Question

Can anyone help answer this: find the volume of the solid E B/t z=(3(x^2+y^2))^(1/2) a cone and z=(1-x^2-y^2)^(1/2) by v=triple integral dv

Answer 1

\[z_1=\sqrt{3(x^2+y^2)}\quad,\quad z_2=\sqrt{1-(x^2+y^2)}\]\[z_1=z_2\quad\Rightarrow\quad x^2+y^2=(1/2)^2\quad(D)\]\[V=\iint\limits_D (z_2-z_1)dxdy=\]\[=\iint\limits_D \left(\sqrt{1-(x^2+y^2)}-\sqrt{3(x^2+y^2)}\right)dxdy=\]\[=\iint\limits_{D }\left(\sqrt{1-\rho^2}-\sqrt{3\rho^2}\right)\rho d\rho d\phi=\]\[=\int\limits_0^{2\pi}d\phi\cdot\int\limits_{0}^{1/2}\left(\sqrt{1-\rho^2}-\rho\sqrt{3}\right)\rho d\rho=\]\[=2\pi\cdot\left(-\frac{1}{3}(1-\rho^2)^{3/2}-\frac{\sqrt{3}}{3}\rho^3\right)_0^{1/2}=\]\[=-\frac{2\pi}{3}\cdot\left((1-\rho^2)^{3/2}+\rho^3\sqrt{3}\right)_0^{1/2}=-\frac{2\pi}{3}\cdot\left(\frac{3\sqrt{3}}{8}-1+\frac{\sqrt{3}}{8}\right)=\]\[=-\frac{2\pi}{3}\cdot\left(\frac{\sqrt{3}}{2}-1\right)=\frac{\pi}{3}(2-\sqrt{3})\]