Use Stokes' theorem to evaluate
$$ \iint_S \operatorname{curl} F \cdot \hat{n}\, dS $$
where $F =\langle xyz, x, e^{xy} \cos(z)\rangle$
$S$ is the hemisphere $x^2+y^2+z^2=25$ for $z ≥ 0$ oriented upward.
I know how to compute the curl of the vector field. I don't know how to get the normal. I'm a bit confused about what it is.
Once I have the dot product of the $\operatorname{curl} F$ and the normal then I can redefine the sphere in terms of $\theta$ and $\phi$ (spherical coordinates) and I can compute the integral, no?
I specifically want to complete this problem using stokes' theorem.
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$\begingroup$What Stokes' Theorem tells you is the relation between the line integral of the vector field over its boundary $\partial S$ to the surface integral of the curl of a vector field over a smooth oriented surface $S$:
$$\oint\limits_{\partial S} \mathbf{F} \cdot d\mathbf{r} = \iint\limits_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \tag{1}\label{1}$$
Since the prompt asks how to calculate the integral using Stokes' Theorem, you can find a good parametrization of the boundary $\partial S$ and calculate the "easier" integral of the LHS of (1).
Note that the boundary of $S$ is given by: $$\partial S=\{(x,y,z)\in \mathbb R^3 : x^2 +y^2 =25, z=0\},$$
so a good parametrization to use for $\partial S$ could be: $$\sigma :[0,2\pi] \subseteq \mathbb{R} \rightarrow \mathbb R^3$$ $$\sigma(\theta)=(5\cos(\theta),5\sin(\theta),0)$$ and finally the integral to calculate ends up being: $$\begin{align}\oint\limits_{\partial S} \mathbf{F} \cdot d\mathbf{r}&= \iint\limits_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \\ &=\int_{0}^{2\pi}(0,5\cos(\theta),e^{25\cos(\theta)\sin(\theta)})\cdot(-5\sin(\theta),5\cos(\theta),0)\, d\theta,\\&=\int_{0}^{2\pi}25\cos^2(\theta)\,d\theta\end{align}$$
and from here you can use trigonometric identities to calculate the last integral.
I hope that helps!
$\endgroup$ $\begingroup$The OP suggests some significant gaps in his knowledge.
$\nabla \times F = \det \begin{bmatrix}\mathbf i& \mathbf j& \mathbf k\\\frac {\partial F}{\partial x}&\frac {\partial F}{\partial y}&\frac {\partial F}{\partial z}\\xyz&z&e^{xy}\cos z \end{bmatrix} = xe^{xy}\cos z\mathbf i -(ye^{xy}cos z-xy)\mathbf j + (1-xz)\mathbf k$
$dS =(-\frac {\partial x}{\partial z},- \frac {\partial x}{\partial z}, 1)$
or $dS = \frac {(x,y,z)}{z} = (\frac {x}{\sqrt{25-(x^2 + y^2)}},\frac {y}{\sqrt{25-(x^2 + y^2)}}, 1) $
Stokes theorem:
$\oint f(x,y,z) \cdot dr = \iint \nabla \times f(x,y,z) \ dS$
This implies that if $S$ and $A$ have the same contour.
$\iint \nabla \times f(x,y,z) \ dS = \iint \nabla \times f(x,y,z) \ dA$
Let $S$ be the hemisphere $x^2 + y^2 + z^2 = 25, z \ge 0,$ And $A,$ the disk $x^2+y^2 = 25, z= 0$
$dA = (0,0,1)$
$\iint \nabla \times f(x,y,z) \ dA = \iint (xe^{xy}\cos z, -ye^{xy}cos z+xy, 1-xz)\cdot(0,0,1) \ dA = \iint 1 \ dA = 25\pi$
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