$\iint_Sz^2d\sigma$ where $S$ is an area of the cone $z=\sqrt{x^2+y^2}$ between planes $z=0$ and $z=1.$

Question

What is the value of $\iint_Sz^2d\sigma$ where $S$ is an area of the cone $z=\sqrt{x^2+y^2}$ between planes $z=0$ and $z=1.$

To solve this by using polar integration I think the integral transforms to

$$\int_0^{2\pi}\int_0^1 r^2 rdrd\theta$$ But the answer in the book says it should be $$\sqrt2\int_0^{2\pi}\int_0^1 r^2 rdrd\theta$$ I'm wondering why they multiplied the integral by the factor$\sqrt2$ ?

Answer 1

The problem in your approach is the fact $\displaystyle \iint_{S}f(x,y,z)\, {\rm d}\sigma$ is a surface integral. You don't have $\displaystyle \iint_{D}f(x,y)\, {\rm d}A$ that is a double integral in the usual sense.

Here you need to use the following definition, $$\iint_{S}f(x,y,z)\, {\rm d}\sigma=\iint_{D}f(\gamma(u,v))\cdot ||\gamma_{u}\times \gamma_{v}||\, {\rm d}A,$$ where $\gamma$ is a parametrization of surface $S$. In this case, we can parameterize directly $z=z(x,y)$ with $x=x$, $y=y$ and $z=z(x,y)$ a we get $$\iint_{S}f(x,y,z)\, {\rm d}\sigma=\iint_{D}f(x,y,z(x,y))\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}}\, {\rm d}A$$ Then using that fact, we get $${\rm d}\sigma=\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}}=\sqrt{2}{\rm d}A$$ Now, you have $$\iint_{D}\sqrt{x^{2}+y^{2}}\cdot \sqrt{2}\, {\rm d}A$$ which is a double integral with solution

$$\sqrt{2}\int_{0}^{2\pi}\int_{0}^{1}r^{2}\cdot r\, {\rm d}r\,{\rm d}\theta=\boxed{ \frac{\sqrt{2}\pi}{2}}$$

Answer 2

The factor $\sqrt2$ arises from the area element of the given cone $$ dS = \sqrt{1+( z_x’)^2+ ( z_y ’)^2} \ dxdy= \sqrt2\ dxdy $$