Surface Integral: Transform, Calculate and Add Up Results?

[coverband]

As you know surface integrals are integrated with respect to dS. We then tranform the integral into one in dxdy. Is this the end of the problem or must we calculate it for dxdz and dydz as well and if so do you add up all results at the end!?

 

[JukkaVayrynen]

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.

 

[coverband]

Just to clarify...

Thanks for reply


Just to clarify, if asked to evaluate
doub_int[f(x,y) dS]

We just solve in xy plane?

Thanks again...

 

[JukkaVayrynen]

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.

 

[HallsofIvy]

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.