Bounds for double integral I have an assignment to find the surface of a figure created by the intersection of two cylinders:
$$ 
x^2 + z^2 = 4
$$
$$
x^2 + y^2 = 4
$$
I've made it to this form:
$$
x = \sqrt{4 - z^2}
$$
$$
y = z
$$
But I have no idea how to find the bounds neccesary for the integrals, nor how to find the function to integrate.
 A: HINT: Inner points of cylinder $x^2 + z^2 = 4$ satisfy $ -2<x<2 \land -\sqrt{4-x^2} < z < \sqrt{4-x^2}$. Inner points of cylinder $x^2 + y^2 = 4$ satisfy $-2 < x <2 \land -\sqrt{4-x^2} < y < \sqrt{4-x^2}$.
The following figure might help you:

A: Consider a point $P = (x,y,z) = (x,\sqrt{4-x^2},z)$ on the surface of the second cylinder. Suppose that $y\ge|z|$, so that $P$ is between the planes $y=z$ and $y=-z$. Then $z^2\le y^2$, so $x^2+z^2\le x^2+y^2=4$, and hence $P$ is automatically within the first cylinder. In other words, every point $(x,y,z)$ that 


*

*lies on the surface of the second cylinder;

*is between the planes $y=z$ and $y=-z$; and

*has $y\ge 0$ is on the surface of the intersection of the two cylinders.
This is a cylindrical face of the intersection; it contains vertical straight line segments that lie on the surface of the cylinder $x^2+y^2=4$, and it has curved top and bottom edges that are symmetric in the $xy$-plane. The mirror image of this face in the $xz$-plane is another face of the intersection, and there are two more such faces; they are also mirror images of each other, but in the $xy$-plane, and they contain straight line segments that lie on the surface of the cylinder $x^2+z^2=4$.
These four faces are pairwise congruent, so you need only get the area of one of them and multiply it by $4$.
