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 🙂
Best Answer
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. $I=\pi$.