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I randomly decided to derive the volume of a sphere.

The area of a circle is $\pi r^2$.

So the volume, I thought, should be $\int \pi r^2 dr = \frac{\pi r^3}{3} $, summing up the area of many discs.

Shouldn't there be a $4$ in there? Why isn't there?

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    $\begingroup$ A factor $4$ applies to the surface area of a sphere. Your approach to finding the volume is at best incomplete. What are the limits of integration (needed for a definite integral)? $\endgroup$
    – hardmath
    Feb 2, 2016 at 14:49
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    $\begingroup$ You can't just throw an integral in there and expect it to work out. First, can you explain (carefully) why you thought that integral would give you the enclosed volume of a sphere? $\endgroup$
    – Deepak
    Feb 2, 2016 at 14:49
  • $\begingroup$ Try taking a spherical shell as the basic element instead of disks. $\endgroup$
    – GoodDeeds
    Feb 2, 2016 at 14:52
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    $\begingroup$ HInt. You can find this volume by summing disks, but you need to be careful The disks aren't all the same size, so $r$ isn't constant. $\endgroup$ Feb 2, 2016 at 14:55
  • $\begingroup$ @EthanBolker That's why I was integrating over $r$ $\endgroup$
    – AJJ
    Feb 2, 2016 at 15:04

2 Answers 2

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The method you tried to apply actually works like this: $$ V = \int_{-r}^r \pi y^2\ dx\qquad (1) $$ where $x^2+y^2=r^2$. Plugging the Pythagorean identity in $(1)$ gives $$ V = \int_{-r}^r \pi (r^2-x^2)\ dx = \pi\left[r^2x-\frac{x^3}{3}\right]_{x=-r}^r=\frac{4}{3}\pi r^3 $$ For more details see this derivation.

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The missing $4$ actually comes from the area of spherical shell with radius $r$.

Let $(x,y)=r(\cos t, \sin t)$. Then $(\dot{x},\dot{y})=r(-\sin t, \cos t)$.

\begin{align*} S &= \int_{-r}^{r} 2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^{2}} dx \\ &= 2\pi\int_{-\pi}^{\pi} y\sqrt{\dot{x}^{2}+\dot{y}^{2}} dt \\ &= 2\pi \int_{-\pi}^{\pi} r^{2} \sin t \, dt \\ &= 4\pi r^{2} \\ V &= \int_{0}^{R} 4\pi r^{2} \, dr \\ &= \frac{4}{3} \pi R^{2} \end{align*}

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