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Disclaimer: I'm basing my information off of this link in the case where I do not provide enough context. All calculus I've learned has been through curiosity, digging up old textbooks and google (I haven't taken any formal courses yet).

$$\displaystyle V = 2\pi \int_0^r (r^2 - y^2) dy$$ $$V = 2\pi \left[ r^2y - \dfrac{y^3}{3} \right]_0^r$$ $$V = 2\pi \left[ \left(r^3 - \dfrac{r^3}{3}\right) - \left(0 - \dfrac{0^3}{3}\right) \right]$$

1) What is the thought process when transforming these equations? Knowing the process would be extremely helpful and appreciated, as well as knowing about some specific notations/conventions...

2) What does the $[...]_0^r$ in the second equation represent? Do the $0$ and $r$ act as limits for something?

3) In the second equation, what happened to delta $y$ ($dy$)? Did we simply "rename" it $y$ in the second equation?

Please use elementary terms when possible in your answer; it will enhance my understanding. Thank you!

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  • $\begingroup$ This is calculus. You need to take a whole class on it. $\endgroup$
    – user223391
    Jul 23 '16 at 0:10
  • $\begingroup$ @ZacharySelk Basic differentiation/integration/limits are understandable, are the basics not applicable here...? $\endgroup$ Jul 23 '16 at 0:14
  • $\begingroup$ The $[\cdots]_0^r$ notation means "evaluate at the top and bottom things and subtract." It's what you do when you apply the fundamental theorem of calculus to calculate a definite integral. This should be covered in most calculus classes (I would hope). The thought process is just "integrate the things it's telling us to integrate." The whole thing here is mechanical, there's no cleverness or creativity to it (except setting up the integral in the first place to it), and it's something you should have learned in a calculus class. $\endgroup$ Jul 23 '16 at 0:17
  • $\begingroup$ @arctictern I thought I pointed out that I haven't taken a formal class yet; regardless thank you for the clarification. $\endgroup$ Jul 23 '16 at 0:25
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Upon applying $$\int_0^r \ldots dy$$

you get an expression like this $$[F(y)]_0^r$$

which is the same thing as $F(r)-F(0)$, where $F$ is any antiderivative of $f$.

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1) To obtain the first equation, the author basically thought of $x$ as a function of $y$, noting that $x^2+y^2=r^2$ (this comes from the equation of a circle). They simply wrote $x^2$ explicitly in terms of $y$ so that they could calculate the integral.

2) $[g(y)]_a^b$ is a shorter way of writing $g(b)-g(a)$. Expressions like this are common when utilizing the fundamental theorem of calculus, which (in part) states that if $f$ is continuous and $F'(t) = f(t)$, then $$\int_a^bf(t)dt = F(b)-F(a) = [F(t)]_a^b$$

3) The second equation was obtained by applying the fundamental theorem of calculus above. The integrand was $f(y) = r^2-y^2$. Then you want to find a function $F(y)$ whose derivative is $f(y)$; using basic rules of differentiation, this turns out to be $F(y) = r^2y-\frac{y^3}{3}$. Then the value of the integral is just $F(r)-F(0)$.

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The problem is to derive the volume of a sphere w with radius r. One way of going about it is to "sum" over the surface areas of smaller spheres.

The first equation says to "add" the surface areas of all spheres with radius y (where 0 < y < r) all together, thereby filling the volume.

The second equation gives the anti-derivative for the surface areas, together with the bounds of integration.

Thirdly, using the Fundamental Theorem of Calculus (which you should read up on), we evaluate the integral, and establish the desired identity

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