How to solve this integral for a hyperbolic bowl?

$$\iint_{s} z dS $$ where S is the surface given by $$z^2=1+x^2+y^2$$ and $1 \leq(z)\leq\sqrt5$ (hyperbolic bowl)

Solution 1:

Use polar cylindrical coordinates. Here, the surface is given by $r(z) = \sqrt{z^2-1}$. The surface integral is then

$$2 \pi \int_1^{\sqrt{ 5}}dz \, z\, r(z) \sqrt{1+\left(\frac{dr}{dz}\right)^2}$$

I leave it to the reader to derive the final form of the integral to be evaluated. I get

$$\pi \int_1^5 du \, \sqrt{2 u-1} = \frac{26 \pi}{3}$$

ADDENDUM

Note that this answer agrees with the parametrization approach:

$$z=1+x^2+y^2 \implies \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2} = \sqrt{\frac{1+2 r^2}{1+r^2}}$$

so that the surface integral is

$$2 \pi \int_0^2 dr \, r z \sqrt{\frac{1+2 r^2}{1+r^2}} = 2 \pi \int_0^2 dr \, r\, \sqrt{1+r^2} \sqrt{\frac{1+2r^2}{1+r^2}}$$

Do a little algebra and see that the surface integral is

$$\pi \int_0^2 du \sqrt{2 u+1}$$

which agrees with the above result.

Solution 2:

A related problem. Note that,

$$ z=\sqrt{ 1+x^2+y^2 } \implies z_x=\frac{x}{\sqrt{ 1+x^2+y^2 }},\quad z_y=\frac{y}{\sqrt{ 1+x^2+y^2 }} $$

$$ \iint_{s} z dS = \iint_{D} z \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2} dA $$

$$ = \iint_{D} \sqrt{1+(x^2+y^2)}\sqrt{{\frac {1+2\,{x}^{2}+2\,{y}^{2}}{1+{x}^{2}+{y}^{2}}}}\,dxdy $$

$$ =\iint_{D} \sqrt{{{1+2\,{x}^{2}+2\,{y}^{2}}{}}}\,dx dy $$

Now, $D\equiv \left\{ x^2+y^2 \leq 4 \right\}$. To see this notice that

$$ 1 \leq z\leq\sqrt5 \implies 1 \leq \sqrt{1+x^2+y^2} \leq\sqrt5 \implies x^2+y^2\leq 4. $$

So, we can use polar coordinates as

$$ = \int_{0}^{2\pi}\int_{0}^{2} \sqrt{1+2 r^2}\,r\, dr d\theta = \frac{26\pi}{3} . $$

Added: if you want to parametrize the surface, you go this way,

$$ x=r\cos(\phi),\quad y = r\sin(\phi), \quad z^2 = 1+x^2+y^2= 1+r^2 \implies z=\sqrt{1+r^2}. $$

You can write it in a vector form as

$$ \textbf{T}(r,\phi)= r\cos(\phi)\textbf{i}+ r\sin(\phi)\text{j}+ \sqrt{1+r^2}\, \text{k} $$

$$ \implies T_r = \cos(\phi)\textbf{i}+ \sin(\phi)\text{j} + \frac{r}{\sqrt{1+r^2}} \text{k}, $$

$$ T_\phi = -r\sin(\phi)\textbf{i} + r\cos(\phi)\text{j}+ 0 \,\text{k}.$$

Solution 3:

The shape of your bowl $S$ is given by $$z(r)=\sqrt{1+r^2}\quad (0\leq r\leq2)\ ,\tag{1}$$ where $r:=\sqrt{x^2+y^2}$. The bowl can be considered as a union of "infinitesimal lampshades" of area $$dS=2\pi r\>ds\ ,$$ where $ds$ denotes arclength along the shape curve $(1)$. Therefore $$ds=\sqrt{1+z'^2(r)}={1\over z(r)}\>\sqrt{1+2r^2}\ dr\ .$$ In this way your integral ($=: J)$ becomes $$J=2\pi\int_0^2 z(r)\> r\ ds=2\pi\int_0^2 r\>\sqrt{1+2r^2}\ dr={\pi\over3}\bigl(1+2r^2\bigr)^{3/2}\biggr|_0^2={26\pi\over3}\ .$$