Find the surface area of the part of the paraboloid $ z=5-(x^2 + y^2)$ that lies between the planes $z=0$ and $z=1$. I have the following math question:
Find the surface area of the part of the paraboloid $ z=5-(x^2 + y^2)$ that lies between
the planes $z=0$ and $z=1$.
So far i have computed
$\sqrt{fx^2+fy^2+1}$ 
to be
$\sqrt{4x^2+4y^2+1}$ 
and have come to: 
$\int \int \sqrt{4x^2+4y^2+1} dx dy$
over a region, R. Is this the right way to go about setting up this equation? I know the region over which i must integrate is the reflection of $ z=5-(x^2 + y^2)$ ( with $z=0$ and $z=1$) on the xy plane, which is hard for me to find. I still have issues with xyz sketching. Can someone please help me sketch this/ give me steps on sketching?  
 A: \begin{align}
{\rm d}A
&=
\left\vert
{\partial\vec{r} \over \partial x}\times{\partial\vec{r} \over \partial y}
\right\vert
\,{\rm d}x\,{\rm d}y
=
\left\vert
\left(\hat{x} - 2x\,\hat{z}\right)\times\left(\hat{y} - 2y\,\hat{z}\right)
\right\vert
\,{\rm d}x\,{\rm d}y
=
\left\vert\,\hat{z} + 2y\,\hat{y} + 2x\,\hat{x}\,\right\vert\,{\rm d}x\,{\rm d}y
\\[3mm]&=
\sqrt{4\left(x^{2} + y^{2}\right) + 1\, }\
\,{\rm d}x\,{\rm d}y
\end{align}
\begin{align}
&
\\[5mm]
A
&=
\int_{0\ <\ z\ <1}\sqrt{4\left(x^{2} + y^{2}\right) + 1\,}\ \
{\rm d}x\,{\rm d}y
=
\int_{2}^{\sqrt{5}}\sqrt{1 + 4\rho^{2}\,}\ \ \rho\ {\rm d}\rho
\int_{0}^{2\pi}{\rm d}\theta
\\[3mm]&=
\pi\int_{4}^{5}\sqrt{1 + 4z\,}\ {\rm d}z
=
\pi\,\left\lbrack%
{\left(1 + 4\times 5\right)^{3/2} \over 6}
 -
 {\left(1 + 4\times 4\right)^{3/2} \over 6} 
\right\rbrack
=
{\large{1 \over 6}\,\pi\,\left(21^{3/2} - 17^{3/2}\right)}
\end{align}
A: For sketching:
sketch the paraboloid $z=x^2+y^2$ and then take the symmetry of the paraboloid with respect to xy plane, translate it $5$ unit along $+z$ axis then sketch the planes $z=0$ and $z=1$.

