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
First find the intersection points by setting the two equations together: $\displaystyle y(2-y) = -y \ \Leftrightarrow \ 0 = y^2 - 3y \ \Leftrightarrow \ 0 = y(y-3)$
These points will be your limits of integration.
Now, recall: $\displaystyle A = \int_a^b \left[f(y) - g(y)\right]dy$ where $\displaystyle f(y) \geq g(y)$ over $\displaystyle a \leq y \leq b$
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 $\displaystyle f(y)$
Hello, holly123!
Did you make a sketch?
Determine the area bounded by: .$\displaystyle \begin{array}{c}x\:=\:y(2-y) \\x\:=\:-y \end{array}$Code:\ | * \(-3,3) | o | \::*. | \:::::*. \::::|::* \:::|::::* \::|:::::* \:|::::* \|::* - - - - - - - - o - - - - - - * |\ * | \ * | \ |
The graphs intersect at (0,0) and (-3,3).
We'd best integrate "sideways": .$\displaystyle A \;=\;\int^b_a\left(x_{\text{right}} - x_{\text{left}}\right)\,dy $
. . $\displaystyle A \;=\;\int^3_0\bigg[y(2-y) - (-y)\bigg]\,dy \;=\;\int^3_0\bigg[3y-y^2\bigg]\,dy $
Got it?