Question

Find the area of the part of the cylinder \(x^2+z^2=a^2\) that lies within the cylinder \(r^2=x^2+y^2=a^2\)

Answer 1

let me type out what I have done so far

Answer 2

\[z(x,y)=\sqrt{a^2-x^2}\]\[z_x=-\frac{x}{\sqrt{a^2-x^2}}\]\[z_y=0\]\[dS=\sqrt{1+(z_x)^2+(z_y)^2}\,dA=\frac{a}{\sqrt{a^2-x^2}}dA\]

Answer 3

so I get lost from here. my solution manual says that the area is\[A=2\iint_D\frac{a}{\sqrt{a^2-x^2}}dA\]Where D is the disk in which the vertical cylinder meets the xy-plane

Answer 4

where is that 2 coming from?

Answer 5

oh ok because I let \(z(x,y)=\sqrt{a^2-x^2}\) and it's really \(z(x,y)=\pm\sqrt{a^2-x^2}\) so by symmetry the area will be twice that?