Question

Find the mass of the solid bounded by z = 1 and

z = x² + y² the density function 

being δ(x,y,z)=|x|

Answer 1

$mass=density × volume\newline \delta=|x|\newline mass=\int \int \int \delta dV\newline Where \delta=|x|\newline mass=\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^ {\sqrt{1-x^2}}\int_{x^2+y^2}^1|x| dV \newline =\int_{-1}^{0} \int_{-\sqrt{1-x^2}}^ {\sqrt{1-x^2}}\int_{x^2+y^2} ^1(-x)dzdydx+\int_{0}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{x^2+y^2}^1xdzdydx \newline Let I_1=\int_{-1}^{0} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{x^2+y^2} ^1(-x)dzdydx\newline and \newline I_2=\int_{0}^{1} \int_{-\sqrt{1-x^2}}^ {\sqrt{1-x^2}}\int_{x^2+y^2}^1xdzdydx \newline Solving I_1,\newline I_1=-\int_{-1}^{0} \int_{-\sqrt{1-x^2}}^ {\sqrt{1-x^2}}x(1-x^2-y^2)dydx\newline =-\int_{-1}^{0}(xy-x^3y-\frac{xy^3}{3})|_{-\sqrt{1-x^2}}^ {\sqrt{1-x^2}}dx\newline =-\int_{-1}^{0}(x\sqrt{1-x^2}-x^3\sqrt{1-x^2}-\frac{x(1-x^2)\sqrt{1-x^2}}{3})-(-x\sqrt{1-x^2}+x^3\sqrt{1-x^2}+\frac{x(1-x^2)\sqrt{1-x^2}}{3})dx\newline =-2\int_{-1}^{0}(x\sqrt{1-x^2}-x^3\sqrt{1-x^2}-\frac{x(1-x^2)\sqrt{1-x^2}}{3})dx\newline Simplify,\newline I_1=-\frac{4}{3}\int_{-1}^{0}x(1-x^2)\sqrt{1-x^2})dx\newline Put 1-x^2=t^2 \implies -xdx=tdt\newline I_1=\frac{4}{3}\int_{0}^{1}t^4dt\newline =\frac{t^5}{5}|_0^1 \newline =\frac{4}{15}\newline Similarly, solving for I_2, we get\newline I_2=\frac{4}{15}\newline Therefore, mass=I_1+I_2 =\frac{8}{15}=0.533$