Calculate area between pair of curves $y+x=2$ and $y^2=2(2-x)$. please help

Question

I need help in solving this area problem. since limits are not provided.

I need to calculate area between pair of curves $y+x=2$ and $y^2=2(2-x)$.

Please help.

I tried solving it by finding its points of intersection but after that the integrating the two functions is confusing.

Answer 1

Hint:

This is the graph of your functions and the intersection points are the solutions of the system $$ \begin{cases} x=2-y\\ y^2=2(2-x) \end{cases} $$

can you do form this?

Answer 2

$y=-x+2$

$y=\pm \sqrt{4-2x}$

But, $y=-\sqrt{4-2x}$ and $y=-x+2$ do not have any area in between them, so you need to find the intersection points of $y=\sqrt{4-2x}$ and $y=-x+2$.

$\sqrt{4-2x}=-x+2$

$4-2x=x^2-4x+4$

$x^2-2x=0$

$x=0,2$

None of which are extraneous.

So, you need to calculate the area, $$I=\int^2_0\left[\sqrt{4-2x}-(-x+2)\right]dx$$

Answer 3

Hint: Try to calculate $\displaystyle \int |x_2(y)-x_1(y)|dy=\int_{y_1}^{y_2} \left|(2-y)-(2-y^2/2)\right| dy=\int |y^2/2-y|$

Note: Integration with respect to $x$ is also ok but takes more time comparing to integration with respect to $y$ but the answers will be the same because of the nature of integration.

Answer 4

Referring to Emilio's drawing:

Points of intersection: $(0,2)$, and $(2,0)$.

Area of parabola,$A_p$, in the first quadrant :

$A_p = \displaystyle \int_{0}^{2}\sqrt{2(2-x)}dx.$

Area of triangle formed by line

$y+x =2$:

$A_{\triangle} :=(1/2)2 \cdot 2 = 2.$

Desired area: $A_p$ - $A_{\triangle}$.

Left to do: Find $A_p$ by integration.