I want to calculate the volume between $c:\ 4x^2+9y^2+z^2=36$ where $x>0,y>0,z>0$ and its tangent plane at the point $(x_0,y_0,z_0)$.
To find perpendicular vector $\vec{a}$ to the plane let $\vec{p}=4x_0^2+9y_0^2+z_0^2-36 \Rightarrow \vec{a}=\nabla\vec{p}=(8x_0,18y_0,2z_0)$ This means that the equation of the plane is:
$E:\ 8x_0(x-x_0)+18y_0(y-y_0)+2z_0(z-z_0)=0$ and since $(x_0,y_0,z_0)$ belong to $c$
$E:\ 4x_0x+9y0y+z_0z=36$
What more do I need to calculate the requested volume? Can it be done somehow through the following integral?
$$\int^{z_0}_0\int^{y_0}_0\int^{x_0}_0 (4x^2+9y^2+z^2-36)\ \ dxdydz ?$$
I will assume that $x_0,y_0,z_0>0$.
The region from the first octant bounded by $4x_0x+9y_0y+z_0z=36$ is$$\int_0^{\frac9{x_0}}\int_0^{\frac{36-4xx_0}{9y_0}}\int_0^{\frac{36-4x_0x-9y_0y}{z_0}}1\,\mathrm dz\,\mathrm dy\,\mathrm dx=\frac{216}{x_0y_0z_0}.$$On the other hand, the volume of the region of the first octant bounded by $4x^2+9y^2+z^2=36$ can be computed doing the substitution $X=2x$, $Y=3y$, and $Z=z$. It becomes $6$ times the volume of the region of the first octant which belongs to the sphere centered at $(0,0,0)$ with radius $6$; that is, it is equal to $1728\pi$. So, the volume that you are interested in is$$\frac{216}{x_0y_0z_0}-1728\pi.$$