How to evaluate Area of $B:= \{(x,y,z) \in A | z \le 1 \}$ with $A:=\{(x,y,z) \in \mathbb{R}^3 | x^2 + y^2 = z\} = \frac{\pi}{6}(5 \sqrt{5}-1) $?

Question

I have following problem:

Let $$A:=\{(x,y,z) \in \mathbb{R}^3 | x^2 + y^2 = z\} \\ B:= \{(x,y,z) \in A | z \le 1 \}. $$

Compute the area $\mu_2(B) $.

First, I thought $\mu_2(B) $ would just be the area of the unit circle, $\pi$. However, that's not right. The solution says that $\mu_2(B) = \frac{\pi}{6}(5 \sqrt{5}-1)$. I tried to use polar coordinates, but I still just get $\pi$ with $\int_C r d\theta dr $ and $C:=\{(x,y,z) \in \mathbb{R}^3 | x^2 + y^2 \le 1\}$.

Can anybody explain me how to get to $\frac{\pi}{6}(5 \sqrt{5}-1)$ and what must be defined differently so that $\mu_2(B) = \pi$?

Answer 1

Hint for the first question: You are looking at a "cap" (sliced off portion) of a paraboloid of revolution that is caused by rotating the graph of the curve $z=x^2$ for $0\le x\le 1$ around the $z$ axis. So use the method to find the area of a solid of revolution.

Hint for the second question: The "cap" is the unit circle in the $xy$ plane, curved up to make a paraboloid. The flat unit circle has the area $\pi$. Curving it up can only increase the area, so your desired area must be greater than $\pi$. Some ways to get an area of $\pi$ are to leave the circle flat (so it is no longer a paraboloid) and reducing the limit on $z$ to something less than one to reduce the area to what you want. I suppose you could also change the equation to $x^2+y^2=cz$ for some constant $c>1$ and leave $z\le 1$ to reduce the area to what you want. You would need to figure out the desired constants in the last two methods.