# Evaluate the double integral. 4xy^2 dA, d is enclosed by x=0 and x=4−y^2 d.

In this question, we have to find the double Integration of the given function $4 x y^2$ by first integrating $x$, and then we will integrate the function with the given limits of $y$.

The basic concept behind this question is the knowledge of double integration, limits of integration, and where to write the limits of the first variable and limits of the second variable in the integral.

Given function:

$4x y^2$

Here, region $D$ is bounded by a double integral in which it is enclosed by:

$x = 0 \space ; \space x = {4 – y^2 }$

And then with another:

$y = -1 \space ; \space y = 1$

So the domain $D$ is given by:

$D = \{ x, y \}\ , -1 \le y \le 1 \space ; \space 0 \le x \le {4-y^2}$

Now to solve the given function in a double integration, we have to identify the limits of integration carefully. As given the limits of integral $y$ varies from $- 1$ to $1$ which can be represented as:

$= \int_{-1}^{1}$

And the limits of $x$ goes from $0$ to ${4-y^2}$ so we can write the function as:

$= \int_{0}^{ {4-y^2} }$

And our function is:

$= {4 x\ y^2 dA}$

Now as $dA$ is enclosed by variable $x$ and variable $y$ so writing the differential in terms of the variable $x$ as well as the variable $y$ we will get it:

$= {4x\ y^2} dx\ dy\$

By putting both the limits together we get:

$= \int_{-1}^{1}{\int_{0}^{{4-y^2}}{4x\ y^2} dx\ dy\ }\$

Now for solving the above equation, first we will solve the integration part of the variable $x$ which will give the equation in terms of variable $y$ as clearly indicated by the limits of variable $x$. Thus solving integral gives:

$=\int_{-1}^{1} \left[ 4\dfrac{x^2}{ 2}\right]_{0}^{{4-y^2}}{ y^2} dy\ \$

Putting the limits of variable $x$ in the above equation we get:

$=\int_{-1}^{1} \left[ \dfrac{4({4-y^2})^2}{2} – \dfrac{4{(0)}^2}{2} \right] { y^2} dy\ \$

Solving the equation by taking a square and simplifying we have:

$=\int_{-1}^{1} \left[ \dfrac{4(y^4- 8y^2+16)}{2} – \dfrac{0}{2} \right] { y^2} dy\ \$

$=\int_{-1}^{1} \left[ \dfrac{4(y^4- 8y^2+16)}{2} – 0 \right] { y^2} dy\ \$

$=\int_{-1}^{1} \left[ \dfrac{4(y^4- 8y^2+16)}{2} \right] { y^2} dy\ \$

$=\int_{-1}^{1} \left[ {2(y^4- 8y^2+16)} \right] { y^2} dy\ \$

Multiplying $2$ inside the brackets:

$=\int_{-1}^{1} \left[ {2y^4- 16y^2+ 32)} \right] { y^2} dy\ \$

Multiplying $y^2$ inside the square brackets:

$=\int_{-1}^{1} {2y^6- 16y^4+ 32 y^2}dy$

Solving for $y$ integral:

$=\left[\dfrac{2y^7}{ 7}-16\dfrac{y^5}{5} +32\dfrac{y^3}{3} \right]_{-1}^{1}$

Now solving the above equation and putting values of the limit, we get:

$=\dfrac{1628}{105}$

$=15.50$

## Numerical Results

$=\dfrac{1628}{105}=15.50$

## Example

Integrate the double integral:

$\int_{0}^{1}{\int_{0}^{y}}{x\ y} dx\ dy$

Solution:

$=\int_{0}^{1} \left[\dfrac{x^2}{ 2}\right]_{0}^{y}{ y}dy$

Putting the limit of $x$:

$=\int_{0}^{1} \left[ \dfrac{y^2}{2}-\dfrac{{0}^2}{2}\right]{y}dy$

$=\left[\dfrac{y^3}{6}\right]_{0}^{1}$

$=\dfrac{1}{6}$