Calculate the iterated integral. ∫03∫014xy(x2+y2)dydx

Consider the iterated integral

${\int }_0^3{\int }_0^{1}4xy(\sqrt{x^2+y^2})dydx$

The objective is to calculated the iterated integral

${\int }_0^3{\int }_0^{1}4xy(\sqrt{x^2+y^2})dydx$

To evaluate the iterated integral, first find the integral with respect to "y" and then apply the integral with respect to "x"${\int }_0^3{\int }_0^{1}4xy(\sqrt{x^2+y^2})dydx={\int }_0^3{\int }_0^1(2x)(2y)(\sqrt{x^2+y^2})dydx$

$={\int }_0^3{\int }_0^1(2x)(\sqrt{4})dudx$

$={\int }_0^3{\int }_0^1(2x)(4^{\frac{1}{2}})dudx$

$={\int }_0^3(2x){\left(\frac{4^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right)}_0^{1}dx$

$={\int }_0^3(2x){\left(\frac{4^{\frac{3}{2}}}{\frac{3}{2}}\right)}_0^{1}dx$

$={\int }_0^3(2x)\cdot \frac{2}{3}(4^{\frac{3}{2}}{)}_0^{1}dx$

$={\int }_0^3\frac{4x}{3}((x^2+y^2{)}^{\frac{3}{2}}{)}_0^{1}dx$

$={\int }_0^3\frac{4x}{3}((x^2+(1{)}^2{)}^{\frac{3}{2}}-(x^2+(0{)}^2{)}^{\frac{3}{2}}dx$

${\int }_0^3{\int }_0^{1}4xy(\sqrt{x^2+y^2})dydx={\int }_0^3\frac{4x}{3}((x^2+1{)}^{\frac{3}{2}}-(x^2{)}^{\frac{3}{2}})dx$

Continuous from the last step

${\int }_0^3{\int }_0^{1}4xy(\sqrt{x^2+y^2})dydx={\int }_0^3\frac{4x}{3}((x^2+1{)}^{\frac{3}{2}}-(x^2{)}^{\frac{3}{2}})dx$

$={\int }_0^3\frac{4x}{3}((x^2+1{)}^{\frac{3}{2}})dx-{\int }_0^3\frac{4x}{3}((x^2{)}^{\frac{3}{2}})dx$

$={\int }_0^3\frac{2}{3}\cdot 2x((x^2+1{)}^{\frac{3}{2}})dx-{\int }_0^3\frac{4x}{3}((x^2{)}^{\frac{3}{2}})dx$

$={\int }_0^3\frac{2}{3}\cdot 2x((x^2+1{)}^{\frac{3}{2}})dx-{\int }_0^3\frac{4x}{3}(x^3)dx$