To get the mass, we integrate the density function over the given region.
In cylindrical coordinates, the sphere has equation
[tex]x^2 + y^2 + (z-1)^2 = 1 \implies (z-1)^2 = 1-r^2 \implies z = 1 \pm \sqrt{1-r^2}[/tex]
The region of interest lies below the top half of the sphere, so that [tex]z\le1+\sqrt{1-r^2}[/tex].
At the lower end, the cone has equation, and hence [tex]z[/tex] has lower limit,
[tex]z = \sqrt{x^2+y^2} = \sqrt{r^2} = r[/tex]
The upper hemisphere and cone meet when
[tex]1 + \sqrt{1 - r^2} = r \implies 1-r^2 = (r-1)^2 \implies 2r^2 - 2r = 0 \\\\ \implies r=0\text{ or } r=1[/tex]
Then the mass of the given solid is
[tex]\displaystyle \mathrm{mass} = \iiint_E z \sqrt{x^2+y^2} \, dV \\\\ ~~~~~~~~ = \int_0^{2\pi} \int_0^1 \int_r^{1+\sqrt{1-r^2}} r^2 z \, dz \, dr \, d\theta \\\\ ~~~~~~~~ = \pi \int_0^1 r^2 \left(\left(1+\sqrt{1-r^2}\right)^2 - r^2\right) \, dr \\\\ ~~~~~~~~ = 2\pi \int_0^1 (r^2 - r^4) \, dr + 2\pi \int_0^1 r^2 \sqrt{1-r^2} \, dr \\\\ ~~~~~~~~ = \frac{4\pi}{15} + 2\pi \int_0^1 r^2 \sqrt{1-r^2} \, dr[/tex]
Integrate by parts.
[tex]u = r \implies du = dr \\\\ dv = r \sqrt{1 - r^2} \, dr \implies v = -\dfrac13 (1-r^2)^{3/2}[/tex]
[tex]\implies \displaystyle \int_0^1 r^2 \sqrt{1-r^2} \, dr = \frac13 \int_0^1 (1 - r^2)^{3/2} \, dr[/tex]
Substitute [tex]r=\sin(s)[/tex] and [tex]dr=\cos(s)\,ds[/tex].
[tex]\displaystyle \int_0^1 (1-r^2)^{3/2} \, dr = \int_0^{\pi/2} \cos^4(s) \, ds \\\\ ~~~~~~~~ = \int_0^{\pi/2} \left(\frac{1 + \cos(2s)}2\right)^2 \, ds \\\\ ~~~~~~~~ = \frac14 \int_0^{\pi/2} (1 + 2\cos(2s) + \cos^2(2s)) \, ds \\\\ ~~~~~~~~ = \frac14 \int_0^{\pi/2} \left(1 + 2\cos(2s) + \frac{1+\cos(4s)}2\right) \, ds \\\\ ~~~~~~~~ = \frac14 \int_0^{\pi/2} (1 + 2\cos(2s)) \, ds + \frac18 \int_0^{\pi/2} (1 + \cos(4s)) \, ds \\\\ ~~~~~~~~ = \frac\pi8 + \frac\pi{16} = \frac{3\pi}{16}[/tex]
Putting it all together, the mass is
[tex]\displaystyle \mathrm{mass} = \frac{4\pi}{15} + 2\pi \int_0^1 r^2 \sqrt{1-r^2} \, dr \\\\ ~~~~~~~~ = \frac{4\pi}{15} + \frac{2\pi}3 \int_0^1 (1 - r^2)^{3/2} \, dr \\\\ ~~~~~~~~ = \frac{4\pi}{15} + \frac{2\pi}3 \cdot \frac{3\pi}{16} = \boxed{\frac{4\pi}{15} + \frac{\pi^2}8}[/tex]
