Calculating Line Integral in xy-Plane

• KUphysstudent
In summary, the line integral is calculated by parameterizing x and y in terms of t and taking the derivative of x and y with respect to t. Then, the dot product of the vector field v and position vector dr is taken and integrated with respect to t from the lower bound of -1 to the upper bound of 2. The final substitution of x = t and y = t3 should be corrected to x = t and y = t^3.
KUphysstudent

Homework Statement

Calculate the line integral ° v ⋅ dr along the curve y = x3 in the xy-plane when -1 ≤ x ≤ 2 and v = xy i + x2 j.

Note: Sorry the integral sign doesn't seem to work it just makes a weird dot, looks like a degree sign, ∫.2. The attempt at a solution
I have to write something along with the calculations for exams so I do that here as well, it is probably not correct.

parameterize x and y in terms of t because we need to take the integral of the dot product of the position vector dr with the vector field v. In this case we can say that x = t which in return gives us y = x3 = t3. Now taking the derivative with of x and y with respect to t, dx/dt = 1 dy/dt = 3t2 which we can put into the equation for dr, which is dr = dx i+ dy j+ dz k → dr = 1 dt i+ 3t2 dt j + 0 dt k
insert in (integralsign)∫ v ⋅ dr → (integralsign)∫ xy i+x2 j ⋅ 1 i+ 3t2 j , with the lower bound of integration being t = -1 and the upper bound t = 2.
The dot product of [a, b, c] ⋅ [d, e, f] = [ad, be, cf]. therefore (integralsign)∫ v ⋅ dr = xy + x21 dt i+ 3t2 dt j = xy dt + 3t2 x2 dt

3. Relevant equations

But now I'm completely lost and don't understand what I'm supposed to do with x's and y's, shouldn't they have been in terms of t's as well? My book has one example which is much easier than this one since the vector field is just v = yi in that one.

Anyone who can teach me what to do please let me know.

Do I just substitute xy and x2 with my parameter x = t and y = t3, so I get xy = t3 and x2 = t2 → (integralsign)∫ t3 dt + 3t2 t2 dt ?

KUphysstudent said:

Homework Statement

Calculate the line integral ° v ⋅ dr along the curve y = x3 in the xy-plane when -1 ≤ x ≤ 2 and v = xy i + x2 j.

Note: Sorry the integral sign doesn't seem to work it just makes a weird dot, looks like a degree sign, ∫.2. The attempt at a solution
I have to write something along with the calculations for exams so I do that here as well, it is probably not correct.

parameterize x and y in terms of t because we need to take the integral of the dot product of the position vector dr with the vector field v. In this case we can say that x = t which in return gives us y = x3 = t3. Now taking the derivative with of x and y with respect to t, dx/dt = 1 dy/dt = 3t2 which we can put into the equation for dr, which is dr = dx i+ dy j+ dz k → dr = 1 dt i+ 3t2 dt j + 0 dt k
insert in (integralsign)∫ v ⋅ dr → (integralsign)∫ xy i+x2 j ⋅ 1 i+ 3t2 j , with the lower bound of integration being t = -1 and the upper bound t = 2.
The dot product of [a, b, c] ⋅ [d, e, f] = [ad, be, cf]. therefore (integralsign)∫ v ⋅ dr = xy + x21 dt i+ 3t2 dt j = xy dt + 3t2 x2 dt

3. Relevant equations

But now I'm completely lost and don't understand what I'm supposed to do with x's and y's, shouldn't they have been in terms of t's as well? My book has one example which is much easier than this one since the vector field is just v = yi in that one.

Anyone who can teach me what to do please let me know.

Do I just substitute xy and x2 with my parameter x = t and y = t3, so I get xy = t3 and x2 = t2 → (integralsign)∫ t3 dt + 3t2 t2 dt ?
Yes. Everything you have done looks correct so far except for your final substitution. You have the right idea, but check for errors.

if you calculate the integral with this step, (integralsign)∫ t3 dt + 3t2 t2 dt from -1 to 2 you get 25.25

I forgot to add that if you graph y=x^3 and look at the area between x = 1 and x = 2 it looks like it adds to about 4, which is quite a bit away from 25.25

Last edited:

1. What is a line integral in the xy-plane?

A line integral in the xy-plane is a mathematical tool used to calculate the total effect of a vector field along a given path in the xy-plane. It takes into account both the magnitude and direction of the vector field.

2. How is a line integral in the xy-plane calculated?

A line integral in the xy-plane is calculated by first parameterizing the given path in terms of one variable, usually t. The integral is then evaluated using the parameterization and the vector field function. The result is a single value representing the total effect of the vector field along the path.

3. What is the difference between a line integral and a double integral?

A line integral is a one-dimensional integral, meaning it takes into account the behavior of a function along a specific path. A double integral, on the other hand, is a two-dimensional integral that considers the behavior of a function over a region in the xy-plane. In other words, a line integral is a special case of a double integral.

4. When is it necessary to calculate a line integral in the xy-plane?

A line integral in the xy-plane is often used in physics and engineering to calculate work, flux, or circulation of a vector field. It is also used in the study of complex functions and in the field of differential geometry.

5. Are there any practical applications of calculating line integrals in the xy-plane?

Yes, there are many practical applications of calculating line integrals in the xy-plane. Some examples include calculating the work done by a force on a moving object, finding the electric field along a specific path in an electric field, and determining the amount of fluid flowing through a pipe. It is also commonly used in the fields of mechanics, electromagnetism, and fluid dynamics.

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