## richyw Group Title Find the area of the part of the cylinder $$x^2+z^2=a^2$$ that lies within the cylinder $$r^2=x^2+y^2=a^2$$ one year ago one year ago

1. richyw Group Title

let me type out what I have done so far

2. richyw Group Title

$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$

3. richyw Group Title

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

4. richyw Group Title

where is that 2 coming from?

5. richyw Group Title

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?