Line Integrals of Vector Fields - AP Calculus BC
Card 0 of 25
Evaluate
, where
, and
is the curve given by
,
.
Evaluate , where
, and
is the curve given by
,
.
First we need to evaluate the vector field evaluated along the curve.


Now we need to find the derivative of 

Now we can do the product of
and
.


Now we can put this into the integral and evaluate it.




First we need to evaluate the vector field evaluated along the curve.
Now we need to find the derivative of
Now we can do the product of and
.
Now we can put this into the integral and evaluate it.
Compare your answer with the correct one above
Find the work done by a particle moving in a force field
, moving from
to
on the path given by
.
Find the work done by a particle moving in a force field , moving from
to
on the path given by
.
The formula for work is given by
.
Writing our path in parametric equation form, we have
.
Hence

Plugging this into our work equation, we get


.

The formula for work is given by
.
Writing our path in parametric equation form, we have
.
Hence
Plugging this into our work equation, we get
.
Compare your answer with the correct one above
Evaluate
on the curve
,
, where
.
Evaluate on the curve
,
, where
.
The line integral of a vector field is given by
![\int \vec{F}\cdot d\vec{r} = \int [\vec{F}(\vec{r}(t))\cdot \vec{r'}(t)]dt](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/855983/gif.latex)
So, we must evaluate the vector field on the curve:

Then, we take the derivative of the curve with respect to t:

Taking the dot product of these two vectors, we get

This is the integrand of our integral. Integrating, we get

The line integral of a vector field is given by
So, we must evaluate the vector field on the curve:
Then, we take the derivative of the curve with respect to t:
Taking the dot product of these two vectors, we get
This is the integrand of our integral. Integrating, we get
Compare your answer with the correct one above
Calculate
on the interval
, where
and 
Calculate on the interval
, where
and
To calculate the line integral of the vector field, we must evaluate the vector field on the curve, take the derivative of the curve, and integrate the dot product on the given interval.
The vector field evaluated on the given curve is

The derivative of the curve is given by

The dot product of these is

Integrating this over our given t interval, we get

To calculate the line integral of the vector field, we must evaluate the vector field on the curve, take the derivative of the curve, and integrate the dot product on the given interval.
The vector field evaluated on the given curve is
The derivative of the curve is given by
The dot product of these is
Integrating this over our given t interval, we get
Compare your answer with the correct one above
Evaluate the integral
on the curve
, where
, on the interval 
Evaluate the integral on the curve
, where
, on the interval
The line integral of the vector field is equal to
![\int_{0}^{1}[\vec{F}(\vec{r}(t))\cdot \vec{r'}(t)]dt](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/875547/gif.latex)
The parameterization (using the corresponding elements of the curve) of the vector field is

The derivative of the parametric curve is

Taking the dot product of the two vectors, we get

Integrating this with respect to t on the given interval, we get

The line integral of the vector field is equal to
The parameterization (using the corresponding elements of the curve) of the vector field is
The derivative of the parametric curve is
Taking the dot product of the two vectors, we get
Integrating this with respect to t on the given interval, we get
Compare your answer with the correct one above
Evaluate
, where
, and
is the curve given by
,
.
Evaluate , where
, and
is the curve given by
,
.
First we need to evaluate the vector field evaluated along the curve.


Now we need to find the derivative of 

Now we can do the product of
and
.


Now we can put this into the integral and evaluate it.




First we need to evaluate the vector field evaluated along the curve.
Now we need to find the derivative of
Now we can do the product of and
.
Now we can put this into the integral and evaluate it.
Compare your answer with the correct one above
Find the work done by a particle moving in a force field
, moving from
to
on the path given by
.
Find the work done by a particle moving in a force field , moving from
to
on the path given by
.
The formula for work is given by
.
Writing our path in parametric equation form, we have
.
Hence

Plugging this into our work equation, we get


.

The formula for work is given by
.
Writing our path in parametric equation form, we have
.
Hence
Plugging this into our work equation, we get
.
Compare your answer with the correct one above
Evaluate
on the curve
,
, where
.
Evaluate on the curve
,
, where
.
The line integral of a vector field is given by
![\int \vec{F}\cdot d\vec{r} = \int [\vec{F}(\vec{r}(t))\cdot \vec{r'}(t)]dt](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/855983/gif.latex)
So, we must evaluate the vector field on the curve:

Then, we take the derivative of the curve with respect to t:

Taking the dot product of these two vectors, we get

This is the integrand of our integral. Integrating, we get

The line integral of a vector field is given by
So, we must evaluate the vector field on the curve:
Then, we take the derivative of the curve with respect to t:
Taking the dot product of these two vectors, we get
This is the integrand of our integral. Integrating, we get
Compare your answer with the correct one above
Calculate
on the interval
, where
and 
Calculate on the interval
, where
and
To calculate the line integral of the vector field, we must evaluate the vector field on the curve, take the derivative of the curve, and integrate the dot product on the given interval.
The vector field evaluated on the given curve is

The derivative of the curve is given by

The dot product of these is

Integrating this over our given t interval, we get

To calculate the line integral of the vector field, we must evaluate the vector field on the curve, take the derivative of the curve, and integrate the dot product on the given interval.
The vector field evaluated on the given curve is
The derivative of the curve is given by
The dot product of these is
Integrating this over our given t interval, we get
Compare your answer with the correct one above
Evaluate the integral
on the curve
, where
, on the interval 
Evaluate the integral on the curve
, where
, on the interval
The line integral of the vector field is equal to
![\int_{0}^{1}[\vec{F}(\vec{r}(t))\cdot \vec{r'}(t)]dt](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/875547/gif.latex)
The parameterization (using the corresponding elements of the curve) of the vector field is

The derivative of the parametric curve is

Taking the dot product of the two vectors, we get

Integrating this with respect to t on the given interval, we get

The line integral of the vector field is equal to
The parameterization (using the corresponding elements of the curve) of the vector field is
The derivative of the parametric curve is
Taking the dot product of the two vectors, we get
Integrating this with respect to t on the given interval, we get
Compare your answer with the correct one above
Evaluate
, where
, and
is the curve given by
,
.
Evaluate , where
, and
is the curve given by
,
.
First we need to evaluate the vector field evaluated along the curve.


Now we need to find the derivative of 

Now we can do the product of
and
.


Now we can put this into the integral and evaluate it.




First we need to evaluate the vector field evaluated along the curve.
Now we need to find the derivative of
Now we can do the product of and
.
Now we can put this into the integral and evaluate it.
Compare your answer with the correct one above
Find the work done by a particle moving in a force field
, moving from
to
on the path given by
.
Find the work done by a particle moving in a force field , moving from
to
on the path given by
.
The formula for work is given by
.
Writing our path in parametric equation form, we have
.
Hence

Plugging this into our work equation, we get


.

The formula for work is given by
.
Writing our path in parametric equation form, we have
.
Hence
Plugging this into our work equation, we get
.
Compare your answer with the correct one above
Evaluate
on the curve
,
, where
.
Evaluate on the curve
,
, where
.
The line integral of a vector field is given by
![\int \vec{F}\cdot d\vec{r} = \int [\vec{F}(\vec{r}(t))\cdot \vec{r'}(t)]dt](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/855983/gif.latex)
So, we must evaluate the vector field on the curve:

Then, we take the derivative of the curve with respect to t:

Taking the dot product of these two vectors, we get

This is the integrand of our integral. Integrating, we get

The line integral of a vector field is given by
So, we must evaluate the vector field on the curve:
Then, we take the derivative of the curve with respect to t:
Taking the dot product of these two vectors, we get
This is the integrand of our integral. Integrating, we get
Compare your answer with the correct one above
Calculate
on the interval
, where
and 
Calculate on the interval
, where
and
To calculate the line integral of the vector field, we must evaluate the vector field on the curve, take the derivative of the curve, and integrate the dot product on the given interval.
The vector field evaluated on the given curve is

The derivative of the curve is given by

The dot product of these is

Integrating this over our given t interval, we get

To calculate the line integral of the vector field, we must evaluate the vector field on the curve, take the derivative of the curve, and integrate the dot product on the given interval.
The vector field evaluated on the given curve is
The derivative of the curve is given by
The dot product of these is
Integrating this over our given t interval, we get
Compare your answer with the correct one above
Evaluate the integral
on the curve
, where
, on the interval 
Evaluate the integral on the curve
, where
, on the interval
The line integral of the vector field is equal to
![\int_{0}^{1}[\vec{F}(\vec{r}(t))\cdot \vec{r'}(t)]dt](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/875547/gif.latex)
The parameterization (using the corresponding elements of the curve) of the vector field is

The derivative of the parametric curve is

Taking the dot product of the two vectors, we get

Integrating this with respect to t on the given interval, we get

The line integral of the vector field is equal to
The parameterization (using the corresponding elements of the curve) of the vector field is
The derivative of the parametric curve is
Taking the dot product of the two vectors, we get
Integrating this with respect to t on the given interval, we get
Compare your answer with the correct one above
Evaluate
, where
, and
is the curve given by
,
.
Evaluate , where
, and
is the curve given by
,
.
First we need to evaluate the vector field evaluated along the curve.


Now we need to find the derivative of 

Now we can do the product of
and
.


Now we can put this into the integral and evaluate it.




First we need to evaluate the vector field evaluated along the curve.
Now we need to find the derivative of
Now we can do the product of and
.
Now we can put this into the integral and evaluate it.
Compare your answer with the correct one above
Find the work done by a particle moving in a force field
, moving from
to
on the path given by
.
Find the work done by a particle moving in a force field , moving from
to
on the path given by
.
The formula for work is given by
.
Writing our path in parametric equation form, we have
.
Hence

Plugging this into our work equation, we get


.

The formula for work is given by
.
Writing our path in parametric equation form, we have
.
Hence
Plugging this into our work equation, we get
.
Compare your answer with the correct one above
Evaluate
on the curve
,
, where
.
Evaluate on the curve
,
, where
.
The line integral of a vector field is given by
![\int \vec{F}\cdot d\vec{r} = \int [\vec{F}(\vec{r}(t))\cdot \vec{r'}(t)]dt](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/855983/gif.latex)
So, we must evaluate the vector field on the curve:

Then, we take the derivative of the curve with respect to t:

Taking the dot product of these two vectors, we get

This is the integrand of our integral. Integrating, we get

The line integral of a vector field is given by
So, we must evaluate the vector field on the curve:
Then, we take the derivative of the curve with respect to t:
Taking the dot product of these two vectors, we get
This is the integrand of our integral. Integrating, we get
Compare your answer with the correct one above
Calculate
on the interval
, where
and 
Calculate on the interval
, where
and
To calculate the line integral of the vector field, we must evaluate the vector field on the curve, take the derivative of the curve, and integrate the dot product on the given interval.
The vector field evaluated on the given curve is

The derivative of the curve is given by

The dot product of these is

Integrating this over our given t interval, we get

To calculate the line integral of the vector field, we must evaluate the vector field on the curve, take the derivative of the curve, and integrate the dot product on the given interval.
The vector field evaluated on the given curve is
The derivative of the curve is given by
The dot product of these is
Integrating this over our given t interval, we get
Compare your answer with the correct one above
Evaluate the integral
on the curve
, where
, on the interval 
Evaluate the integral on the curve
, where
, on the interval
The line integral of the vector field is equal to
![\int_{0}^{1}[\vec{F}(\vec{r}(t))\cdot \vec{r'}(t)]dt](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/875547/gif.latex)
The parameterization (using the corresponding elements of the curve) of the vector field is

The derivative of the parametric curve is

Taking the dot product of the two vectors, we get

Integrating this with respect to t on the given interval, we get

The line integral of the vector field is equal to
The parameterization (using the corresponding elements of the curve) of the vector field is
The derivative of the parametric curve is
Taking the dot product of the two vectors, we get
Integrating this with respect to t on the given interval, we get
Compare your answer with the correct one above