 necessaryh

2021-05-23

Calculate the iterated integral.
${\int }_{0}^{3}{\int }_{0}^{1}4xy\left(\sqrt{{x}^{2}+{y}^{2}}\right)dydx$ Velsenw

Heres Jeffrey Jordon

Consider the iterated integral

${\int }_{0}^{3}{\int }_{0}^{1}4xy\left(\sqrt{{x}^{2}+{y}^{2}}\right)dydx$

The objective is to calculated the iterated integral

${\int }_{0}^{3}{\int }_{0}^{1}4xy\left(\sqrt{{x}^{2}+{y}^{2}}\right)dydx$

To evaluate the iterated integral, first find the integral with respect to "y" and then apply the integral with respect to "x"${\int }_{0}^{3}{\int }_{0}^{1}4xy\left(\sqrt{{x}^{2}+{y}^{2}}\right)dydx={\int }_{0}^{3}{\int }_{0}^{1}\left(2x\right)\left(2y\right)\left(\sqrt{{x}^{2}+{y}^{2}}\right)dydx$

$={\int }_{0}^{3}{\int }_{0}^{1}\left(2x\right)\left(\sqrt{4}\right)dudx$

$={\int }_{0}^{3}{\int }_{0}^{1}\left(2x\right)\left({4}^{\frac{1}{2}}\right)dudx$

$={\int }_{0}^{3}\left(2x\right){\left(\frac{{4}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right)}_{0}^{1}dx$

$={\int }_{0}^{3}\left(2x\right){\left(\frac{{4}^{\frac{3}{2}}}{\frac{3}{2}}\right)}_{0}^{1}dx$

$={\int }_{0}^{3}\left(2x\right)\cdot \frac{2}{3}\left({4}^{\frac{3}{2}}{\right)}_{0}^{1}dx$

$={\int }_{0}^{3}\frac{4x}{3}\left(\left({x}^{2}+{y}^{2}{\right)}^{\frac{3}{2}}{\right)}_{0}^{1}dx$

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

${\int }_{0}^{3}{\int }_{0}^{1}4xy\left(\sqrt{{x}^{2}+{y}^{2}}\right)dydx={\int }_{0}^{3}\frac{4x}{3}\left(\left({x}^{2}+1{\right)}^{\frac{3}{2}}-\left({x}^{2}{\right)}^{\frac{3}{2}}\right)dx$

Continuous from the last step

${\int }_{0}^{3}{\int }_{0}^{1}4xy\left(\sqrt{{x}^{2}+{y}^{2}}\right)dydx={\int }_{0}^{3}\frac{4x}{3}\left(\left({x}^{2}+1{\right)}^{\frac{3}{2}}-\left({x}^{2}{\right)}^{\frac{3}{2}}\right)dx$

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

$={\int }_{0}^{3}\frac{2}{3}\cdot 2x\left(\left({x}^{2}+1{\right)}^{\frac{3}{2}}\right)dx-{\int }_{0}^{3}\frac{4x}{3}\left(\left({x}^{2}{\right)}^{\frac{3}{2}}\right)dx$

$={\int }_{0}^{3}\frac{2}{3}\cdot 2x\left(\left({x}^{2}+1{\right)}^{\frac{3}{2}}\right)dx-{\int }_{0}^{3}\frac{4x}{3}\left({x}^{3}\right)dx$

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