# Thread: changing order of integration

1. ## changing order of integration

∫∫x^3 y/ 1-y^2 dy dx

x= 0
x=1
y= x^2
y=x

By changing the order of integration show that the value of the integral is 1/16

2. $\int_{0}^{1}\int_{y}^{\sqrt{y}}\left(\frac{x^{3}y} {1-y^{2}}\right)dxdy$

3. dude the y integral is not the square root of x. it is x squared!

4. Originally Posted by stuwy
dude the y integral is not the square root of x. it is x squared!
stop shouting.

$y = x^2 \implies x = \pm \sqrt{y}$

after sketching the region, galactus realize he needed to take the positive square root

5. Hello, duude!

$\int^1_0\int^x_{x^2}\frac{x^3y}{1-y^2}\,dy\,dx$

By changing the order of integration, show that the value of the integral is $\frac{1}{16}$
The limits are: . $\begin{Bmatrix}x = 1 \\ x=0\end{Bmatrix} \text{ and }\begin{Bmatrix}y = x \\ y = x^2\end{Bmatrix}$

The region looks like this:
Code:
        |
1+ - - - - - *
|         **:
|       *:: :
|     *:::* :
|   *::::*  :
| *::::*    :
- **- - - - - + - -
|           1

Reversing the limits, we have: . $\begin{Bmatrix}y = 1 \\ y = 0\end{Bmatrix}\text{ and }\begin{Bmatrix}x = \sqrt{y} \\ x = y\end{Bmatrix}$

And the integral is: . $\int^1_0\int^{\sqrt{y}}_y\frac{x^3y}{1-y^2}\,dx\,dy$

The inner integral is: . $\int^{\sqrt{y}}_y\frac{x^3y}{1-y^2}\,dx \;\;=\;\;\frac{1}{4}\cdot\frac{x^4y}{1-y^2}\,\bigg|^{\sqrt{y}}_y \;\;=\; \;\frac{1}{4}\,\left[\frac{(\sqrt{y})^4y}{1-y^2} - \frac{y^4\cdot y}{1-y^2}\right]$

. . $= \;\;\frac{1}{4}\left[\frac{y^3}{1-y^2} - \frac{y^5}{1-y^2}\right] \;\;=\;\;\frac{1}{4}\left[\frac{y^3(1-y^2)}{1-y^2}\right] \;\;=\;\;\frac{1}{4}y^3$

Then we have: . $\int^1_0\frac{1}{4}y^3\,dy \;\;=\;\;\frac{1}{16}y^4\,\bigg|^1_0 \;\;=\;\;\frac{1}{16}$

6. Thanks dude!
That was great!

Cheers!