Calculating Area of Hemisphere Cut by Cylinder

[beowulf.geata]

Homework Statement

Find the area of the portion of the cylinder x^2 + y^2 = 2x that lies inside the hemisphere x^2 + y^2 + z^2 = 4, z $$\geq$$ 0. Hint: Project onto the xz-plane.

Homework Equations

I want to use the formula for surface area

$$\int\int\frac{|\nabla f|}{|\nabla f\bullet\vec{p}|}dA$$

The Attempt at a Solution

I'm going to consider only the surface in the first octant (for reasons of symmetry). I get

$$\frac{|\nabla f|}{|\nabla f\bullet\vec{p}|} = \frac{1}{y}$$

hence:

$$\int\int\frac{|\nabla f|}{|\nabla f\bullet\vec{p}|}dA = \int\int\frac{1}{\sqrt{2x-x^2}}dzdx$$


and using

sqrt(4-x^2) and 0 as limits of integration for z

and 2 and 0 as limits of integration for x, I get

$$\int\sqrt{\frac{2+x}{x}}dx$$

(with 2 and 0 as limits of integration for x)

The problem is that this integral doesn't evaluate to 4, which I know is the correct answer (I do get this result by evaluating the integral

$$\int h ds$$

where h is the altitude of the cylinder and ds is the element of arc length on the circle x^2 + y^2 = 2x in the xy-plane)

Could you please tell me where I'm going wrong?

Many thanks in advance!

 

[beowulf.geata]

I've just realized that there is a mistake in my limits of integration for z.

Apart from 0, the other limit is sqrt(4 - (x^2 + y^2)), not sqrt(4 - x^2), and, since this z also belongs to the cylinder, z = sqrt(4 - 2x) and this leads to the correct result, i.e. 4.