Question #180531

Find the mass of the solid bounded by z = 1 and , 2 2 z = x + y the density function

being d (x, y, z) = | x | .


Expert's answer

Let Ω\Omega be the region bounded by z = 1 and z = x2 + y2 .

Then ∫Ω∣x∣dxdydz=∫01∫−zz∫−z−y2z−y2∣x∣dxdydz=∫01∫−zz∫0z−y22∣x∣dxdydz=∫01∫−zzx2∣0z−y2dydz=∫01∫−zz(z−y2)dydz=∫01(zy−y3/3)∣−zzdydz=∫0143z2/3dz=4335z5/3∣01=4/5\int\limits_{\Omega}|x|dx dy dz=\int\limits_0^1\int\limits_{-\sqrt{z}}^{\sqrt{z}}\int\limits_{-\sqrt{z-y^2}}^{\sqrt{z-y^2}}|x|dxdydz=\int\limits_0^1\int\limits_{-\sqrt{z}}^{\sqrt{z}}\int\limits_{0}^{\sqrt{z-y^2}}2|x|dxdydz=\int\limits_0^1\int\limits_{-\sqrt{z}}^{\sqrt{z}}x^2|_0^{\sqrt{z-y^2}}dydz=\int\limits_0^1\int\limits_{-\sqrt{z}}^{\sqrt{z}}(z-y^2)dydz=\int\limits_0^1(zy-y^3/3)|_{-\sqrt{z}}^{\sqrt{z}}dydz=\int\limits_0^1\frac{4}{3}z^{2/3}dz=\frac{4}{3}\frac{3}{5}z^{5/3}|_0^1=4/5

Answer. The mass of the solid equals to 4/5


LATEST TUTORIALS
APPROVED BY CLIENTS