# A Determine the flux of the vector field trough the surface

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1. Aug 6, 2017

From my drawings it seems to be half of hemisphere. Am I right? How can I solve this task?

Determine the flux of the vector field $$f=(x,(z+y)e^x,-xz^2)^T$$ through the surface $Q(u,w)$, which is defined in the follwoing way:

1) the two boundaries are given by $$\delta Q_1=\{(x,y,z):x^2+y^2=1,z=0,y\ge0\}$$ and $$\delta Q_2=\{(x,y,z):x^2+z^2=1,y=0,z\ge0\}$$

2) the points on the two arcs $$\delta Q_1$$ and $$\delta Q_2$$ are connected by straight lines, which are parallel to the plane $$x=0$$

2. Aug 6, 2017

### andrewkirk

It's not a hemisphere. There are lines on the surface that are straight, and one cannot draw a straight line on a hemisphere.

Also, the surface lies in only two octants, whereas a hemisphere lies in four of the eight octants. It is a surface that lies over half of the plane $z=0$.

To solve this, you calculate
$$\int_{0}^1 \int_{-1}^1 \vec f(x,y,g(x,y)) \cdot \vec n(x,y) \,dx\,dy$$
where $g(x,y)$ is the $z$ coordinate of the point on the surface with $x,y$ coordinates $(x,y)$; and $\vec n(x,y)$ is the normal to the surface at that point.

Note that, because of the last statement in the OP, $\vec n(x,y)$ is constant over $y$ and can be obtained from the gradient of the line connecting
• the point on the ring $\delta Q_2$ with specified $x$ coordinate and $z\geq 0$, to
• the point on the ring $\delta Q_1$ with the same $x$ coordinate and $y\geq 0$
The equation of that line can also be used to calculate the $z$ coordinate $g(x,y)$.

Last edited: Aug 7, 2017
3. Aug 7, 2017