Question

${\mathbf{\int }}_{\mathbf{y}\mathbf{=}\mathbf{0}}^{\mathbf{1}}{\mathbf{\int }}_{\mathbf{x}\mathbf{=}\mathbf{0}}^{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}}\mathbf{ydxdy}$

Short answer

The required solution is$\frac{1}{3}$

Step-by-step solution

Definition of double integral

The double integral of f(x,y)over the area Ain the (x,y)plane as the limit of this sum, and we write it as$\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{dxdy}$

Drawing the area bounded by the curve

The area of integration or bounded region is shown below,

Calculation of the area under the curve in an integral way

At first, the integral is the way it is set up in the problem:

$$\begin{gathered} \mathrm{l}={\int }_0^{1}ydy{\int }_0^{\sqrt{1-y^2}}dx \\ ={\int }_0^{1}ydy\sqrt{1-y^2} \end{gathered}$$

Let's assume,

$$\begin{gathered} y=\sin u \\ dy=\cos udu \end{gathered}$$

Therefore,

$$\begin{gathered} \mathrm{l}={\int }_0^{\pi /2}\cos udu\sin u\cos u \\ =-{\int }_0^{\pi /2}{\cos }^{2}ud\left(\cos u\right) \\ ={\overline{)-\frac{1}{3}{\cos }^{3}u}}_0^{\pi /2} \\ =\frac{1}{3} \end{gathered}$$

Calculation of the area under the curve changing boundaries 

The other way, changing the boundaries of the given integration:

$x:0\to 1,\mathrm{and}y:0\to \sqrt{1-x^2}$

So, the integration

$$\begin{gathered} \mathrm{l}={\int }_0^{1}dx{\int }_{y^2}^{\sqrt{1-x^2}}ydy \\ ={\overline{)\frac{1}{2}{\int }_0^{1}dx{y}^2}}_0^{\sqrt{1-x^2}} \\ =\frac{1}{2}{\int }_0^{1}dx\left(1-x^2\right) \\ ={\overline{)\frac{1}{2}\left[x-\frac{1}{3}{x}^3\right]}}_0^1 \\ =\frac{1}{3} \end{gathered}$$

Therefore, both way gives the same results of the integration that is,$\frac{1}{3}$