Work of the vector field-Stokes' Theorem

In summary, we can use the Stokes' Theorem to calculate the work of a given vector field $\overrightarrow{F}$ at the circumference of a circle $x^2+y^2=9$ that is traversed in a counter-clockwise direction. The work is equal to the surface integral of the curl of $\overrightarrow{F}$ multiplied by the normal vector, which can be found by finding the area of the circle. Alternatively, we can switch to polar coordinates and integrate over the disk to find the same result.
  • #1
mathmari
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Use the Stokes' Theorem to calculate the work of the vector field $\overrightarrow{F}=2y \hat{i}+3x\hat{j}-z^2\hat{k}$ at the circumference of the circle $x^2+y^2=9$ that is traversed in a counter-clockwise direction.

$\text{ Work } =\oint_C{\overrightarrow{F}}d\overrightarrow{R}= \iint_S {\nabla \times \overrightarrow{F} \cdot \hat{n}} d \sigma$

$\nabla \times \overrightarrow{F}=\hat{k}$

To find the vector $\hat{n}$:
Since the circle is on the plane $xy$, a perpendicular vector to the plane is a vector on the z-axis, so $\hat{n}=\hat{k}$. Is this correct??

$d\sigma=dxdy$

So $\text{ Work } = \iint_S{1}d \sigma=\text{ area of the circle with radius } 3=9 \pi ^2$.
Is this right? Is using the formula of the area of the circle the only way to calculate the last integral?
 
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  • #2
Yes, that is correct. Of course, you don't have use the formula for the area of the circle. You could actually integrate 1 over the disk. Of course, it would be foolish to do that extra work when you know perfectly well that the integral of 1 over a disk is the area of the disk.
 
  • #3
mathmari said:
So $\text{ Work } = \iint_S{1}d \sigma=\text{ area of the circle with radius } 3=9 \pi ^2$.

Hi! :)

I would make it $9 \pi$ instead. :eek:
Except for that it is all correct!
Is this right? Is using the formula of the area of the circle the only way to calculate the last integral?

Typical alternative is to switch to polar coordinates:
$$\iint_S d\sigma = \int_0^3 \int_0^{2\pi} r d\theta dr = \int_0^3 2\pi r d\theta = \pi r^2 \Big|_0^3 = 9\pi$$
 
  • #4
HallsofIvy said:
Yes, that is correct. Of course, you don't have use the formula for the area of the circle. You could actually integrate 1 over the disk. Of course, it would be foolish to do that extra work when you know perfectly well that the integral of 1 over a disk is the area of the disk.

Nice! Thanks a lot! :eek:

- - - Updated - - -

I like Serena said:
Hi! :)

I would make it $9 \pi$ instead. :eek:
Except for that it is all correct!

Oh yes..you're right!
Great! :eek:
I like Serena said:
Typical alternative is to switch to polar coordinates:
$$\iint_S d\sigma = \int_0^3 \int_0^{2\pi} r d\theta dr = \int_0^3 2\pi r d\theta = \pi r^2 \Big|_0^3 = 9\pi$$

Shouldn't it be:
$\iint_S d\sigma = \int_0^3 \int_0^{2\pi} r d\theta dr = \int_0^3 2\pi r $$dr$ $= \pi r^2 \Big|_0^3 = 9\pi$ ?
 
  • #5
mathmari said:
Shouldn't it be:
$\iint_S d\sigma = \int_0^3 \int_0^{2\pi} r d\theta dr = \int_0^3 2\pi r $$dr$ $= \pi r^2 \Big|_0^3 = 9\pi$ ?

Oh yeah. You're right! :eek:
 

FAQ: Work of the vector field-Stokes' Theorem

What is a vector field?

A vector field is a mathematical concept used to describe the behavior of a vector quantity, such as force or velocity, at various points in space. It is represented graphically by arrows indicating both direction and magnitude.

How does a vector field relate to Stokes' Theorem?

Stokes' Theorem is a mathematical theorem that relates the integral of a vector field over a surface to the line integral of the vector field along the boundary of that surface. In other words, it allows us to calculate the flux of a vector field through a surface by looking at the circulation of the vector field around the boundary of that surface.

What is the significance of Stokes' Theorem in physics?

Stokes' Theorem is used in physics to calculate the work done by a vector field, which can represent things like gravitational or electromagnetic forces. It is an important tool in the study of fluid dynamics, electromagnetism, and other physical phenomena.

How is Stokes' Theorem applied in real-world situations?

Stokes' Theorem is used in many real-world situations, such as calculating the flow of fluids in pipes or around objects, determining the force of a magnetic field on a moving charged particle, or analyzing the motion of air around an airplane wing. It is also used in engineering and design to optimize the shape and performance of various structures.

What are some common misconceptions about Stokes' Theorem?

One common misconception is that Stokes' Theorem only applies to flat surfaces, when in fact it can be applied to surfaces of any shape. Another misconception is that it only applies to two-dimensional vector fields, but it can also be used for three-dimensional vector fields. Finally, some people may mistakenly assume that Stokes' Theorem is only used in advanced mathematics or physics, when in reality it has many practical applications and uses in various fields of study.

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