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Actually that's the answer that I got as well. I thought about it for a while and it sorta makes sense to me. If you think about when the density function is integrated for the bounds of $-r$ to $r$ it creates a situation where the bounds are the same and one subtracts from the other. I'll edit this post a little better later.

Actually that's the answer that I got as well. I thought about it for a while and it sorta makes sense to me. If me when I drew out the graphs on my paper. When I followed through the problem I got:$$\int_0^{2\pi}\int_0^1\int_{(-r)}^{(r)}z\delta zr\delta r\delta \theta$$ Which is equal to saying:$$2\pi\int_0^1\int_{(-r)}^{(r)}z\delta z r \delta r$$ Following through the integration you think about when the density function get:$$2\pi\int_0^1\frac{z^2}{2}\biggr|_{(-r)}^{(r)}r\delta r$$ Which is integrated for the bounds of $-r$ to $r$ it creates a situation where the bounds equal to:$$\pi\int_0^1z^2\biggr|_{(-r)}^{(r)}r\delta r$$ Ending up with:$$\pi\int_0^1(r)^2-(-r)^2r\delta r$$ Which is also:$$\pi\int_0^1(0)r\delta r$$ Which is also:

$$\biggr.-$$$$\biggr.=$$$$\biggr.(0)$$ Which are the same graphs of $(r^2)$ and one subtracts $(-r)^2$ and from examination it can be noticed that you are subtracting the other. I'll edit this post a little better later.function from itself, ending up with $\biggr.{(0)}$ and then following through with the rest of the integral which is multiplied by $\biggr.{(0)}$, ending up with $\biggr.{(0)}.

Actually that's the answer that I got as well. I thought about it for a while and it sorta makes sense to me when I drew out the graphs on my paper. When I followed through the problem I got:$$\int_0^{2\pi}\int_0^1\int_{(-r)}^{(r)}z\delta zr\delta r\delta \theta$$ Which is equal to saying:$$2\pi\int_0^1\int_{(-r)}^{(r)}z\delta z r \delta r$$ Following through the integration you get:$$2\pi\int_0^1\frac{z^2}{2}\biggr|_{(-r)}^{(r)}r\delta r$$ Which is equal to:$$\pi\int_0^1z^2\biggr|_{(-r)}^{(r)}r\delta r$$ Ending up with:$$\pi\int_0^1(r)^2-(-r)^2r\delta r$$ Which is also:$$\pi\int_0^1(0)r\delta r$$ Which is also:

$$\biggr.-$$$$\biggr.=$$$$\biggr.(0)$$ Which are the graphs of $(r^2)$ and $(-r)^2$ and from examination it can be noticed that you are subtracting the function from itself, ending up with $\biggr.{(0)}$ and then following through with the rest of the integral which is multiplied by $\biggr.{(0)}$, ending up with $\biggr.{(0)}.$\biggr.{(0)}$.

Actually that's the answer that I got as well. I thought about it for a while and it sorta makes sense to me when I drew out the graphs on my paper. When I followed through the problem I got:$$\int_0^{2\pi}\int_0^1\int_{(-r)}^{(r)}z\delta zr\delta r\delta \theta$$ Which is equal to saying:$$2\pi\int_0^1\int_{(-r)}^{(r)}z\delta z r \delta r$$ Following through the integration you get:$$2\pi\int_0^1\frac{z^2}{2}\biggr|_{(-r)}^{(r)}r\delta r$$ Which is equal to:$$\pi\int_0^1z^2\biggr|_{(-r)}^{(r)}r\delta r$$ Ending up with:$$\pi\int_0^1(r)^2-(-r)^2r\delta with:$$\pi\int_0^1((r)^2-(-r)^2)r\delta r$$ Which is also:$$\pi\int_0^1(0)r\delta r$$ Which is also:

$$\biggr.-$$$$\biggr.=$$$$\biggr.(0)$$ Which are the graphs of $(r^2)$ and $(-r)^2$ and from examination it can be noticed that you are subtracting the function from itself, ending up with $\biggr.{(0)}$ and then following through with the rest of the integral which is multiplied by $\biggr.{(0)}$, ending up with $\biggr.{(0)}$.

Actually that's the answer that I got as well. I thought about it for a while and it sorta makes sense to me when I drew out the graphs on my paper. When I followed through the problem I got:$$\int_0^{2\pi}\int_0^1\int_{(-r)}^{(r)}z\delta zr\delta r\delta \theta$$ Which is equal to saying:$$2\pi\int_0^1\int_{(-r)}^{(r)}z\delta z r \delta r$$ Following through the integration you get:$$2\pi\int_0^1\frac{z^2}{2}\biggr|_{(-r)}^{(r)}r\delta r$$ Which is equal to:$$\pi\int_0^1z^2\biggr|_{(-r)}^{(r)}r\delta r$$ Ending up with:$$\pi\int_0^1((r)^2-(-r)^2)r\delta r$$ Which is also:$$\pi\int_0^1(0)r\delta r$$ Which is also:

$$\biggr.-$$$$\biggr.=$$$$\biggr.(0)$$ $$\Huge-$$$$\Huge=$$$$\Huge(0)$$ Which are the graphs of $(r^2)$ and $(-r)^2$ and from examination it can be noticed that you are subtracting the function from itself, ending up with $\biggr.{(0)}$ and then following through with the rest of the integral which is multiplied by $\biggr.{(0)}$, ending up with $\biggr.{(0)}$.