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I was recently studying a question based on Electrostatics. Here is the link to the question (along with the answer below). I haven't learned integration yet. But my question here is how did we get $x^2\text dx$ as the radius of the Gaussian shell?

Link to the question and answer: https://www.sarthaks.com/40525/let-there-be-a-spherically-symmetric-charge-distribution-with-charge-density

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    $\begingroup$ I haven't learned integration yet. This seems to be the thing you need to do. Basic integration isn't hard and that's all you'll need. $\endgroup$
    – StephenG - Help Ukraine
    Commented Sep 13, 2021 at 3:22

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$x^2dx$ is not radius here. $4\pi x^2$ is the surface area of the spherical surface, multiplying by $dx$ gives you the differential volume.

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  • $\begingroup$ Thank you, Brother. If you don't mind, whenever you are free, could tell me how exactly the concept of differential volume is used here? What I meant is could you kindly elaborate on that part of the answer and how we derive upon the conclusion of x^2dx. I haven't taken. $\endgroup$
    – Hrishi
    Commented Sep 13, 2021 at 3:14
  • $\begingroup$ $V = \frac{4}{3} \pi x^3$, differentiating we get, $dV = 4 \pi x^2 dx$ $\endgroup$
    – Mechanic
    Commented Sep 13, 2021 at 3:37
  • $\begingroup$ @Mechanic Ahhhh Thanks:) $\endgroup$
    – Hrishi
    Commented Sep 13, 2021 at 4:08
  • $\begingroup$ @Hrishi why do you think $x^2dx$ would be the radius, looking by dimension clearly it is not. $\endgroup$
    – Sumit Gupta
    Commented Sep 13, 2021 at 6:04

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