Use a triple integral and an appropriate change of coordinates to calculate the volume of the region. I am a math major undergraduate who loves learning math very much. I am confused by a calculus problem today and want to ask for help.
Use a triple integral and an appropriate change of coordinates to calculate the volume of the region bounded by the xy-plane, the surface z = ${x}^2$ + ${y}^2$ and the cylinder over ${x}^2$/9 + ${y}^2$/4 = ${1}$.
What would the volume be? Thank you all for your answers and help!
 A: 
So I've added a picture so that you can better understand what we are dealing with.
There are usually 2 options for integrating something like this:


*

*Integrate by using a tripple integral, each going into the x,y and z direction respectively. For this, one needs to calculate the bounds of the integrals with respect to the outter integrals:
$$\int_{x_0}^{x_1}\int_{y_0(x)}^{y_1(x)}\int_{z_0(x,y)}^{z_1(x,y)}1 \ dz \ dy \ dz$$
Note that the integrals can be swapped. If this happens, the integral bounds need to be changed aswell. This is possible in any case(also in this one) but not recommended

*Integrate by using a tripple integral and a change of coordinates. There are multiple ways of transforming your coordinates. The most commong are spherical coordinates and cylindrical coordinates. Basically you can use cylindrical coordinates for anything that is kinda symetric around on axis. 
For cylindrical coordinates we do not use $(x,y,z)$ anymore but $(\phi, r, h)$
The integral looks basically the same with one major change:
$$\int_{\phi_0}^{\phi_1}\int_{r_0(\phi)}^{r_1(\phi)}\int_{h_0(\phi,r)}^{h_1(\phi,r)}1\cdot r \ dz \ dy \ dz$$
Note the $r$ in the integral. This is due to the coordinate change. e.g. if you use spherical coordinates, you would use $r^2 \cdot sin(\theta)$ (but this isnt needed here so we will stick with $r$).

So let's do the transformation:
   $$(x,y,z)^T \rightarrow (r \ sin(\phi), r \ cos(\phi), h)^T$$
Next, we need to find the boundary. Let's start with $\phi$:
Because the volume goes around the z axis without cutting out a piece (like a cake), our $\phi$ goes all the way around: $\phi \in [0,2\pi]$
Next, we will talk about the bounds of the radius with respect to to $phi$. Note that it varies with height so we need to talk about the height first:
Now this is tricky because the intersection of the cylinder and the cone is not at a constant height:
$$z = x^2 + y^2 = (r \ cos(\phi)) ^2 + (r \ cos(\phi)) ^2 = r^2$$
$$1 = \dfrac{(r \ cos(\phi)) ^2}{9} + \dfrac{(r \ sin(\phi)) ^2}{4}$$
Now, plugging the first equation into the second:
$$1 = \dfrac{z \cdot cos(\phi) ^2}{9} + \dfrac{z \cdot sin(\phi)^2}{4} = z \left( \dfrac{cos(\phi) ^2}{9} + \dfrac{sin(\phi)^2}{4}\right)$$
$$\Rightarrow h = z = \dfrac{1}{\left( \dfrac{cos(\phi) ^2}{9} + \dfrac{sin(\phi)^2}{4}\right)}$$
Let me be honest and this point because this looks to complicated but I will keep going. I might have made a mistake.
Therefor: $$z \in \left[0, \dfrac{1}{\left( \dfrac{cos(\phi) ^2}{9} + \dfrac{sin(\phi)^2}{4}\right)} \right]$$
Now the last thing we need to evaluate is the range for $r$. This one depends on the  height and the angle. It goes from the parabola to the cylinder:
$$r_0 = \sqrt{h} \qquad \text{(this comes from)} z = x^2 + y^2 = (r \ cos(\phi)) ^2 + (r \ sin(\phi)) ^2 = r^2$$
$r_1$ is the radius of the outter cylinder with respect to the angle:
$$r_1 = \sqrt{\dfrac{1}{\left( \dfrac{cos(\phi) ^2}{9} + \dfrac{sin(\phi)^2}{4}\right)}}$$
Now the integral looks like this:
$$\int_{0}^{2\pi}\int_{0}^{\frac{1}{\frac{cos(\phi)^2}{9} + \frac{sin(\phi)^2}{4}}} \int_{\sqrt{h}}^{\sqrt{\frac{1}{\left( \frac{cos(\phi) ^2}{9} + \frac{sin(\phi)^2}{4}\right)}}} r \ dz \ dy \ dz$$
Feel free solving this but I would not. I might have done a mistake. If that's the case, I am sorry. Hope you understand how it works.
