Basically, I'm stuck in this exercise for 2 hours and the manual doesn't give the answer, plus it seems like a easy one, but I can't get it. So basically I have to demonstrate that: $$Area(S)=\int \int_D \frac{dA}{|cos\phi|}$$ where $$\phi$$is is the angle between the normal unit vector of the surface and the vertical which is the unit vector $$k$$ So from the formula I have in my books for area of a surface, is $$ Area(S)=\int \int_D ds =\int \int_D ||ru \times rv|| dA$$ And my idea is that $$n\cdot k = |n||k|cos \phi$$ since k is a unit vector than |k| = 1 and we can then isolate |n| to get $$|n|=\frac{n\cdot k}{cos\phi}$$ and since $$||ru \times rv|| = |n|$$ so $$Area(S)=\int \int_D \frac{n \cdot k\ast dA}{cos\phi}$$ But then I don't how I can remove the n * k.
Thanks in advance
The element area of the surface dS projects on xOy as $dA = dS \, cos \phi$
with dS the domain is defined on the surface.
with dA the domain is defined in the xOy plane.
example : surface of a half sphere
integration on the sphere : $dS = R d\theta \, R cos\theta\,d\phi$
$S = \int dS = R^2 \int_0^{\pi/2} d\theta \int_0^{2\pi} d\phi \, cos\theta = 2 \pi R^2$
integration on the xOy plane : $dA = r d\phi \, dr $ and $dA = dS \, cos \theta$ (in this example)
$S = \int dS = \int_0^{R} dr \int_0^{2\pi} d\phi \, r/cos \theta $
as $r = R \, cos \theta$ : $S = \int_0^{R} dr \int_0^{2\pi} d\phi \, R = 2\pi R^2 $