Question

# Calculate the iterated integral. \int_{0}^{3} \int_{0}^{1} 4xy(\sqrt{x^2+y^2})dy dx

Integrals
Calculate the iterated integral.
$$\int_{0}^{3} \int_{0}^{1} 4xy(\sqrt{x^2+y^2})dy dx$$

2021-05-24

2021-09-08

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"$$\int_0^3\int_0^14xy(\sqrt{x^2+y^2})dydx=\int_0^3\int_0^1(2x)(2y)(\sqrt{x^2+y^2})dydx$$

$$=\int_0^3\int_0^1(2x)(\sqrt{4})dudx$$

$$=\int_0^3\int_0^1(2x)(4^{\frac{1}{2}})dudx$$

$$=\int_0^3(2x)\left(\frac{4^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right)_0^1dx$$

$$=\int_0^3(2x)\left(\frac{4^{\frac{3}{2}}}{\frac{3}{2}}\right)_0^1dx$$

$$=\int_0^3(2x)\cdot\frac{2}{3}(4^{\frac{3}{2}})_0^1dx$$

$$=\int_0^3\frac{4x}{3}((x^2+y^2)^{\frac{3}{2}})_0^1dx$$

$$=\int_0^3\frac{4x}{3}((x^2+(1)^2)^{\frac{3}{2}}-(x^2+(0)^2)^{\frac{3}{2}}dx$$

$$\int_0^3\int_0^1 4xy(\sqrt{x^2+y^2})dydx=\int_0^3\frac{4x}{3}((x^2+1)^{\frac{3}{2}}-(x^2)^{\frac{3}{2}})dx$$

Continuous from the last step

$$\int_0^3\int_0^1 4xy(\sqrt{x^2+y^2})dydx=\int_0^3\frac{4x}{3}((x^2+1)^{\frac{3}{2}}-(x^2)^{\frac{3}{2}})dx$$

$$=\int_0^3\frac{4x}{3}((x^2+1)^{\frac{3}{2}})dx-\int_0^3\frac{4x}{3}((x^2)^{\frac{3}{2}})dx$$

$$=\int_0^3\frac{2}{3}\cdot2x((x^2+1)^{\frac{3}{2}})dx-\int_0^3\frac{4x}{3}((x^2)^{\frac{3}{2}})dx$$

$$=\int_0^3\frac{2}{3}\cdot2x((x^2+1)^{\frac{3}{2}})dx-\int_0^3\frac{4x}{3}(x^3)dx$$

$$=\int_0^3\frac{2}{3}\cdot2x((x^2+1)^{\frac{3}{2}})dx-\int_0^3\frac{4}{3}(x^4)dx$$

$$=\int_0^3\frac{2}{3}\cdot2x((x^2+1)^{\frac{3}{2}})dx-\frac{4}{3}(\frac{x^5}{5})_0^3$$

$$=\int_0^3\frac{2}{3}\cdot2x((x^2+1)^{\frac{3}{2}})dx-\frac{4}{15}(x^5)_0^3$$

$$=\int_0^3\frac{2}{3}\cdot2x((x^2+1)^{\frac{3}{2}})dx-\frac{4}{15}((3)^5-(0^5))$$

$$=\int_0^3\frac{2}{3}\cdot2x((x^2+1)^{\frac{3}{2}})dx-\frac{4}{15}(243-0)$$

$$=\int_0^3\frac{2}{3}\cdot2x((x^2+1)^{\frac{3}{2}})dx-\frac{972}{15}$$

$$=\int_0^3\frac{2}{3}((4)^{\frac{3}{2}})du-\frac{972}{15}$$

$$\int_0^3\int_0^14xy(\sqrt{x^2+y^2})dydx=\frac{2}{3}\left(\frac{4^{\frac{3}{2}+1}}{\frac{3}{2}+1}\right)_0^3-\frac{972}{15}$$

$$=\frac{2}{3}\left(\frac{4^{\frac{5}{2}}}{\frac{5}{2}}\right)_0^3-\frac{972}{15}$$

$$=\frac{2}{3}\cdot\frac{2}{5}(u^{\frac{5}{2}})_0^3-\frac{972}{15}$$

$$=\frac{4}{15}(u^{\frac{5}{2}})_0^3-\frac{972}{15}$$

$$=\frac{4}{15}((x^2+1)^{\frac{5}{2}})_0^3-\frac{972}{15}$$

$$=\frac{4}{15}(((3)^2+1)^{\frac{5}{2}})_0^3-\frac{972}{15}$$

$$\int_0^3\int_0^1 4xy(\sqrt{x^2+y^2})dydx=\frac{4}{15}((10)^{\frac{5}{2}}-(1)^\frac{5}{2})-\frac{972}{15}$$

$$=\frac{4}{15}(100\sqrt{10}-1)-\frac{972}{15}$$

$$=\frac{5\cdot80}{5\cdot3}\sqrt{10}-\frac{976}{15}$$

$$=\frac{80}{3}\sqrt{10}-\frac{976}{15}$$

Hence, the value for iterated integral is

$$\int_0^3\int_0^1 4xy(\sqrt{x^2+y^2})dydx=\frac{80}{3}\sqrt{10}-\frac{976}{15}$$