I'm having trouble understanding visually how a surface integral works/calculates. For a standard double integral, function $f(x,y)$ and a rectangular region $U=[a,b]\times[c,d]$ then the double integral: $$\iint_{U}f(x,y)dxdy=\int_{a}^{b}dx\int_{c}^{d}dy f(x,y)$$ simply calculates the volume underneath the graph $f$. My problem now is when you aren't integrating over a flat region of space but a curved surface $S\subseteq\mathbb{R}^3$. Now I can't visualize this as a volume below the graph $f$ as there is another restraint.

However someone explained surface integrals in this way to me. Imagine I am calculating the mass of a sheet of metal represented as a surface $S\subseteq\mathbb{R}^3$, with a function $f(x,y,z)$ defining its density at any point. In this case the mass will be obtained by integrating the density-function over the surface.

i.e I am still finding the surface area of the surface but multiplying that area by its value of $f(x,y,z)$ at every point. So if I want to find the surface area itself, I use $f(x,y,z)=1$. This would be consistent with my description of volume in the standard flat space integral as I'm simply multiplying the area $(U)$ by the height at that individual point $(f(x,y))$.

Now I'm wondering whether for the curved space model whether the volume is actually the space between the two graphs if plotted on the same axis.

So my question is this. Do my ideas make any sense? Also you have any ideas on how I can visualize these integrals please share them? It really helps my problem solving when I can physically visualize the problem in my head.



Short answer: Yes, your ideas do make sense.

Long answer: I see that the only conflict you have is about visualizing volume between 2 curved surfaces as a double integral.

For 2 surfaces represented by $u=f(x,y)$ and $v=g(x,y)$, the volume of the space between them on a region $R$ can be represented as: \begin{equation} V=\iint_{R} |\ u-v\ |\ dxdy \end{equation}

Now, all that remains is the sense of region $R$ in the terms of one of the surfaces itself. I see that as just a transformation of the coordinates, albeit a non linear one.

Does that answer your question?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.