Recall the for polar coordinate
[tex]x^2+y^2=r^2 \\ \\ dA=rdrd\theta[/tex]
Given that solid is bounded by [tex]z=\sqrt{x^2+y^2}[/tex] and [tex]z=\sqrt{50-x^2-y^2}[/tex]
[tex]\sqrt{x^2+y^2}=\sqrt{50-x^2-y^2} \\ \\ \Rightarrow x^2+y^2=50-x^2-y^2 \\ \\ \Rightarrow2(x^2+y^2)=50 \\ \\ x^2+y^2=25 \\ \\ \Rightarrow r=5[/tex]
[tex]V= \int\limits_{-5}^{5} \int\limits^{\sqrt(25-x^2}_{-\sqrt{25-x^2}} {\sqrt{50-x^2-y^2}-\sqrt{x^2+y^2}} \, dydx \\ \\ =\int\limits_{0}^{2\pi} \int\limits^{5}_{0}} {(\sqrt{50-r^2}-r)} \, rdrd\theta=\int\limits_{0}^{2\pi}{ \frac{250(\sqrt{2}-1)}{3} }d\theta=\left.\frac{250(\sqrt{2}-1)}{3} \theta}\right|^{2\pi}_0 \\ \\ =\frac{500(\sqrt{2}-1)}{3} [/tex]
