# View factor in concentric spheres

Consider a hollow sphere $$A$$ of radius $$1$$, containing a concentric sphere $$B$$ of radius $$r < 1$$. I am trying to calculate the view factor $$F_{A \rightarrow B}$$, but different approaches give me different results.

### Approach 1

The apparent radius of $$B$$, as seen from any point of $$A$$, is: $$\theta = \arcsin(r)$$.
The solid angle occupied by $$B$$ from that point of view is: $$\Omega = 2\pi(1 - \cos(\theta))$$.
The point of $$A$$ emits radiation in every directions uniformly across the hemisphere, so the fraction that hits $$B$$ is: $$\Omega / 2\pi = 1 - \cos(\arcsin(r))$$.
Since this is true of all points of $$A$$, we have $$F_{A \rightarrow B} = 1 - \cos(\arcsin(r))$$

### Approach 2

The areas of $$A$$ and $$B$$ are $$A_A = 4\pi$$ and $$A_B = 4\pi r^2$$, respectively.
Since B is convex, we have $$F_{B \rightarrow B} = 0$$ and therefore $$F_{B \rightarrow A} = 1$$.
By the reciprocity theorem, we have $$A_A F_{A \rightarrow B} = A_B F_{B \rightarrow A}$$ and therefore $$F_{A \rightarrow B} = r^2$$

Obviously these results are different. I assume the theorem is correct and my first approach is flawed. Where did I go wrong?

The problem in the first method is that you are not calculating the view factor exactly by calculating the solid angle, which is because solid angle is the ratio of the area of the object that it's projection on a unit sphere subtends. So here what you are calculating (for a particular point) is the area of the white part of the smaller sphere, while what we actually need is the red area of the smaller sphere : (the projection of the area of the smaller sphere on which the radiation is hitting. That is the effective area for the total power captured by the smaller sphere, as power is taken to be perpendicular to the given area). Like in the sun and earth question we use the area of the earth on which the radiation is hitting to be π$$r^2$$. So in this question also while calculating the solid angle, you will have to take into account the projection of area and then you would have to integrate it, so the reciprocity theorem is better.