# Line integral problems

• EdisT
In summary, the conversation discussed how to calculate a line integral along the curve y=x3 in the xy-plane. The expert summarizer provided guidance on how to decompose a vector and set up the integral expression using the curve's components. They also reminded the speaker to include the j component in their calculations. Finally, the expert confirmed the correct integral expression to be used.

#### EdisT

Thread moved from the technical forums, so no HH Template is shown.
I'm used to parameterizing however I'm not sure how to solve these types of problems, any help would be much appreciated.

1) Calculate the line integral ∫v⋅dr along the curve y=x3 in the xy-plane when -1≤x≤2 and v=xyi+x2j

2) a) Find the work that the force F = (y2+5)i+(2xy-8)j carries out along two paths ABC and ADC which are composed of perpendicular lines between the points A,B,C,D. A,B,C,D are corners in a square and the corners have the coordinates (0,0),(1,0),(1,1),(0,1) respectively

b) Calculate the work done along a straight line from A to C.

c) Since the work done appears to be independent of the path taken, we expect that the force can be written as the gradient to a potential V. Find the potential-function and show that the difference between the potential at A and C are equivalent to the work done.

Attempt at reaching a solution:

1) I substituted y=x3 into the equation then did a normal integral with respect to x. It doesn't seem right to me however I'm used to dS and t so these problems just messed my thinking up.

2) Similar thinking to 1) is messing me up I think. Did normal integrals with respect to x and y.

EdisT said:
I'm used to parameterizing however I'm not sure how to solve these types of problems, any help would be much appreciated.

1) Calculate the line integral ∫v⋅dr along the curve y=x3 in the xy-plane when -1≤x≤2 and v=xyi+x2j

Here, r is a vector between the origin and a point (x,y) on the curve y = x3.

Do you know how to set up the components of the vector r to describe y = x3 ?

Once you find r, you can calculate dr and substitute into the line integral expression. After calculating v ⋅ dr, then you can start to treat the line integral like a regular integral.

• EdisT
SteamKing said:
Here, r is a vector between the origin and a point (x,y) on the curve y = x3.

Do you know how to set up the components of the vector r to describe y = x3 ?

Once you find r, you can calculate dr and substitute into the line integral expression. After calculating v ⋅ dr, then you can start to treat the line integral like a regular integral.
No I have no idea how to split it up, the dot product should be easy enough but I'm more used to x(t), y(t) being given.. How would you split it up?

EdisT said:
No I have no idea how to split it up, the dot product should be easy enough but I'm more used to x(t), y(t) being given.. How would you split it up?
Do you know how to decompose a vector into its components?

• EdisT
SteamKing said:
Do you know how to decompose a vector into its components?
It's been a while... I don't quite remember

EdisT said:
It's been a while... I don't quite remember
You can look up how on the internet. That's what it's for.

Remember, i, j , and k are the unit vectors for a general vector r, such that r = x i + y j + z k

You have a curve described by y = x3. Care to take a guess as to how this curve could be turned into a vector expression?

• EdisT
SteamKing said:
You can look up how on the internet. That's what it's for.

Remember, i, j , and k are the unit vectors for a general vector r, such that r = x i + y j + z k

You have a curve described by y = x3. Care to take a guess as to how this curve could be turned into a vector expression?
r=x^3i+yj ?

EdisT said:
r=x^3i+yj ?
What's the independent variable in y = x3 ?

• EdisT
SteamKing said:
What's the independent variable in y = x3 ?
x would be the independent variable, so it would be x^3j?

EdisT said:
x would be the independent variable, so it would be x^3j?
Correct. r = x i + x3 j + 0 k

Can you calculate dr now?

• EdisT
SteamKing said:
Correct. r = x i + x3 j + 0 k

Can you calculate dr now?
dr = (i + 3x2) dx

The dot product would be v⋅dr = xy+3x4 dx

The integral would then be: ## \int_{-1}^2 xy+3x^4 \, dx ## ?

EdisT said:
dr = (i + 3x2) dx

Don't forget the j component for the 3x2.
The dot product would be v⋅dr = xy+3x4 dx

The integral would then be: ## \int_{-1}^2 xy+3x^4 \, dx ## ?

What does y equal when traversing the curve y = x3 ?

• EdisT
SteamKing said:
Don't forget the j component for the 3x2.

What does y equal when traversing the curve y = x3 ?

Of course, the integral becomes x^4+3x^4 dx. thank you!