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I agree with Mr. Spiff, except for one thing. I believe that the integral should be:

$$ \int_0^{2\pi} \int_0^{\pi/4} \int_0^2 \rho^4 \sin(\varphi) \ d\rho \ d\varphi \ d\theta $$

Note the extra $ \sin(\varphi) $. I also have a more rigorous reasoning for the $\pi/4$ radian angle:

We know that $z = \sqrt{x^2 + y^2}$, which means that in cylindrical coordinates, $z = r$. By transforming $z$ and $r$ into their spherical form ($z = \rho \cos(\varphi), r = \rho \sin (\varphi) $), we obtain:

$$ \rho \cos(\varphi) = \rho \sin (\varphi) $$

Assuming that $ \rho > 0 $, we can divide by $\rho$ on both sides.

$$ \cos(\varphi) = \sin (\varphi) $$

If you can't tell what $\varphi$ is yet, it may be helpful to isolate it by dividing by $\cos(\varphi)$ on both sides of the equation and then taking the arctangent of both sides. It is easier to just watch the process than to explain:

$$ \frac{\cos(\varphi)}{\cos(\varphi)} = \frac{\sin (\varphi)}{\cos(\varphi)} $$ $$ 1 = \tan(\varphi) $$ $$ \varphi = \tan^{-1}(1) $$

We use the restriction $0 \leq \varphi \leq \pi$, we conclude that $\varphi = \frac{\pi}{4}$ a.k.a. $45^{\circ}$.

I agree with Mr. Spiff, except for one thing. I believe that the integral should be:

$$ \int_0^{2\pi} \int_0^{\pi/4} \int_0^2 \rho^4 \sin(\varphi) \ d\rho \ d\varphi \ d\theta $$

Note the extra $ \sin(\varphi) $. I also have a more rigorous reasoning for the $\pi/4$ radian angle:

We know that $z = \sqrt{x^2 + y^2}$, which means that in cylindrical coordinates, $z = r$. By transforming $z$ and $r$ into their spherical form ($z = \rho \cos(\varphi), r = \rho \sin (\varphi) $), we obtain:

$$ \rho \cos(\varphi) = \rho \sin (\varphi) $$

Assuming that $ \rho > 0 $, we can divide by $\rho$ on both sides.

$$ \cos(\varphi) = \sin (\varphi) $$

If you can't tell what $\varphi$ is yet, it may be helpful to isolate it by dividing by $\cos(\varphi)$ and then taking the arctangent on both sides of the equation and then taking the arctangent of both sides. equation. It is easier to just watch the process than to explain:

$$ \frac{\cos(\varphi)}{\cos(\varphi)} = \frac{\sin (\varphi)}{\cos(\varphi)} $$ $$ 1 = \tan(\varphi) $$ $$ \varphi = \tan^{-1}(1) $$

We use Using the restriction $0 \leq \varphi \leq \pi$, we conclude that $\varphi = \frac{\pi}{4}$ a.k.a. $45^{\circ}$.