# Calculate the iterated integral: $\int_{0}^{3} \int_{0}^{1} 4xy (\sqrt{x^2 + y^2}) \, dydx$

Double and Triple Integrals
Double and Triple Integrals

This question aims to find the iterated integral by first finding the integral of $y$ and then $x$ with the given range for $x$ and $y$.

This question uses the concept of Calculus and especially double integrals. The basic idea of integration is to find the surface area of two-dimensional regions and the volume of three-dimensional objects.

The given Iterated integral is as follows:

$\int_{0}^{3} \int_{0}^{1} 4xy (\sqrt{x^2 + y^2}) dydx$

We first need to solve it for $y$ and then for $x$.

$= \int_{0}^{3} \int_{0}^{1} (2x)(2y) (\sqrt{x^2 + y^2}) dydx$

$Assume, u=x^2 + y^2$

$= \int_{0}^{3} \int_{0}^{1} (2x)(\sqrt{u}) dudx$

$= \int_{0}^{3} \int_{0}^{1} (2x)(u^\frac{1}{2}) dudx$

By using the formula: $\int x^n=\frac{x^n+1}{n+1}$

We get:

$= \int_{0}^{3} (2x)\frac{2}{3}\left[(u^\frac{3}{2})\right]_{1}^{0} dudx$

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

So we already know that $u=x^2 +y^2$

$= \int_{0}^{3} \frac{4x}{3}\left [(x^2 +(1)^2)^\frac{3}{2} – (x^2 +(0)^2)^\frac{3}{2} \right]dx$

$= \int_{0}^{3} \frac{4x}{3}\left [(x^2 +1)^\frac{3}{2} – (x^2 )^\frac{3}{2} \right]dx$

$= \int_{0}^{3} \frac{4x}{3}\left [(x^2 +1)^\frac{3}{2}\right]dx – \int_{0}^{3} \frac{4x}{3}\left [(x^2 )^\frac{3}{2}\right]dx$

$= \int_{0}^{3} \frac{4x}{3}\left [(x^2 +1)^\frac{3}{2}\right]dx – \int_{0}^{3} \frac{4x}{3}\left [(x^3)\right]dx$

$= \int_{0}^{3} \frac{4x}{3}\left [(x^2 +1)^\frac{3}{2}\right]dx – \int_{0}^{3} \frac{4}{3}\left [(x^4)\right]dx$

$= \int_{0}^{3} \frac{4x}{3}\left [(x^2 +1)^\frac{3}{2}\right]dx – \int_{0}^{3} \frac{4}{3}\left [(x^4)\right]dx$

$= \int_{0}^{3} \frac{4x}{3}\left [(x^2 +1)^\frac{3}{2}\right]dx – \frac{4}{3}\left [(\frac{x^5}{5})\right]_{0}^{3}$

$= \int_{0}^{3} \frac{4x}{3}\left [(x^2 +1)^\frac{3}{2}\right]dx – \frac{4}{15}\left [(x^5)\right]_{0}^{3}$

$= \int_{0}^{3} \frac{4x}{3}\left [(x^2 +1)^\frac{3}{2}\right]dx – \frac{4}{15}\left [(3)^5-(0)^5\right]_{0}^{3}$

By inserting the integral values, we get:

$= \int_{0}^{3} \frac{4x}{3}\left [(x^2 +1)^\frac{3}{2}\right]dx – \frac{4}{15}(243)$

$= \int_{0}^{3} \frac{4x}{3}\left [(x^2 +1)^\frac{3}{2}\right]dx – \frac{972}{15}$

$= \int_{0}^{3} \frac{2}{3}2x\left [(x^2 +1)^\frac{3}{2}\right]dx – \frac{972}{15}$

Assume $u=x^2+1$, so $du=2x dx$

$= \int_{0}^{3} \frac{2}{3}\left [(u^\frac{3}{2}) \right]du – \frac{972}{15}$

$= \frac{4}{15}\left [(u^\frac{5}{2}) \right]_{0}^{3} – \frac{972}{15}$

As we know that $u=x^2+1$, so:

$= \frac{4}{15}\left [(x^2 +1)^\frac{5}{2}) \right]_{0}^{3} – \frac{972}{15}$

$= \frac{4}{15}\left [(10)^\frac{5}{2} -(1)^\frac{5}{2} \right]_{0}^{3} – \frac{972}{15}$

By inserting the integral values, we get:

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

$= \frac{400}{15}\sqrt{10}-\frac{4}{15}-\frac{972}{15}$

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

## Numeric Result

The iterate integral of given the given expression is as follows:

$\int_{0}^{3} \int_{0}^{1} 4xy (\sqrt{x^2 + y^2}) dydx = \frac{80}{3}\sqrt{10}-\frac{976}{15}$

## Example

Calculate the iterated integral of the expression given below.

$\int_{0}^{3}\int_{0}^{3}\dfrac{8 + 10y}{\sqrt{x}} dx dy$

Simplifying the given expression:

$= \int_{0}^{3}\int_{0}^{3}(8 + 10y)x^{-\frac{1}{2}} dx dy$

$=\int_{0}^{3}(8 + 10y)dy \int_{0}^{3}x^{-\frac{1}{2}} dx$

$= \int_{0}^{3}(8 + 10y)dy \left[ \frac{x^{- \frac{1}{2} + 1}}{\frac{-1}{2} + 1} \right]_{0}^{3}$

$= \int_{1}^{2}(8 + 10y)dy \left[ \frac{x^{\frac{1}{2}}}{\frac{1}{2}} \right]_{0}^{3}$

By inserting the integral values and solving the expression for $dx$ as:

$= \int_{1}^{2}(3 + 5y)dy \left[ 2(9^{\frac{1}{2}} – 4^{\frac{1}{2}}) \right]$

$= \int_{0}^{3}(8 + 10y)dy \left[ 2(3 ) \right]$

$= 3.46\int_{0}^{3}(8 + 10y)dy$

$= 3.46\left[8y + \frac{10y^2}{2} \right]_{0}^{3}$

By inserting the integral values and solving the expression for $dy$ as:

$= 3.46\left[ 3(3) + \frac{10}{2}(3^2) \right]$

$= 3.46\left[ 9 + \frac{90}{2}\right]$

$= 3.46(54)$

$= 186.84$

Hence, the final value we have is:

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