1. ## Surface Integral

I need to compute the surface integral of (curl A).n dS where:

A= (xyz+2)i + xj + ((x^2)z + (y^2))k

and S is defined by z>=0, z=16-(x^2)-(y^2).

(n is the unit normal field).

Thanks.

2. First, I calculated this using Stokes' theorem but since it's been a while since I did these, I wasn't sure of the outcome. Today, I did it without Stokes (just the surface integral) and I found the same, I'll show both here

Stokes states that the surface integral of the curl of the vector field (scalar multiplication with the exterior normal vector) is equal to the line integral of the vector field over the boundary of the surface.

$\iint\limits_S curl \vec v \cdot \vec n} \, \, dS = \oint\limits_C {\vec v\left( {\vec r} \right) \cdot d\vec r}$

In this case we have:
$\vec v = \left( {xyz + 2,x,x^2 z + y^2 } \right)$
$\vec r = \left( {x,y,16 - x^2 - y^2 } \right)$

S is the upper part of a paraboloid, the boundary in the xy-plane is the circle $C:x^2 + y^2 = 4^2$

I'll first do the line integral, we parametrize into polar coordinates:
$
\left\{ \begin{gathered}
x = 4\cos t \Rightarrow dx = - 4\sin t \hfill \\
y = 4\sin t \Rightarrow dy = 4\cos t \hfill \\
\end{gathered} \right.
$

Since we're working in the xy-plane, we have that z = 0.

$
\oint\limits_C {\vec v\left( {\vec r} \right) \cdot d\vec r} = \int\limits_0^{2\pi } {\left( {2,4\cos t,16\sin ^2 t} \right)} \frac{{d\left( {4\cos t,4\sin t,0} \right)}}
{{dt}}dt
$

$
4\int\limits_0^{2\pi } {\left( {\frac{1}
{2},\cos t} \right)} \left( { - \sin t,\cos t} \right)dt = 8\int\limits_0^{2\pi } {2\cos ^2 t - \sin tdt}
$

Elementary integration then yields a result of $16\pi$

---

Now, without Stokes. Since the surface is given as z = f(x,y), we can parametrize in x and y. This simplifies the calculation of the surface integral. We can calculate the flux as follows:

$
\iint\limits_S {curl \vec v \cdot \vec n}\,dS = \iint\limits_g {\left( {curl \vec v \cdot \vec n} \right)}\,\left\| {\frac{{\partial \vec r}}
{{\partial x}} \times \frac{{\partial \vec r}}
{{\partial y}}} \right\|dS
$

Now we integrate over the projection of the paraboloid on the xy-plane which gives the same circle as integration area again. By the definition of the normal vector, we can simplify this to

$
\iint\limits_g {curl \vec v \cdot \left( {\frac{{\partial \vec r}}
{{\partial x}} \times \frac{{\partial \vec r}}
{{\partial y}}} \right)}\,dS
$

With r and v as above, we have

$
\frac{{\partial \vec r}}
{{\partial x}} \times \frac{{\partial \vec r}}
{{\partial y}} = \left( {1,0, - 2x} \right) \times \left( {0,1, - 2y} \right) = \left( {2x,2y,1} \right)
$

$
curl \vec v = \nabla \times \left( {xyz + 2,x,x^2 z + y^2 } \right) = \left( {2y,xy - 2xz,1 - xz} \right)
$

But since we're integrating over the circle in the xy-plane, we have again that z = 0. So the scalar product gives

$
\left( {2x,2y,1} \right) \cdot \left( {2y,xy,1} \right) = 2xy^2 + 4xy + 1
$

Now it's possible to coninue cartesian or to convert to polar coordinates. I'll give the integrals and leave the integration to you, both give the same result.

$
\int\limits_{ - 4}^4 {\int\limits_{ - \sqrt {16 - y^2 } }^{\sqrt {16 - y^2 } } {2xy^2 + 4xy + 1\,dx} dy} = 16\pi
$

$
\int\limits_0^{2\pi } {\int\limits_0^4 {\left( {128\sin ^2 t\cos t + 32\sin 2t + 1} \right)rdr} dt} = 16\pi
$