Question 162738
You'll need calculus for this one. If you plot the given function you see that it is a parabola with vertex at (0,9) and x intercepts at (-3,0) and (3,0). So let's find the volume in the solid between x=0 and x=3. Then we just need to double that to get the entire volume

{{{graph(400,400,-10,10,-10,10, 9-x^2)}}}

When you rotate that curve around the x axis, you'll get a solid that you can slice into 'disks'. Imagine the disks are standing on end. As you make the disk thinner and thinner, it approaches becoming a whole bunch of 'circles' with a width of deltax. The radius of each thin circle is given by {{{9-x^2}}} from 0 to 3. So the area of each circle is {{{pi*r^2}}} = {{{pi*((9-x^2))^2}}}
{{{pi*((9-x^2))^2}}}
{{{pi*(81-18x^2 + x^4)}}}

Now we need to take that Area and integrate if over from 0 to 3

{{{int( A[x], dx, 0, 3 )}}}


{{{int( pi*(81-18x^2 + x^4), dx, 0, 3 )}}}
pi*(81x - 6x^3 + x^5/5) from 0 to 3
pi*(81*3 - 6*3^3 + (3^5)/5) - pi*(0+0+0)
{{{pi*(243 - 162 + 48.6)}}}
{{{pi*129.6}}}
Remember to double it to pick up the volume on the left side of the y-axis
{{{2*pi*129.6}}}
{{{814.3}}}
{{{814}}} to three sig figs