Use cylindrical coordinates to evaluate the triple integral , where is the solid bounded by the circular paraboloid and the xy-plane.
I'm having trouble deciding what the bounds for r would be.
Use cylindrical coordinates to evaluate the triple integral , where is the solid bounded by the circular paraboloid and the xy-plane.
I'm having trouble deciding what the bounds for r would be.
Let z = 0 to find the domain in the xy plane
$\displaystyle 0 = 9 -16 (x^2 + y^2) $
$\displaystyle 9/16 = x^2 + y^2 $
When we transform to polar this becomes
$\displaystyle \sqrt{9/16} = r $
Therefore
$\displaystyle 0 \le r \le 3/4 $
If you recall that in polar
$\displaystyle dA = rdr d \theta dz$
Then we have
$\displaystyle V= \int_0^{2 \pi} d \theta \int_0^{ 3/4} r^2dr \int_0^{9-16r^2} dz $