Calculate the iterated integral∫01∫014xyx2+y2dxdy

${\displaystyle {\int }_0^1{\int }_0^{1}4xy\sqrt{x^2+y^2}dxdy}$

Evaluate ${\displaystyle {\int }_0^{1}4xy\sqrt{x^2+y^2}dx}$.

Since ${\displaystyle 4y}$ is constant with respect to ${\displaystyle x}$, move ${\displaystyle 4y}$ out of the integral.

${\displaystyle {\int }_0^{1}4y{\int }_0^{1}x\sqrt{x^2+y^2}dxdy}$

Let ${\displaystyle u_1=x^2+y^2}$. Then ${\displaystyle d{u}_1=2xdx}$, so ${\displaystyle \frac{1}{2}d{u}_1=xdx}$. Rewrite using ${\displaystyle u_1}$ and ${\displaystyle d}$${\displaystyle u_1}$.

${\displaystyle {\int }_0^{1}4y{\int }_{y^2}^{1+y^2}\sqrt{u_1}\frac{1}{2}d{u}_{1}dy}$

Combine ${\displaystyle \sqrt{u_1}}$ and ${\displaystyle \frac{1}{2}}$.

${\displaystyle {\int }_0^{1}4y{\int }_{y^2}^{1+y^2}\frac{\sqrt{u_1}}{2}d{u}_{1}dy}$

Since ${\displaystyle \frac{1}{2}}$ is constant with respect to ${\displaystyle u_1}$, move ${\displaystyle \frac{1}{2}}$ out of the integral.

${\displaystyle {\int }_0^{1}4y\left(\frac{1}{2}{\int }_{y^2}^{1+y^2}\sqrt{u_1}d{u}_1\right)dy}$

Simplify the expression.

${\displaystyle {\int }_0^{1}2y{\int }_{y^2}^{1+y^2}{u_1}^{\frac{1}{2}}d{u}_{1}dy}$

By the Power Rule, the integral of ${\displaystyle {u_1}^{\frac{1}{2}}}$ with respect to ${\displaystyle u_1}$ is ${\displaystyle \frac{2}{3}{u_1}^{\frac{3}{2}}}$.

${\displaystyle {\int }_0^{1}2y\frac{2}{3}{u_1}^{\frac{3}{2}}{]}_{y^2}^{1+y^2}dy}$

Combine ${\displaystyle \frac{2}{3}}$ and ${\displaystyle {u_1}^{\frac{3}{2}}}$.

${\displaystyle {\int }_0^{1}2y\frac{2{u_1}^{\frac{3}{2}}}{3}{]}_{y^2}^{1+y^2}dy}$

Substitute and simplify.

${\displaystyle {\int }_0^1\frac{2}{3}y(2{(1+y^2)}^{\frac{3}{2}}-2y^3)dy}$

Evaluate ${\displaystyle {\int }_0^1\frac{2}{3}y(2{(1+y^2)}^{\frac{3}{2}}-2y^3)dy}$.

Combine 23${\displaystyle \frac{2}{3}}$ and y${\displaystyle y}$.

${\displaystyle {\int }_0^1\frac{2y}{3}(2{(1+y^2)}^{\frac{3}{2}}-2y^3)dy}$

Let ${\displaystyle u_2=1+y^2}$. Then ${\displaystyle d{u}_2=2ydy}$, so ${\displaystyle \frac{1}{2y}d{u}_2=dy}$. Rewrite using ${\displaystyle u_2}$ and ${\displaystyle d}$${\displaystyle u_2}$.

${\displaystyle {\int }_1^2\frac{1}{3}(2{u_2}^{\frac{3}{2}}-2{\sqrt{u_2-1}}^3)d{u}_2}$

Rewrite ${\displaystyle {\sqrt{u_2-1}}^3}$ as ${\displaystyle {\left({(u_2-1)}^3\right)}^{\frac{1}{2}}}$.

${\displaystyle {\int }_1^2\frac{1}{3}(2{u_2}^{\frac{3}{2}}-2\sqrt{{(u_2-1)}^3})d{u}_2}$

Since ${\displaystyle \frac{1}{3}}$ is constant with respect to ${\displaystyle u_2}$, move ${\displaystyle \frac{1}{3}}$ out of the integral.

${\displaystyle \frac{1}{3}{\int }_1^{2}2{u_2}^{\frac{3}{2}}-2\sqrt{{(u_2-1)}^3}d{u}_2}$

Split the single integral into multiple integrals.

${\displaystyle \frac{1}{3}({\int }_1^{2}2{u_2}^{\frac{3}{2}}d{u}_2+{\int }_1^2-2\sqrt{{(u_2-1)}^3}d{u}_2)}$

Since ${\displaystyle 2}$ is constant with respect to ${\displaystyle u_2}$, move ${\displaystyle 2}$ out of the integral.

${\displaystyle \frac{1}{3}(2{\int }_1^2{u_2}^{\frac{3}{2}}d{u}_2+{\int }_1^2-2\sqrt{{(u_2-1)}^3}d{u}_2)}$

By the Power Rule, the integral of ${\displaystyle {u_2}^{\frac{3}{2}}}$ with respect to 2${\displaystyle u_2}$ is ${\displaystyle \frac{2}{5}{u_2}^{\frac{5}{2}}}$.

${\displaystyle \frac{1}{3}(2\left(\frac{2}{5}{u_2}^{\frac{5}{2}}{]}_1^2\right)+{\int }_1^2-2\sqrt{{(u_2-1)}^3}d{u}_2)}$

Combine ${\displaystyle \frac{2}{5}}$ and ${\displaystyle {u_2}^{\frac{5}{2}}}$.

${\displaystyle \frac{1}{3}(2\left(\frac{2{u_2}^{\frac{5}{2}}}{5}{]}_1^2\right)+{\int }_1^2-2\sqrt{{(u_2-1)}^3}d{u}_2)}$

Since ${\displaystyle -2}$ is constant with respect to ${\displaystyle u_2}$, move ${\displaystyle -2}$ out of the integral.

${\displaystyle \frac{1}{3}(2\left(\frac{2{u_2}^{\frac{5}{2}}}{5}{]}_1^2\right)-2{\int }_1^2\sqrt{{(u_2-1)}^3}d{u}_2)}$

Let ${\displaystyle u_3=u_2-1}$. Then ${\displaystyle d{u}_3=d{u}_2}$. Rewrite using ${\displaystyle u_3}$ and ${\displaystyle d}$${\displaystyle u_3}$.

${\displaystyle \frac{1}{3}(2\left(\frac{2{u_2}^{\frac{5}{2}}}{5}{]}_1^2\right)-2{\int }_0^1\sqrt{{u_3}^3}d{u}_3)}$

Use ${\displaystyle \sqrt[n]{a^x}=a^{\frac{x}{n}}}$ to rewrite ${\displaystyle \sqrt{{u_3}^3}}$ as ${\displaystyle {u_3}^{\frac{3}{2}}}$.

${\displaystyle \frac{1}{3}(2\left(\frac{2{u_2}^{\frac{5}{2}}}{5}{]}_1^2\right)-2{\int }_0^1{u_3}^{\frac{3}{2}}d{u}_3)}$

By the Power Rule, the integral of ${\displaystyle {u_3}^{\frac{3}{2}}}$ with respect to ${\displaystyle u_3}$ is ${\displaystyle \frac{2}{5}{u_3}^{\frac{5}{2}}}$.

${\displaystyle \frac{1}{3}(2\left(\frac{2{u_2}^{\frac{5}{2}}}{5}{]}_1^2\right)-2\left(\frac{2}{5}{u_3}^{\frac{5}{2}}{]}_0^1\right))}$

Combine ${\displaystyle \frac{2}{5}}$ and ${\displaystyle {u_3}^{\frac{5}{2}}}$.

${\displaystyle \frac{1}{3}(2\left(\frac{2{u_2}^{\frac{5}{2}}}{5}{]}_1^2\right)-2\left(\frac{2{u_3}^{\frac{5}{2}}}{5}{]}_0^1\right))}$

Substitute and simplify.

${\displaystyle \frac{2(2^{\frac{7}{2}}-2)-4}{15}}$

Simplify the numerator.

${\displaystyle \frac{2^{\frac{9}{2}}-8}{15}}$

The result can be shown in multiple forms.

Exact Form:

${\displaystyle \frac{2^{\frac{9}{2}}-8}{15}}$

Decimal Form:

$0.97516113\dots$