# Vector area

The vector area is given by the integral $\vec a = \int_s d\vec a$ for a surface s.

I have three problems to prove based on this:

1] Find the vector area of a hemispherical bowl of radius R.

I solved a bit of this.
The area element for a hemisphere is given by $$d\vec a = R^2\sin \theta d\theta d\phi \hat r$$.
So,$$\vec a = R^2\int_{0}^{\frac{\pi}{2}}\sin \theta d\theta \int_{0}^{2\pi} d\phi \hat r$$

How am I supposed to integrate over the unit vector $\hat r$?

2] Show $$\vec a = 0$$ for any closed surface.

3] Show $$\vec a$$ is the same for all surfaces sharing the same boundary.

Tide
Homework Helper
Hint:

$$\hat r = \frac {x \hat i + y \hat j + z \hat k}{r}$$

Does this mean I have to convert all the factors into Cartesian coordinates, since I have used spherical polar in this case?

Tide
Homework Helper
No, but you should now convert x, y and z to their spherical forms.

The basic problem is that the direction of the unit vector $\hat r$ varies with position. The Cartesian unit vectors, however, have fixed directions.

Tide said:
No, but you should now convert x, y and z to their spherical forms.

The basic problem is that the direction of the unit vector $\hat r$ varies with position. The Cartesian unit vectors, however, have fixed directions.

I converted them and after making the necessary evalutions to the integral, I got:
$$\vec a = \int_s d\vec a = \pi R^2\hat z$$
What does this equation mean?
How is the vector area different from the scalar area particularly in this case?
Why is $\vec a = 0$ for closed surfaces?

HallsofIvy