1. ## Area Between Curves

x=y(2-y) and x=-y

Determine the area of the region bound

I can solve it but I just need to know how to set up the integral.

Thanks

2. first find the intersection points to determine the limits of integration ...

solve $y(2-y) = -y$ , you will get two solutions for y.

$A = \int_c^d (right \, curve) - (left \, curve) \, dy$

3. thank you but how do i graph them? i dont know how to solve for y in
x=y(2-y)

4. First find the intersection points by setting the two equations together: $y(2-y) = -y \ \Leftrightarrow \ 0 = y^2 - 3y \ \Leftrightarrow \ 0 = y(y-3)$

These points will be your limits of integration.

Now, recall: $A = \int_a^b \left[f(y) - g(y)\right]dy$ where $f(y) \geq g(y)$ over $a \leq y \leq b$

So, find the larger function (i.e. the one that is more to the right since we're integrating with respect to y) and this will be our $f(y)$

5. Hello, holly123!

Did you make a sketch?

Determine the area bounded by: . $\begin{array}{c}x\:=\:y(2-y) \\x\:=\:-y \end{array}$
Code:
        \         |
*    \(-3,3)  |
o       |
\::*.  |
\:::::*.
\::::|::*
\:::|::::*
\::|:::::*
\:|::::*
\|::*
- - - - - - - - o - - - - - -
*   |\
*       | \
*             |  \
|

The graphs intersect at (0,0) and (-3,3).

We'd best integrate "sideways": . $A \;=\;\int^b_a\left(x_{\text{right}} - x_{\text{left}}\right)\,dy$

. . $A \;=\;\int^3_0\bigg[y(2-y) - (-y)\bigg]\,dy \;=\;\int^3_0\bigg[3y-y^2\bigg]\,dy$

Got it?

6. thanks so much i get it now