# Question about the use of integration in physics

I've always thought of integration as a way to solve differential questions. I'd solve physics problems involving calculus by finding the change in the function $df(x)$when I increment the independent variable (say $x$ of $f(x)$) by an infinitesimal amount $dx$, attaching some physical significance to $df = f(x+dx) - f(x)$, writing a differential equation relating $df$ and $dx$. This would give me $f'(x)$ which I would integrate with respect to $x$ to give me the function I was looking for.

I want to know why I can also look at integration as summing up of infinitesimal elements.

For e.g If I want to find the mass of a ring of mass M(redundant, I know) , I can do that by "summing up" infinitesimal elements of mass $dm => \int {dm}$ from 0 to M or summing up $\rho Rd\theta$ from $0$ to $2\pi$

Or if I want to find the total gravitational potential due to a ring: I could write the small potential $dV$ for an element $dm$ and sum up these $dV$'s using integration to give me the total $V$ due to the ring.

But I want to know why this works/makes sense, mathematically. Because I think of integration either a way to solve differential equations or as the area under a curve, and I don't understand why the area under the curve(and which curve?) would represent the total potential $V$ or mass $M$

(sorry if this is too elementary for this site)

• Because (Riemann) integration is defined to be the limiting case of a sum, see the Riemann integral. The fact that integration is the reverse of differentiation (see the FTC) was a discovery that came later. Feb 24 '17 at 11:38

I want to know why I can also look at integration as summing up of infinitesimal elements.

Because that's the definition.

It's actually stranger that you find no difficulty accepting that integration is the inverse operation to differentiation. Most people have trouble accepting that (and it's harder to prove). That connection is called the fundamental theorem of calculus.

I think you need to work your way through the definitions and the fundamental theorem to connect these things in your head.

• could you show how I could apply the Riemann sum for the two examples i mentioned at the end? (mass and gravitational potential) Feb 24 '17 at 13:05

The area under the curve is only one interpretation of integration. The fundamental interpretation is a Riemann Sum : the total is approximated by the sum of arbitrarily small parts. As the size of those parts shrinks to zero, becoming infinitesimal, the sum becomes an integral. The Riemann Sum is just the same as the integral.

In physics integration by summation usually works because of the Principle of Superposition which applies for Linear Systems. For a system consisting of several masses, the gravitational potential at a point is the algebraic sum of the potential due to each mass. This principle applies whether the system consists of 2 or 10 or 100 point masses, or an infinite number of point particles of infinitesimal mass spread throughout a finite volume.

• Could you show how I could apply the Riemann sum for the two examples i mentioned at the end? (mass and gravitational potential) Feb 25 '17 at 1:55
• I think you mean, can I solve the problem for you? No, sorry. That is against site policy. Your question was why integration works in physics?. That is on-topic here. Asking us to solve the problem for you is off-topic. You have already described how you would do it. What is your difficulty? Feb 25 '17 at 2:02
• But this is exactly what I was looking for. Thanks for that. Just show me how to apply a Riemann sum to those two problems. I don't know how to set up the index for the sigma Feb 25 '17 at 2:02
• I know $M = \Sigma \Delta m$ and $V = \Sigma G\Delta M/y$ I don't know how to put in an index for summation Feb 25 '17 at 2:04
• What do you want an index for? Write an expression for the potential at P due to a point mass situated at any point on the ring, then integrate. (Actually for this problem you don't even need to integrate, because the potential is exactly the same for every element of mass on the ring.) Feb 25 '17 at 2:07

The infinitesmal parts dx are taken as subintervals of the domain of the curve f(x). Each x you plug in outputs a y that describes the curve. Then you plug in x + dx to find y1, x + 2 dx to find y2, etc. Each output y = [y0, y1, etc] describes the curve. Integrating means you add up all the y's from a = y0 to b = y_n over a subinterval where _n --> upper bound. This is why it represents the area under the curve; you add up x0 * y0 + x1 * y1 + etc. Just as xy where y > 0 is the area of a rectangle, the sum of xy's where y = f(x) gives the area of the shape.

In elementary calculus Differential is about a shape of the volume, Integral is about volume of that shape. ( volume of two-dimensional region is an area)

Divide your ring in small pieces of length $ad\phi$. One such piece would contain a small mass $dm=\lambda rd\phi$, where $\lambda=m/(2\pi a)$ is the linear mass density of your ring, assumed uniform. This $dm$ will produce a small potential $dV$, and the net potential would be obtained by "summing the $dm$'s", i.e. integrating over $\phi$.

In principle, $dm$ should come with the continuous label $\phi$ to locate the small piece on the ring; $\phi$ is a continuous rather than discrete index. But since all $dm$'s are the same there is no need to keep track of $\phi$. You would need to keep track of $\phi$ (i.e. you would need the continuous equivalent of a discrete index) if the mass distribution of the ring was not uniform, so that pieces at two different values of $\phi$ had two different masses.

'hope this is enough, as I'd rather not go further.