Finding limits of integration using spherical coordinates I would like to integrate some function $f:\mathbb{R}^3\to\mathbb{R}$ over $C_1\cap C_2$ where
$$C_1=\{(x,y,z)\in\mathbb{R}^3:x^2+4y^2+9z^2\leq1\}$$
$$C_2=\{(x,y,z)\in\mathbb{R}^3:x^2+4y^2+9z^2\leq 6z\}$$
My method was to find parametric equations for $C_1,C_2$ using spherical coordinates. However, I am having trouble finding the limits of integration. I attempted to use some geometric intuition but I can't seem to get the right result. Any help would be appreciated.
I used the following parametrization for $C_1$
$$x=r\sin(a)\cos(b),\quad y= r \sin(a) \sin(b)/2,\quad z= r \cos(a)/3$$
For $C_2$ I used
$$x=r\sin(a)\cos(b),\quad y= r \sin(a) \sin(b)/2,\quad z= r \cos(a)/3+1/3$$
In both cases, $0\leq r\leq 1, 0\leq a \leq \pi, 0\leq b\leq 2\pi$.
 A: The projection of $C_1 \cap C_2 $ in the $xy$ plane is the ellipse $E$ with equation
$$
x^2 + 4y^2 \le \frac{3}{4}
$$
So you could write the integral as 
$$
I = \iint_{E} \left(\int_{z=g(x,y)}^{h(x,y)} f(x,y,z)\; dz \right) dS
$$
where $g(x,y)$ is the bottom part of $x^2+4y^2+9z^2 = 6z$ and $h(x,y)$ the upper part of $x^2+4y^2+9z^2 = 1$.
A: Here's an idea, which may or may not work depending on what function you're integrating. First of all, I would let $(u,v,w)=(x,2y,3z)$, to get $u^2+v^2+w^2 \le 1$ and $u^2 + v^2 + (w-1)^2 \le 1$. So the region is the intersection of two balls, and for each fixed $w$, the slice through the region is a disk whose radius depends on $w$ (and it will be different above and below the plane $w=1/2$, since that determines which one of the balls that gives the active constraint). Then I would write the integral as
$$
\int_{w=0}^{1/2} \left( \iint_{\text{some disk}} \dots \right) dw
+
\int_{w=1/2}^{1} \left( \iint_{\text{some disk}} \dots \right) dw
.
$$
You could use origin-centered spherical coordinates too ($u=\dots,\, v=\dots,\, w=r \cos\theta$), since the second ball is described by $r \le 2 \cos \theta$. This might be better if the integrand is a function of $r$. But you'd have to split the integral over $\theta$ into two ranges, depending on which one of the balls that gives the active constraint. (I would have to draw a picture to explain that, though...)
