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The region looks like an ice cream cone. It is an upside-down circular cone attached to a slice of a sphere. I'm pretty sure the way you are "supposed" to solve it is with spherical coordinates, since all of the numbers become suspiciously convenient.

I know the formula for mass is $$\int \int \int\,\, [\text{density}] \,\,dV$$

I will do $dr\, d\phi \,d\theta$. I think $r$ goes between $0$ and $2\sqrt{2}$, and $\phi$ goes between $0$ and $\pi/4$ (is the angle of the cone $45^\circ$?), and $\theta$ goes between $0$ and $2\pi$.

Also, $$x = r\sin\phi \cos \theta$$ $$y = r\sin\phi \sin \theta$$ $$z = r\cos\phi$$

And that means that the density, which is $x^2 + y^2 + z^2$, is just equal to $r^2$. We also have to multiply by $r^2 \sin \phi$ because it's spherical coordinates.

So the mass is

$$\text{mass} = \int_0^{2\pi}\int_0^{\pi/4}\int_0^{2\sqrt{2}} \,r^2 \left(r^2 \sin \phi\right)\,\,dr\, d\phi \,d\theta$$

I got $$\frac{256 \pi \sqrt{2} - 256 \pi}{5}$$ but that isn't one of the answers.

What did I do wrong? Or is the question wrong?

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    $\begingroup$ See nothing wrong with your working. $\endgroup$
    – Math Lover
    Oct 18, 2020 at 21:33

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The angle of $\phi$ can by determined by substituting function

$$ x^2+y^2=z^2 \to (using spherical coordinates) \to r^2sin^2(\phi)cos^2(\theta)+r^2sin^2(\phi)sin^2(\theta)=r^2cos^2(\phi) $$ $$ \to sin^2(\phi) = \cos^2(\phi) $$ This condition is met for $\phi = \pi/4$ so there is your angle.

I calculated the mass to be $2^6\pi\frac{1}{5}\sqrt{2}^5(1-\frac{\sqrt{2}}{2})$

But it is pretty late so I may have made a mistake so please check it. And also check your result too.

Hopefully it will be helpful. At least your question about the angle being $\pi/4$ is answered.

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