Question 587934
I'm not sure my solution is what your calculus teacher expects, but hopefully it will help.
I like to visualize that paraboloid in x-y-z coordinates (in cm) as the space between
{{{z=10}}} and {{{z=k(x^2+y^2)}}} (I'll find the right {{{k}}} later).
The vertex of that paraboloid is at (0,0,0), with {{{z=0}}}.
At {{{z=10}}} we have the base of the paraboloid, which is a circle.
Its radius, R is such that {{{2*pi*R=16pi}}} --> {{{R=8}}}
If we were allowed, I would search for volume, V, of a paraboloid with height h and base radius a, and find the formula
{{{V=(1/2)pi*a^2h}}} and for your paraboloid {{{V=(1/2)pi*8^2*10=320pi}}}
In some calculus class, I would have some formula for volume of revolution solids, but I d not have it. (My 1973 Tom Apostol Calculus book has accumulated too much dust, and is probably too heavy and abstract to be of use).
So I figure out the radius, r, of a cross section as a function of z.
At {{{z=10}}} , {{{z=k(x^2+y^2)=kr^2}}} is {{{10=k8^2}}} --> {{{k=10/8^2}}}
Now I have {{{z=(10/8^2)r^2}}} --> {{{r^2=(8^2/10)z}}}
Then, the area of a cross section circle is {{{pi*r^2=pi*(8^2/10)z}}} .
Now I'm ready to find the volume as
{{{int(pi*(8^2/10)z, dz, 0, 10 )=pi*(8^2/10)(10^2/2)=pi*64*10/2=320pi}}}