Question

$\mathbf{\iint }{\mathit{y}}^{\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{2}}\mathit{d}\mathit{x}\mathit{d}\mathit{y}$ over the area bounded by $\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{,}\mathit{x}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{2}$, and theaxis.

Short answer

The required solution is 1.438.

Step-by-step solution

Definition of double integral

The double integral of $\mathit{f}\left(x,y\right)$over the areaA in the$\left(x,y\right)$ plane as the limit of this sum, and we write it as$\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\mathit{f}\left(x,y\right)\mathit{d}\mathit{x}\mathit{d}\mathit{y}$.

Drawing the area bounded by the curve

The area is bounded by the curve $y=x^2,x+y=2$and y- axis.

Intersecting points identification

From the figure the curves the intersection points:

$$\begin{gathered} x^2=2-x \\ {x}^2+x-2=0 \\ {x}_{1,2}=\frac{-1\pm \sqrt{1+8}}{2} \\ {x}_{1,2=\left(1,-2\right)} \end{gathered}$$

The integration limit will be x=0 from to x=1.

Calculation of the area under the curve

First, integrate over y:

$$\begin{gathered} \int {\int }_A\frac{dxdy}{\sqrt{y}}={\int }_{x=0}^1\left[{\int }_{y=x^2}^{2-x}\frac{dy}{\sqrt{y}}\right]dx \\ ={\int }_0^1\left(2{\overline{)\sqrt{y}}}_{x^2}^{2-x}\right)dx \\ =2{\int }_0^1\left(\sqrt{2-x}-x\right)dx \end{gathered}$$

Further calculation of the area of the bounded region

Now, integrate over x :

$$\begin{gathered} 2{\int }_0^1\left(\sqrt{2-x}-x\right)dx=2-\left[{\int }_0^1\sqrt{2-}xd\left(2-x\right)-{\overline{)\frac{1}{2}{x}^2}}_0^1\right] \\ =-\frac{4}{3}\left(2-x\right){\overline{){}^{3/2}}}_0^1-1 \\ =-\frac{7}{3}+\frac{\sqrt[8]{2}}{3} \\ \approx 1.438 \end{gathered}$$

Therefore, the value is 1.438.