by Guest » Sun Nov 07, 2021 9:02 pm
You said "below the xy-plane". Did you mean [tex]z = x^2+ y^2- 2[/tex]?
I am really not at all enthused about helping someone who not only copies a problem wrong but can't check back to say if a proposed correction is right!
But since someone else might be interested: Assuming this is actually [tex]z= x^2+ y^2- 2[/tex] then [tex]\frac{\partial z}{\partial x}= 2x[/tex] and [tex]\frac{\partial z}{\partial y}= 2y[/tex] so [tex]dS= \sqrt{4x^2 + 4y^2}dxdy= 2\sqrt{x^2+ y^2}dxdy= 2rdxdy[/tex]. That "r" is because of the "cylindrical symmetry" and tells us it would be smart to convert to cylindrical coordinates. In cylindrical coordinates [tex]x= r cos(\theta)[/tex] and [tex]y= r sin(\theta)[/tex]` so [tex]dx= cos(\theta)dr- r sin(\theta)d\theta[/tex] and [tex]dy= sin(\theta)dr+ r cos(\theta)d\theta[/tex].
So [tex]dxdy= (cos(\theta)dr- r sin(\theta)d\theta)(sin(\theta)dr+ r cos(\theta)d\theta)= cos(\theta)sin(\theta)drdr+ r cos^2(\theta)drd\theta- r sin^2(\theta)d\theta dr- r^2 sin(\theta)cos(\theta)d\theta d\theta[/tex].
But multiplication of differentials is "anti-commutative" so [tex]drdr= d\theta d\theta= 0[/tex] and [tex]drd\theta= -d\theta dr[/tex]. dxy reduces to
[tex]rcos^2(\theta)d\theta dr+ rsin^2(\theta)d\theta dr= rd\theta dr[/tex] and then [tex]dS= 2r dxdy= 2r^2 d\theta dr[/tex].
Now the graph is [tex]z= x^2+ y^2- 2[/tex] and we want the surface area of the part below the xy-plane. The minimum value of z is -2 when x= y= 0 so r= 0 and z is 0 (the xy-plane) when [tex]x^2+ y^2= 2[/tex] so [tex]r= \sqrt{2}[/tex]. Of course with circular symmetry we can take [tex]\theta[/tex] from 0 to [tex]2\pi[/tex].
That means the surface area is given by [tex]2\int_{\theta= 0}^{2\pi}\int_{r= 0}^{\sqrt{2}} r^2 drd\theta[/tex].