OK... so this problem isn't really specific... it's more of a conceptual puzzle. I've recently started using integrals while solving problems in physics (specifically Newtonian Mechanics and other similar problems), and quite often it's painfully hard to "compose" these integrals... so I figure, I'm doing something wrong or some knowledge is missing. Namely, when I should have an integral, sometimes I'm missing the variable under the integrand with respect to which I am integrating, and sometimes it's naturally there:

1st Example - neat and "normal" integration
If I have a rod of length L whose linear density at point x is given by

$$ \lambda(x) = \lambda_0 + Ax $$ where A is a given constant.The mass of the rod is equal to the sum of all of it's little parts, whose width is infinitesimally small - dx. So, I could easily get the mass of the rod by solving the following integral $$ m = \int_0^L \lambda(x) dx $$

2nd Example - I'm doing something wrong
Let's say we have a cylinder, whose base diameter is 2R and height L, dipped in water, so that it's center of mass is under the water level by h. I want to calculate the force acting on one half of the cylinder via the pressure of the surrounding medium. I've done my best to give a graphical representation below, from a birds eye view:

2nd Example - Birds view

So, as we could see the horizontal components cancel out due to symmetry, and all that remains are the normal components F_n. We have the equation: $$ F_n = F \sin \alpha = p_{\text{avg}} S \sin \alpha $$ where $\alpha$ goes from $0$ to $\pi$ (we can easily find $p_{\text{avg}}$ - the average pressure of the water). So trying to integrate this would yield: $$ R = \int dF_n = \int_0^\pi p_{\text{avg}} dS \sin \alpha = p_{\text{avg}}\int_0^\pi dS \sin \alpha $$ I would need to integrate with respect to $\alpha$. You could see that i added the limits, so the reader would have sense of what I'm trying to do... but from here I can't go anywhere. I am missing the variable in respect to which I want to integrate ($d\alpha$), and I don't know where to find it, whereas in the previous example, it just came naturally. Also, doing this by sums I can solve the problem. I guess that is because sums don't need the variable under the integrand with respect to which I am integrating.

Can anybody tell me what am I doing wrong? Is there a "general rule" to avoid these types of can't-do-it-with-integrals situations? Really all it is, is that I fail to write these integrals, respecting the convention, and I feel like I'm going to need them.

  • 1
    $\begingroup$ Hi Knight1805! Your last question seemed to be rather subjective, so I've removed it. $\endgroup$
    – Kyle Kanos
    Jul 27, 2015 at 13:28

2 Answers 2


For this sort of problem it really helps to go back to the Riemann sum. You don't even need to write it down per se; you just need to think about what your $\delta x$ is in the Riemann sum.

So, in this case you're trying to find a "sum of forces" or we might say a "net" force. To find this, it helps a little to think about momentum conservation/Newton's Third law: we break this thing up into little masses of size $\delta m$, then each of those gets accelerated a little by your force field; we sum all of these little forces up into a "net force"; we are not troubled by the internal forces between masses because they all cancel out due to the third law, except possibly at the very boundary.

Setting up our Riemann sum we have $$\bar F_y = \sum \vec a\cdot\hat y~\delta m.$$ Now assuming a uniform mass distribution across the outer circle, we would say that $$\delta m = \frac{M}{2\pi R} ~ (R ~ \delta \alpha)$$On the flip-side you seem to be doing something a little different, with S as some sort of surface area for your cylinder. In that case, when you divide the cylinder into little slivers of angle $\delta\alpha$ you have $\delta S = R ~L~\delta\alpha$ since $R~\delta\alpha$ is the arc-length of that change in angle when the angle is measured in radians, so $R~L~\delta\alpha$ is the surface area available at $\alpha$.

A similar effect happens in electromagnetism: usually you want a force on a charged object, so you have to divide it into little charges $\delta q$ and then sum up all of the little forces on the little charges.

Can you see? In each case we break the situation apart into something little (masses, areas, charges) which we can solve easily, getting a bunch of little forces. The "something little" is the $\delta x$ in our Riemann sum $\int f(x)~dx \approx \sum_i f(x_i)~\delta x_i.$ When we can find that there is a little force $f(x_i)~\delta x_i$ we can sum it up using a Riemann sum and immediately turn that into a corresponding integral.

  • $\begingroup$ Thank you. The answer was very helpful. Now I see that all the confusion was caused because I made unnecessary complications. When I think of dS, I really imagine a point on the side of the cylinder, whereas you imagine a strip so you already know the length of one side of dS - it's L! No reason to complicate, right? I guess I'll learn to avoid these mistakes when I learn to do multiple integrals at once... I'm sure there exists something like that... thanks! $\endgroup$
    – Knight1805
    Jul 27, 2015 at 14:56

Well in this case, you could consider a really thin ring of elemental thickness say t,

then proceed to form the integral as you have done to that point... Now, you might write dS as tdx , dx being an elemental strip along the circumference of the ring which subtends an angle d(theta) at the center.

So you may now rewrite dS as t* R *d(theta) , and proceed with your integration.


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