On the surface of the cylinder $C = \{ x^2 + y^2 = 1, \ 0<z<1 \}$ there is subset A that is measurable in terms of $\lambda_2$.

Additionally, we know that: $A_t = A \cap \{z <t \} , 0<t<1$

Prove that:

$$\int\limits_0^1 \lambda_2(A_t) dt = \iint\limits_{A} (1-z) d \lambda_2$$

I know that such cyliner can be parametrized with cylindrical coordinates, those are:

  • $x = r \cos(\theta) = \cos(\theta)$
  • $y = r \sin(\theta) = \sin(\theta)$
  • $z= z$

for: $\theta \in (0, 2\pi)$ and $r = 1$

But it doesn't help in any way with my inequality. I don't know how to use those subsets of A that appear on the LHS. Any help would be much appreciated.

  • $\begingroup$ What is the integral on the right? Are you missing a $\mathrm{d}z$, and the limits of the outer integral? $\endgroup$ Commented May 18 at 19:39
  • 1
    $\begingroup$ No, the integral on the right is over $A$. That is a 2-dimensional set and the integral is 2-dimensional. I will try to edit my question to put $A$ directly under both integral signs. @stoic-santiago $\endgroup$
    – thefool
    Commented May 19 at 6:24

1 Answer 1


Let $a_{t}={d\lambda_2(A_t)}/{dt}$ represent the 1-dimensional volume of the part of $A$ at height $t$ (defined for almost every $t\in[0,1]$). Then the left hand side is equal to $$\int_{t=0}^1\int_{z=0}^ta_{z}\ dz\ dt$$ Changing the order of the integrals this becomes $$\int_{z=0}^1\int_{t=z}^1a_{z}\ dt\ dz=\int_{z=0}^1(1-z)a_z\ dz$$


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