# Surface integral of a scalar

Evaluate the following integral $$\int \int_S z^2 dS$$ where $S$ is the surface of the cube $[-1,1] \times [-1,1] \times [-1,1]$

My thoughts
I'm quite lost here. How do I know the projection and which vectors do i take to find the normal when there are no vectors? I really am in need of some urgent help please

• – Ataraxia Jun 15 '13 at 17:16
• @ZettaSuro Hi Zetta, thankyou for your response. I read the following post and I am still rather confused. We haven't learnt the divergence thereom yet, only surface integrals and we were given this question, I tend to understand more by seeing the example worked out, if its not too much of a hassle, is it possible if you could show me the working out? – amanda Jun 15 '13 at 17:26
• Yea, I just thought it was worth mentioning. Actually, in that question we determined that the best way to do it is without the divergence theorem :) – Ataraxia Jun 15 '13 at 17:29
• Any feedback is always helpful :) – amanda Jun 15 '13 at 17:38

Do it one surface at a time. To take the surface integral of a scalar field:

$$\iint_Sf(x,y,z)dS=\iint_Df(x,y,g(x,y))\sqrt{g_x(x,y)+g_y(x,y)+1}$$

Where $z=g(x,y)$ defines the surface.

Start with the upper and lower faces ($z=\pm1$).

$$z=1$$

$$g(x,y)=1$$

$$A_1=A_2=\int_{-1}^1\int_{-1}^1{1dx\space{dy}}$$

Since $f(x,y,z)$ is symmetric over the xy plane, $A_2$ will be the same.

Now do the second set of faces (the ones parallel to the xz-plane):

$$y=1$$

$$g(x,z)=1$$

$$A_3=A_4=\int_{-1}^1\int_{-1}^1z^2dz\space{dx}$$

Finally, the last set of faces (the ones parallel to the yz-plane):

$$x=1$$

$$g(y,z)=1$$

$$A_5=A_6=\int_{-1}^1\int_{-1}^1z^2dz\space{dy}$$

The total surface area is the sum $A_1+A_2+A_3+A_4+A_5+A_6$.

• Thankyou so so much, you have made that so clear :)). – amanda Jun 17 '13 at 4:50