I want to find the volume between the $xy$ plane and the paraboloid $z = 4 - x^2 - y^2$.
The points of intersection are of the paraboloid with the plane lie along the circle
$$x^2 + y^2 = 4$$
In addition $0 \le z \le 4$.
Let $x_o = \sqrt{4-y^2}$
I can think of two methods of calculating this volume:
$$\displaystyle \int_{z=0}^{z=4} \int_{y=-2}^{y=2} \int_{-x_o}^{x_o} dxdydz = 16 \pi$$
And the other method:
$$\int_{y=-2}^{y=2} \int_{-x_o}^{x_o} (4-x^2-y^2)dxdy = 8 \pi$$
The results have been verified by Wolfram.
Why do two methods calculating the same thing yield different results?
The volume of a region $U$ is given by the triple integral $$\iiint_U 1\,dV$$ There are actually 6 ways to convert the triple integral into iterated integrals, depending on which order you integrate in:
$$dxdydz, dxdzdy, dydxdz, dydzdx, dzdydx, dzdxdy$$
In this case, because the region is symmetric with respect to exchanging $x$ and $y$, 3 of the integrations are effectively the same as the other three. So I will only cover the examples where $x$ is integrated before $y$:
If we integrate by $z$ then $x$ then $y$, $y$ is integrated over $[-2,2]$. The integration for $x$ is restricted by the value of $y$, so it is over $[-\sqrt{4- y^2},\sqrt{4- y^2}]$, and the integration of $z$ depends on both $x$ and $y$, so is over $[0, 4 - x^2 - y^2]$ $$\int_{-2}^2\int_{-\sqrt{4- y^2}}^{\sqrt{4- y^2}}\int_0^{4 - x^2 - y^2} 1\,dz\,dx\,dy$$But the inner integral is obviously $4 - x^2 - y^2$, so this reduces to your second method: $$\int_{-2}^2\int_{-\sqrt{4- y^2}}^{\sqrt{4- y^2}}4 - x^2 - y^2\,dx\,dy$$
If we integrate by $x$ then $y$ then $z$, this time $z$ takes on every value in $[0,4]$, but $y$ is restricted by $z$ and $x$ by both $y$ and $z$. So we get $$\int_0^4\int_{-\sqrt{4-z}}^{\sqrt{4-z}}\int_{-\sqrt{4-z-y^2}}^{\sqrt{4-z-y^2}} 1\,dx\,dy\,dz$$ Again, the inner integral is trivial so this easily reduces to a double integral.
Finally, if we integrate by $x$, then $z$, then $y$, we get $$\int_{-2}^2\int_0^{4-y^2}\int_{-\sqrt{4-z-y^2}}^{\sqrt{4-z-y^2}} 1\,dx\,dz\,dy$$
It is fairly obvious that the first choice leads to the simplest integration.
But there is another choice that is even easier: convert to using cylindrical coordinates $z, r, \theta$. Then $x = r\cos\theta, y = r\sin\theta$ and $dxdy = rdrd\theta$. This gives $$\int_0^2\int_0^{2\pi}\int_0^{4-r^2}1\cdot r \, dz\,d\theta\,dr = 2\pi\int_0^2(4-r^2)r\,dr$$