### formulajoe

f = 2xy in the x dir, (x^2 - z^2) in the y dir, -3xz^2 in the z dir.
particle travles from (0,0,0) to (2,1,3) along the segments (0,0,0) ->(0,1,0)->(2,1,0)->(2,1,3)

the integral is F dot product with dl.
i can't figure this out.

do i dot product each part individually and evaluate? i mean dot the x component of F and dl and evaluate them on an integral?

You have to cut your integral into 3 parts for the three line segments:

$$\int_{(0,0,0)}^{(0,1,0)}\vec F \cdot d\vec l+\int_{(0,1,0)}^{(2,1,0)}\vec F \cdot d\vec l+\int_{(2,1,0)}^{(2,1,3)}\vec F \cdot d\vec l$$

Since each of the the line segments are parallel to either the x,y or x-axis, getting $d\vec l$ is very simple.

Galileo said:
You have to cut your integral into 3 parts for the three line segments:

$$\int_{(0,0,0)}^{(0,1,0)}\vec F \cdot d\vec l+\int_{(0,1,0)}^{(2,1,0)}\vec F \cdot d\vec l+\int_{(2,1,0)}^{(2,1,3)}\vec F \cdot d\vec l$$

Since each of the the line segments are parallel to either the x,y or x-axis, getting $d\vec l$ is very simple.

so the first integral would be from (0,0,0) to (0,1,0), and i would be evaluating 2xy*dx? that's the dot product of Fx and dl sub x. what do i do with the y?
i tried it this way earlier and i get a final answer after evaluating the three integrals of -22. the book says -50.

Note the direction. If you go from (0,0,0) to (0,1,0) in a straight line, then you only have the y-component to worry about (not the x!)
So the first integral is:

$$\int_0^1 F_y(0,y,0)dy=0$$
since the y component of $\vec f$ is zero along this line.
Do the same for the 2nd and 3rd integral. The second yields 4, the third is -54.

ive got another problem, my electro-magnetics prof didnt get a chance to do any examples on this stuff. D = 2 Ro z^2 in the Ro dir, and Ro cos^2 theta in the z dir.
find closed surface integral of D dot Ds.
0<Ro<5
-1<z<1
0<theta<2pi

is this the same thing or is it different?

this is the same problem but this time they want me to evaulate it on the straight line from (0,0,0) to (2,1,3).
everything i can find on the net wants me to parametrize it, and the book doesn't say anything about that. so I am sure there's an easier way to do it.

the book gives the answer of -39.5

if i make z = 3/2 x and y = 1/2 x and substitute those into the integral of F dot dl i end up with an answer of -27.66667. the way I am doing it makes sense from the things I've seen on the net. but the book gives an answer of -39.5. ??