# evaluate $\iint (x^2+y^2)dS$ where $S$ is the surface of the cone $z^2=3(x^2+y^2)$ bounded by z=0 and z=3

I am a bit confused on how to take the question forward.

Should I write $\mathrm ds=r\,\mathrm dr\,\mathrm dθ$ and hence solve by polar coordinates?

Should I first write $\vec{\mathbf{F}} \cdot \vec{\mathbf{n}}=x^2+y^2$ where $\vec{\mathbf{n}}$ is unit normal vector to the surface and hence first find out $\vec{\mathbf{F}}$ and then apply gauss divergence theorem by converting this double integral into triple integral?

Or should I divide the whole surface into two surfaces, one with $z^2-3x^2-y^2=0$ and other with $z=3$,then add both the values?

I tried all three cases, and answer for all three comes different. I am very badly confused. Please try and help.

I find it easier to start in cartesian and then convert to spherical mid-way.

$dS = $$(-\frac {dz}{dx},-\frac {dz}{dy}, 1)\\ (-\frac {3x}{z},-\frac {3y}{z},1) \;dy\;dx\\ (-\frac {3x}{\sqrt {3(x^2 + y^2)}},-\frac {3y}{\sqrt {3(x^2 + y^2)}},1) \;dy\;dx\\ \|dS\| =$$\sqrt {\frac {9(x^2 + y^2)}{3(x^2+y^2)} + 1} \;dy\;dx\\ 2$

$\iint 2(x^2 + y^2)\;dy\;dx$

Now, lets covert to polar.

$\int_0^{2\pi}\int_0^{\sqrt 3} 2(r^2)(r \;dr)\;d\theta$

• Shouldn't r's limit go from 0 to root 3? And how is dz/dx =6x? For me it is 3x/z..and so ds=2.dx.dy – Parul Jan 9 '17 at 18:50
• you are right, I mucked that up. – Doug M Jan 9 '17 at 18:57
• Nd how about using other methods? – Parul Jan 9 '17 at 20:01