# Area of region bounded by sqrt x, 2 - sqrt x and y = 0

• Sep 22nd 2010, 08:40 AM
darkvelvet
Area of region bounded by sqrt x, 2 - sqrt x and y = 0
I need to find the area of the region bounded by the graphs of:
y = sqrt x,
y = 2 - sqrt x, and
y = 0.

I have sketched the graph and I know where the boundary is.

I intend to integrate with respect to y. Am I on the right track?
$\int_0^2 y^2 - (4 - y^2) dy$

The answer given is 4 sq units but I don't think I will get that answer after integration.
• Sep 22nd 2010, 03:40 PM
Soroban
Hello, darkvelvet!

And I don't agree with their answer.

Quote:

$\text{Find the area of the region bounded by: }\;\begin{Bmatrix}y \:=\:\sqrt{x} \\ y \:=\:2 - \sqrt{x} \\ y\:=\:0.\end{Bmatrix}$

The graph of $y = \sqrt{x}$ is the upper half of a "horizontal" parabola,
. . vertex at (0,0), opening to the right.

The graph of $y =-\sqrt{x}$ is the above graph reflected over the $x$-axis.
The graph of $y \,=\,2 - \sqrt{x}$ is the same graph raised 2 units.

And $\,y = 0$ is the $x$-axis.

The graph looks like this:

Code:

      |     2 *                              *       |                    *       |*            *       | *    *       |  *.       | .:|:::*..       |*::|:::::::::*...       |:::|:::::::::::::::::*.....   - - * - + - - - - - - - - - - - - - * - -       |  1                          4       |

$\displaystyle \text{Area} \;=\;\int^1_0\!\! x^{\frac{1}{2}}\,dx + \int^4_1\!\!\left(2-x^{\frac{1}{2}}\right)\,dx$

$\displaystyle A = \int_0^1 (2-y)^2 - y^2 \, dy = 2$