# Surface Integral

by coverband
Tags: integral, surface
 P: 6 You can do it several ways. If you have some arbitrary surface, the trick is to project the surface to some simpler surface, for example the x-y-plane. With projection we have simpler integral in which we use dA, dA being infinitesimal surface element on our simpler surface, for example on the x-y-plane the area element is dx*dy (could be $$r dr d\phi$$ if we used polar cordinates). You asked wether we calculate it for dxdz or dydz, the answer is: you have to use the plane on which the surface is projected on. If we have some surface f(x,y), we project it on the x-y-plane and this is almost always the case. So we have to evaluate only one integral, in this case one with dxdy.
 P: 6 Surface Integral Yes. Solve in x-y-plane. But the most important thing to remember is the projection! Let da be surface element on f(x,y) and dA a surface element on x-y-plane. Then we have a relation $$da = dA \sqrt{\frac{\partial f(x,y)}{\partial x}^2 + \frac{\partial f(x,y)}{\partial y}^2 +1}$$ (follows from the cosine of the angle between the surface normal and the x-y-plane normal). So when you're doing the surface integral you get $$\int f(x,y)da = \int f(x,y) \sqrt{\frac{\partial f(x,y)}{\partial x}^2 + \frac{\partial f(x,y)}{\partial y}^2 +1} dA$$. For only the surface area you have similar formula, you just have $$\int da$$ and so on.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 More generally, one can have a surface in terms of any 2 parameters. If x= x(u,v), y= y(u,v), z= z(u,v), then we can write the "position vector" of any point on the surface as $x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}$. The two derivatives $\vec{r}_u= x_u\vec{i}+ y_u\vec{j}+ z_u\vec{k}$ and $\vec{r}_v= x_v\vec{i}+ y_v\vec{j}+ z_v\vec{k}$ lie in the tangent plane and their lengths are the differentials of length in that direction. Their cross product, $\vec{r}_u\times\vec{r}_v$ is called the "fundamental vector product" and its length, times dudv, is the differential of surface area. In particular, if z= f(x,y), this gives exactly what JukkaVayrynen said.