# On surface of $C = \{ x^2+y^2=1,0<z<1 \}$ there is subset A, $A_t=A \cap \{z <t \}$. Prove $\int_0^1 \lambda_2(A_t) dt= \iint_A (1-z)d \lambda_2$

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

Additionally, we know that: $$A_t = A \cap \{z

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.

• What is the integral on the right? Are you missing a $\mathrm{d}z$, and the limits of the outer integral? Commented May 18 at 19:39
• 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 Commented May 19 at 6:24

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 $$\int_{z=0}^1\int_{t=z}^1a_{z}\ dt\ dz=\int_{z=0}^1(1-z)a_z\ dz$$