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**.

## Expert Answer

Given function:

\[ 4x y^2\]

Here,Â **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:

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

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:

\[Â = \int_{-1}^{1}{\int_{0}^{{4-y^2}}{4x\ y^2} dx\ dy\ }\ \]

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:

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

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

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

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

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

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

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

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

Multiplying $2$ inside the brackets:

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

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 $x$:

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

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

\[=\dfrac{1}{6}\]