MHB Work of the vector field-Stokes' Theorem

[mathmari]

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?

 

[HallsofIvy]

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.

 

[I like Serena]

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.
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:
$$\iint_S d\sigma = \int_0^3 \int_0^{2\pi} r d\theta dr = \int_0^3 2\pi r d\theta = \pi r^2 \Big|_0^3 = 9\pi$$

 

[mathmari]

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!

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Hi! :)

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

Oh yes..you're right!
Great!

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$ $= \pi r^2 \Big|_0^3 = 9\pi$ ?

 

[I like Serena]

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!