# Confused how to use calculus in physics

1. Sep 8, 2013

### jaydnul

I understand simple concepts, like $\frac{dx}{dt}=v$ and why that is, but when I'm doing, for example, uniform charge distributions, I don't understand what the integral is actually doing. For example:

$$E_x=∫dEcosθ$$

From what I learned in calculus, the dE means with respect to. So when taking an integral you usually have the form $$∫y(x)dx$$ and the interval is [a,b], which are x values.

Why isn't the integral above in that form then? I mean at the very least, $∫dθcosθ$ would make more sense to me.

2. Sep 8, 2013

### mishima

Why isn't it in that form? Because you haven't made it into that form yet, that is your goal. You need to express E in terms of theta, or theta in terms of E, by looking at the geometry of the situation.

Every little point in a charge distribution contributes to the overall electric field. If you just had 2 point charges you would add the fields in accordance with superposition. But now that you have an infinite number of points in a larger distribution, you need to do an integral to add them all up.

3. Sep 10, 2013

### ThomasO

Jd0g33,

∫ means a "sum" over differential amounts.

In ∫dE cosθ the differential amount is dE cosθ

dE is a vector and dE x cosθ is its projection on the x-axis.

Adding up all the projections of every dE, you get E$_{x}$.

When taking ∫y(x)dx, the differential amount being added up is y(x)dx, that is, y(x) times dx.
This is the "area" under the point y(x).

In this explanation, I have used some loose terms, but I hope I could pass the message.