I want to find the difference in volume between two balls, one with radius of x while the other with radius of $x+dx$. I find two answers, $\frac{4}{3}\pi(x+dx)^3-\frac{4}{3}\pi x^3$ and $4πx^2 dx$. Please tell me what I did wrong. Thanks. Pardon my bad English.
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2@user615343 Would you please show your steps and use MathJax? By the way, your English is perfectly fine. – Toby Mak Feb 13 '19 at 10:06
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1The first answer comes from the idea that the volume of big ball minus that of the small one. And the second answer is from the idea that the surface of the small ball times the thickness. – user615345 Feb 13 '19 at 10:25
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Your answers are the same if we treat higher powers of $dx$ as being equal to $0$.
$$\begin{align} \frac{4}{3}\pi(x+dx)^3-\frac{4}{3}\pi x^3 &= \frac{4}{3}\pi(x^3+3x^2dx + 3x dx^2 + dx^3)-\frac{4}{3}\pi x^3 \\ &= \frac{4}{3}\pi(x^3+3x^2dx)-\frac{4}{3}\pi x^3 \\ &=\frac{4}{3}\pi x^3+4 \pi x^2dx-\frac{4}{3}\pi x^3 \\ &= 4 \pi x^2dx \\ \end{align}$$
Essentially, you're using the dual numbers approach to nonstandard analysis.
goblin GONE
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2@user615345. Make $dx=10^{-3}$ and compute its powers. They are very small, aren't they ? – Claude Leibovici Feb 13 '19 at 10:59