Let, $z = 16-x^2-y^2$ be our paraboloid, and let our surface be bounded above by the paraboloid and bounded bellow by the plane $z=6$.
We can parametrize the surface as, $R(x,y) = (x,y,16-x^2-y^2)$ which gives us $R_{x} = (1,0,-2x)$ and $R_{y} = (0,1,-2y)$, and then we get $ \vec{n} = R_{x} \times R_{y} = (2x,2y,1)$.
We know we can obtain the volume of the solid by using the fact that:
$$V = \frac{1}{3}\int_{S} (x,y,z) \cdot \frac{\vec{n}}{||\vec{n}||} dS.$$
From, $z=16-x^2-y^2$ and $z=6$, we get $x^2+y^2=10$; hence,
$$0 \le \theta \le 2\pi, ~ 0 \le r \le \sqrt{10},$$
gives us a parametrization of the projection $S$ makes on the $x,y$ plane; thus,
$$V = \frac{1}{3}\int^{2\pi}_{0}\int^{\sqrt{10}}_{0} (r^2 + 16)r ~ drd\theta,$$
but the previous expression does not give me the correct answer which is $50\pi$, what am I doing wrong? Thanks.
Best Answer
From the work you already have, let's make a few corrections. First, we should use the following vector surface integral instead
$$V = \frac{1}{4}\oint_S(x,y,2z)\cdot d\vec{S}$$
which we can verify has the correct divergence
$$\nabla \cdot \left(\frac{1}{4}(x,y,2z)\right) = \frac{1}{4} + \frac{1}{4} + \frac{1}{2} = 1$$
We will see why this is nice later, but this is a computational simplification choice - your original choice of vector field was correct, too. The only mistake is that you forgot to close your surface and include the integral on the bottom disk. This should look like
$$V = \frac{1}{4}\oint_S(x,y,2z)\cdot d\vec{S} = \frac{1}{4}\int_{S_1}(x,y,2z)\cdot d\vec{S_1} + \frac{1}{4}\int_{S_2}S(x,y,2z)\cdot d\vec{S_2}$$
where $S_1$ denotes the paraboloid and $S_2$ denotes the disk. You already correctly computed the vector Jacobian for the first integral, so let's go ahead and compute that integral
$$\frac{1}{4}\int_{S_1}(x,y,32-2x^2-2y^2)\cdot(2x,2y,1)\:dA = \int_0^{2\pi}\int_0^{\sqrt{10}}8\:r\:dr\:d\theta = 80\pi$$
Next, we have the flat disk. We can compute the vector Jacobian the long way, or note that the disk is flat and horizontal, thus its vector Jacobian will be $1$ times the outward unit normal. In this case, that would be the downward vector $(0,0,-1)$, giving the integral
$$\frac{1}{4}\int_{S_2}(x,y,12)\cdot(0,0,-1)\:dA = \int_0^{2\pi}\int_0^{\sqrt{10}}-3\:r\:dr\:d\theta = -30\pi$$
Therefore the total volume is the sum of the two integrals, $\boxed{50\pi}$