Integral bounds when finding surface are of hyperbolic paraboloid $z=y^2-x^2$ between cylinders by switching to cylindrical coords

1.4k Views Asked by At

enter image description here

I came to this problem, and had to look up cylindrical coords since we didn't cover them in any of my classes. But from everything I've read so far I'm still unsure about how they got the 4 and 2 fro the upper and lower bounds of the inner integral. Can someone please explain this?

1

There are 1 best solutions below

2
On BEST ANSWER

The reason to get the 2 and the 4 in the limits of the new integral is the fact that when changing coordinates, the limits must change in the same way as they do when a simple variable change is done when solving a simple integral. However, this limits are wrong in this problem, they are 1 and 2. If you substitute these values, you will get the final answer, thing that you do not get with the limits established in your picture.

For this specific case, you are asked to do the integral of the function between the cylinders $x^2 + y^2=1$ and the cylinder $x^2 + y^2=4$. The good thing of using cylindric coordinates in problems regarding surfaces inside cyllinders is that the limits are easily got, being the radius limits the radius of the small cyllinder to the one of the big one, and then the $\theta$ angle is from zero to $2\pi$ so that you cover the full cillinder. Intuitively what happens is that the radius limits rotate a phase defined by the angle, an so you get the area between those cyllinders. As the general equation for a circle is given by $x^2+y^2=r^2$, you can get the radiuses easily, and so for your case you'll get 1 and 2.