I have to evaluate the given surface integrals $\displaystyle \iint s {z}^{2 } d\sigma$ where S is the portion of the cone $\displaystyle z = \sqrt{{x}^{2 } +{y}^{2 } } $ between the planes z = 1 and z = 2
I dont know where to start
The top and bottom are circles or radius 2 and 1 respectivily and can be parametrized using cylindrical polar coordinates.
Note that on the side the azimuthal angle is fixed at
$\displaystyle \theta=\frac{\pi}{4}$
So the surface can be parameterized as follows
$\displaystyle x=r\sin\left(\frac{\pi}{4} \right)\cos(\phi) \quad y=r\sin\left(\frac{\pi}{4} \right)\sin(\phi) \quad z=r\cos\left(\frac{\pi}{4} \right)$
So as a vector we get
$\displaystyle \mathbf{r}(r,\phi)=[r\sin\left(\frac{\pi}{4} \right)\cos(\phi)]\mathbf{i}+[r\sin\left(\frac{\pi}{4} \right)\sin(\phi)]\mathbf{j}+[r\cos\left(\frac{\pi}{4} \right)]\mathbf{k}$
Can you finish from here?
You can use the standard method
$\displaystyle \displaystyle\iint_S F(x,y,z)dS=\displaystyle\iint_DF(x,y,f(x,y))\sqrt{ 1+\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial y}}\;dxdy$
In this case
$\displaystyle F(x,y,z)=z^2,\;z=f(x,y)=\sqrt{x^2+y^2},\;D\equiv 1\leq x^2+y^2 \leq 4$
Edited: Sorry, I didn't see the TheEmptySet's post.
You have the cone
$\displaystyle z=\sqrt{x^2+y^2}=r$
From spherical coordinates we know that
$\displaystyle z=r\cos(\theta) \iff \theta =\cos^{-1}\left( \frac{z}{r}\right)$
but from the equation of the cone we know that
$\displaystyle z=r$
So we get
$\displaystyle \theta =\cos^{-1}\left( \frac{r}{r}\right) =\cos^{-1}\left( 1\right)=\frac{\pi}{4}$
If we have the vector form of a surface its surface area is given by
$\displaystyle \iint \bigg|\frac{\partial \mathbf{r}}{\partial r} \times \frac{\partial \mathbf{r}}{\partial \theta} \bigg| dr d\theta$