1. ## shells

A hole of diameter a is bored through the centre of a sphere of radius a. Show that the remaining volume is root3/2.pi.a(cubed).

2. Hello, dannyshox!

A hole of diameter $\displaystyle a$ is bored through the centre of a sphere of radius $\displaystyle a.$
Show that the remaining volume is: $\displaystyle \frac{\sqrt{3}}{2}\pi a^3$
Code:
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* * *
*:::::|:::::*
- * - - - + - - - * -   y = a/2
*|       |       |*
|       |       |
* |       |       | *
----*-+-------+-------+-*----
* |       |       | *a
|       |       |
*|       |       |*
* - - - | - - - *
*     |     *
* * *
|

The equation of the circle is: .$\displaystyle x^2+y^2\:=\:a^2 \quad\Rightarrow\quad y^2 \:=\:a^2-x^2$

The equation of the horizontal line is: .$\displaystyle y \:=\:\frac{a}{2}$

Revolve the shaded region about the $\displaystyle x$-axis.
That is the desired volume.

The graphs intersect for: .$\displaystyle \begin{array}{cc}x^2+y^2 \:=\:a^2 & [1] \\ y \:=\:\frac{a}{2} & [2]\end{array}$

. . Substitute [2] into [1]: .$\displaystyle x^2 + \left(\frac{a}{2}\right)^2 \:=\:a^2 \quad\Rightarrow\quad x^2 \:=\:\frac{3a^2}{4} \quad\Rightarrow\quad x \:=\:\pm\frac{\sqrt{3}}{2}a$

Formula: .$\displaystyle V \;=\;\pi\int^b_a\bigg(y_2^2 - y_1^2\bigg)\,dx$

Due to symmetry: .$\displaystyle V \;=\;2 \times \pi\int^{\frac{\sqrt{3}}{2}a} _0 \bigg[(a^2-x^2) - \left(\frac{a}{2}\right)^2\bigg]\,dx$ .$\displaystyle = \;2\pi\int^{\frac{\sqrt{3}}{2}a}_0\bigg[\frac{3}{4}a^2 - x^2\bigg]\,dx$

Got it?

thanks!