# How would one work this out (surface integral of a function)

Given the definition of the surface integral:

$$Area=\iint_{S}{\mathbf{F}\cdot d\mathbf{S}} = \iint_{D}{f(\mathbf{r}(u,v))\cdot \left |\mathbf{r_u} \times \mathbf{r_v} \right |dA}$$

Where $\mathbf{r}(u,v)$ is a vector function to describe $\mathbf{S}$. (Note the bold typesetting indicates vectors for F, r, S). $\mathbf{r}_u$ means the partial differential of r with respect to u.

Now the question given is:

Evaluate the surface integral $\iint_{S}{x^2z^2}dS$ S is the part of the cone $z^2 = x^2 + y^2$ that lies between the planes $z=1$ and $z = 2$

Now I tried this and came to a totally wrong solution. Looking up the solution in the solution manual gave me a different to approach the problem. The manual parameterized the function to $x,y$ by describing "S is the part of the surface $z = \sqrt{x^2 + y^2}$

And then it used the standard approach of $$\iint_{S}{\mathbf{F}\cdot d\mathbf{S}} = \iint_{D}{f(x,y,g(x,y))\sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial x}\right)^2+1}dA}$$ Which results in $\frac{384\sqrt{2}}{3}\pi$

I tried a complete different approach in my work around. Instead of parameterizing the surface by simply stating "z" I tried to parameterize it using the fact that z is a cone: $$z = z, 1 < z < 3$$ $$x = \cos(\theta), 0 < \theta < 2\pi$$ $$y = \sin(\theta)$$ $$\mathbf{r}(z,\theta) = \cos(\theta)\mathbf{i}+\sin(\theta)\mathbf{j}+z\mathbf{k}$$

working the above problem out I could find $$\left |\mathbf{r_z} \times \mathbf{r_\theta} \right | = \sqrt{ z^2\cos^2(\theta) + z^2\cos^2(\theta)} = z$$ $$f(\mathbf{r}(z,\theta)) = \cos^2(\theta)z^2$$ $$A=\int\limits_0^{2\pi}{\int\limits_1^3{\cos^2(\theta)z^3}dz}d\theta= \int\limits_0^{2\pi}{\cos^2(\theta)}d\theta \int\limits_1^3{z^3}dz$$ $$A = 20 \int\limits_0^{2\pi}{\tfrac{1}{2} + \tfrac{1}{2}\cos(2\theta)}d\theta$$ $$A=20\left(\pi+\left[ \sin(4\pi)-\sin(0)\pi\right]\right)=20\pi$$

Which is a total different solution. What did I do wrong, where did I take the wrong decision in this answer?

• @GitGud Hmm I blame mathjax for that :P - simple typos I made thanks to not seeing the overview when typing tex based formulas.. Updating now. What did you mean with "forget to multiple z"? Jun 20 '13 at 15:32
• As it stands the parametrization isn't correct. For instance when $z=2$, you don't have $x^2+y^2=z^2$. Jun 20 '13 at 15:40
• @GitGud Ah, ok! That at least shows the problem. Though than the question becomes: how would I parametrize this? In the way that it leads to the "cylindrical coordinates" EDIT: scratch that, it seems I should say: $r(u,i) = z\cos(\theta)i + z\sin(\theta)j + zk$ - recalculating Jun 20 '13 at 15:52
• It seems to me that you mix two definition of surface integral, that of a (scalar) function and that of a vector field, also said "flux". Jun 20 '13 at 16:11
• @paul23 Try $r(z,\theta)=(z\cos (\theta), z\sin (\theta), z)$. Also notice enzotib's comment there's something wrong with what you wrote, you seem be writting $\mathbf F$ and $f$ interchangeably even though they seem to be different functions. Jun 20 '13 at 16:27

You parameterized a cylinder. You can use spherical coordinates to parameterize a cone. The equation of the cone is $\phi = \frac{\pi}{4}$, so that the surface $S$ is parameterized by $$\mathbf r(\rho,\theta) = \langle \frac{\sqrt{2}}2 \rho \cos \theta, \frac{\sqrt{2}}2 \rho \sin \theta, \frac{\sqrt{2}}2 \rho \rangle,\ 0 \le \theta \le 2\pi,\ \sqrt{2} \le \rho \le 2 \sqrt{2}.$$