"Find the mass of the region bounded by the xy-plane and the hemisphere $z = \sqrt{100 − x^2 − y^2}$, if the mass density of the region is given by the function $f(x,y,z) = \frac{z}{\sqrt{x^2+y^2+z^2}}$".
So originally I tried graphing this problem but when it comes to determining the bounds for the triple integral is where i fall. I started by attempting to find the individual traces that I can analyze for bounds,
i end up getting:
Hint. To find the bounds for the hemisphere, note that the flat face is a circle in the $xy$-plane. Now looking at the expression for $z$ and imposing the condition that $z$ be real gives $$100-x^2-y^2\ge 0,$$ which implies $$ x^2+y^2\le 10^2.$$ Thus the radius of the flat face, and hence the hemisphere is $10.$ Solving for $y^2$ in the last inequality gives $y^2\le 10^2-x^2,$ so that $|y|\le \sqrt{10^2-x^2},$ or in other words we have that $$-\sqrt{10^2-x^2}\le y\le \sqrt{10^2-x^2}.$$ Now imposing the condition that $y^2$ be nonnegative gives $0\le 10^2-x^2\implies x^2\le 10^2\implies |x|\le 10,$ or in other words $$-10\le x\le 10.$$ Finally we obtain the bounds for $z$ as varying from $0$ to $\sqrt {100-x^2-y^2}.$
Hence the differential of mass is $$\frac{z}{\sqrt{x^2+y^2+z^2}}\mathrm dx\,\mathrm dy\,\mathrm dz,$$ so that the desired mass is $$\int_{-10}^{10}\int_{-\sqrt{10^2-x^2}}^{\sqrt{10^2-x^2}}\int_0^{\sqrt {100-x^2-y^2}}\frac{z}{\sqrt{x^2+y^2+z^2}}\mathrm dz\,\mathrm dy\,\mathrm dx=\frac12\int_{-10}^{10}\int_{-\sqrt{10^2-x^2}}^{\sqrt{10^2-x^2}}\int_0^{\sqrt {100-x^2-y^2}}\frac{2z}{\sqrt{x^2+y^2+z^2}}\mathrm dz\,\mathrm dy\,\mathrm dx=\int_{-10}^{10}\int_{-\sqrt{10^2-x^2}}^{\sqrt{10^2-x^2}}\int_0^{\sqrt {100-x^2-y^2}}\frac12\frac{1}{\sqrt{x^2+y^2+z^2}}\mathrm d(x^2+y^2+z^2)\,\mathrm dy\,\mathrm dx,$$ and the innermost integral is simply $\sqrt{x^2+y^2+z^2},$ which you need to evaluate between the limits $0\le z\le \sqrt {100-x^2-y^2}.$ Then proceed.