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There is a hemisphere, radius $1$, centred at $(0,0,0)$, where the vector field is $$\vec F = \Big(x^3+\frac{z^4}{4}\Big) \hat i + 4x \hat j + (xz^3+z^2) \hat k$$

Verify Stokes' theorem for this hemisphere.

I'm attempting to compute $$\iint_S (\vec \nabla \times \vec F) \ \cdot \ \mathrm d \vec S$$

So far my attempt is:

$$(\vec \nabla \times \vec F) = \begin{vmatrix} \hat i & \hat j & \hat k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^3+\frac{z^4}{4} & 4x & xz^3+z^2 \end{vmatrix} = (0,0,4)$$

Parametrise the hemisphere surface: $$\Phi(x,y) = \Big(x,y,\sqrt{1-x^2-y^2}\Big)$$

Finding the normal vector: $$\vec T_x = \Big(\frac{\partial x}{\partial x}, \frac{\partial y}{\partial x}, \frac{\partial z}{\partial x} \Big) = \Bigg(1,0,\frac{-x}{\sqrt{1-x^2-y^2}} \Bigg)$$

$$\vec T_y = \Big(\frac{\partial x}{\partial y}, \frac{\partial y}{\partial y}, \frac{\partial z}{\partial y} \Big) = \Bigg(0,1,\frac{-y}{\sqrt{1-x^2-y^2}} \Bigg)$$

$$\vec T_x \times \vec T_y = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 0 & \frac{-x}{\sqrt{1-x^2-y^2}} \\ 0 & 1 & \frac{-y}{\sqrt{1-x^2-y^2}}\end{vmatrix} = \Bigg(\frac{x}{\sqrt{1-x^2-y^2}}\Bigg)\hat i + \Bigg(\frac{y}{\sqrt{1-x^2-y^2}}\Bigg)\hat j + \hat k$$

and $$\mathrm d \vec S = (\vec T_x \times \vec T_y) \ \mathrm dx \mathrm dy$$

So $$\iint_S (\vec \nabla \times \vec F) \ \cdot \ \mathrm d \vec S = \iint_S (0,0,4) \cdot \Bigg(\frac{x}{\sqrt{1-x^2-y^2}}, \frac{y}{\sqrt{1-x^2-y^2}}, 1\Bigg) \mathrm dx \mathrm dy $$

$$ = \iint_S 4 \mathrm dx \mathrm dy$$ $$ = 4 \iint_S \mathrm dx \mathrm dy$$ $$ = 4 \times \text{(Surface area of upper hemisphere radius 1)}$$ $$ = 8\pi$$

But the answer is $4\pi$

So I'm not sure exactly where I went wrong but any help would be appreciated, thanks :)

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    $\begingroup$ Surface of upper hemisphere is $4 \pi / 2$ right? $\endgroup$ – Stockfish Oct 23 '18 at 13:04
  • $\begingroup$ hmm good point but still off by a factor of 2 $\endgroup$ – Patrick Jankowski Oct 23 '18 at 13:05
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Maybe i'm mistaking, but your integral $$ I = \iint_S dxdy $$ runs over the corresponding parameter space (note the integration variables), which is the $\textbf{base}$ of the hemisphere, having the surface of a circle of radius one, i.e. $I=\pi$.

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  • $\begingroup$ ohh so i'm not integrating over the surface but over the base? $\endgroup$ – Patrick Jankowski Oct 23 '18 at 13:24
  • $\begingroup$ Yes, after all this is how your parametrization $\Phi$ works. $\endgroup$ – denklo Oct 23 '18 at 13:25
  • $\begingroup$ ah I see what you mean about the integration variables, yeah that's absolutely an integral in the x-y plane only, thanks man for the help :) $\endgroup$ – Patrick Jankowski Oct 23 '18 at 13:26

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