\[yz\]
plane, passes through the line \[y=0, \; z=a\]
and has a chord in common with the circle \[x^2+y^2=a^2, \; z=0\]
, As the circle moves it generates a volume.To meet the conditions at
\[x=0\]
the circle must have radius \[a\]
and centre \[(0,0)\]
and to meet the conditions at \[x= \pm a\]
it must have radius \[r= \frac{a}{2}\]
and centre at \[z= \frac{a}{2}\]
so the circle must obey an equation of the form \[y^2+(z-p)^2=r^2\]
where \[p, \; r\]
are both functions of \[x\]
.For the stationary circle
\[(x_1,y_1)\]
satisfy \[y_1= \sqrt{a^2-x^2_1}\]
and this is equal to half a chord. \[(x_1,y_1)\]
is also a point on the moving circle with \[z=0\]
so we can write \[y^2_1+p^2=r^2\]
.The moving circle passes through
\[y=0, \; z=a\]
so \[a^2-2ap+[^2=r^2\]
.Solving these two equations simultaneously, we have
\[y^2_1=a^2-2ap\]
from which \[p=\frac{a^2y^1_1}{2a}= \frac{a^2-(a^2-x^2_1)}{2a}=\frac{x^2_1}{2a}\]
then \[r^2=y^2_1+p^2=a^2-x^2_1+ \frac{x^4_1}{4a^2}\]
.A disc of radius
\[r\]
and thickness \[\Delta x\]
at \[x_1\]
has volume \[\Delta V = \pi r^2 \Delta x = \pi (a^2-x^2_1+ \frac{x^4_1}{4a^2}) \Delta x \]
and the total volume traced by the moving circle is \[V = \pi \int^a_{-a} (a^2-x^2_1+ \frac{x^4_1}{4a^2}) dx =\frac{43 \pi a^3}{30}\]
.