Integrating $\vec{F}$ Along C

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In summary, the given equation and line segment are used to find the integral \int\limits_C \vec{F} d\vec{r}. The line is parametrized using t and the integral is set up using the given equation. The final answer is \frac{9}{5} - 4e^6.
  • #1
-EquinoX-
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Homework Statement



say I have [tex] \vec{F} = ye^x\vec{i} + e^x\vec{j} [/tex] and C the line segment from the point (1,4,-2) to the point (6,7,-2).I need to find:

[tex] \int\limits_C \vec{F} d\vec{r} [/tex]

Homework Equations


The Attempt at a Solution



First I parametrize C as:

[tex] \vec{r} = 6 + 5t \vec{i} + 7+3t \vec{j} -2 \vec{k} [/tex]

Then I set up the integral as:

[tex] \int_{-1}^{0} (7+3t)e^{6+5t} \vec{i} + e^{6+5t} \vec{j}) \cdot (5\vec{i} + 3\vec{j}) dt [/tex]

is this correct so far?
 
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  • #2
-EquinoX- said:

Homework Statement



say I have [tex] \vec{F} = ye^x\vec{i} + e^x\vec{j} [/tex] and C the line segment from the point (1,4,-2) to the point (6,7,-2).


I need to find:

[tex] \int\limits_C \vec{F} d\vec{r} [/tex]

Homework Equations





The Attempt at a Solution



First I parametrize C as:

[tex] \vec{r} = 6 + 5t \vec{i} + 7+3t \vec{j} -2 \vec{k} [/tex]
I assume you mean [itex]\vec{r}= (6+ 5t)\vec{i}+ (7+ 3t)\vec{j}- 2\vec{k}[/itex].
When t= 0 this is [itex]\vec{r}= 6\vec{i}+ 7\vec{j}- 2\vec{k}[/itex] which is one of the given points. When t= -1 this is [itex]\vec{r}= \vec{i}+ 4\vec{j}- 2\vec{k}[/itex] which is the other point. Yes, this is a vector equation for the line- although the choice of t= -1 for one of the is peculiar!

Then I set up the integral as:

[tex] \int_{-1}^{0} (7+3t)e^{6+5t} \vec{i} + e^{6+5t} \vec{j}) \cdot (5\vec{i} + 3\vec{j}) dt [/tex]

is this correct so far?
Yes, that is correct. You will probably wnat to "separate" [itex]e^{6+ 5t}[/itex] as [itex]e^6 e^{5t}[/itex].
 
  • #3
and so after doing all the computation I got:

[tex] \frac{12(y^2-x^2)}{(x^2+y^2)^2} [/tex] being evaluated from -1 to 0. Is this correct? Reason I ask is because the integral was kind of complex
 
  • #4
Why did you change from t to x and y?
 
  • #5
oops.. I guess you're right
 
  • #6
I got [tex] \frac{9}{5} - 4e [/tex]
 

1. What does it mean to integrate a vector field along a curve?

Integrating a vector field along a curve means finding the total effect of the vector field along that specific curve. This involves finding the line integral of the vector field, which takes into account both the magnitude and direction of the vector field along the curve.

2. Why is it important to integrate a vector field along a curve?

Integrating a vector field along a curve allows us to determine the work done by the vector field along that curve, which can have practical applications in physics and engineering. It also helps us understand the behavior of the vector field in a specific region.

3. What is the difference between integrating a scalar field and a vector field?

Integrating a scalar field involves finding the total effect of a scalar quantity (such as temperature or pressure) along a certain path. Integrating a vector field, on the other hand, involves finding the total effect of a vector quantity (such as force or velocity) along a certain path.

4. Can you integrate a vector field along any curve?

No, you cannot integrate a vector field along any curve. The curve must be smooth and the vector field must be continuous in order for the integration to be valid. Additionally, the curve must be within the domain of the vector field.

5. How do you calculate the line integral of a vector field along a curve?

To calculate the line integral, you must first parameterize the curve and then plug the parameterization into the formula for the line integral. This involves taking the dot product of the vector field and the tangent vector of the curve and integrating over the given limits of the curve.

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