Question

Express the integral$\mathit{I}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{1}}\mathit{d}\mathit{x}{\mathbf{\int }}_{\mathbf{0}}^{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}}\mathbf{}{\mathit{e}}^{\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}\mathit{d}\mathit{y}$as an integral in polar coordinates $\left(r,\theta \right)$ and so evaluate it.

Short answer

The value of Integral is$\mathrm{I}=\frac{\mathrm{\tau \tau }}{4}\left(1-{\mathrm{e}}^{-1}\right)$

Step-by-step solution

Given Integral

The given integral is, $\mathrm{I}={\int }_0^1\mathrm{dx}{\int }_0^{\sqrt{1-{\mathrm{x}}^2}}{\mathrm{e}}^{-{\mathrm{x}}^2-{\mathrm{y}}^2}\mathrm{dy}$

Concept of converting integral to polar coordinates

Use $\mathit{x}\mathbf{=}\mathit{r}\mathit{c}\mathit{o}\mathit{s}$, $\mathit{r}\mathbf{=}\mathbf{sin}\mathit{\theta }$and$\mathit{d}\mathit{A}\mathbf{=}\mathit{r}\mathit{d}\mathit{r}\mathit{d}\mathit{\theta }$to convert an integral in rectangular coordinates to an integral in polar coordinates.

Evaluate the integral

It is required to express the integral as an integral in polar coordinates $\left(r,\theta \right)$and evaluate it.

The area of integration is the quarter-circle in the first quadrant, of unit radius.

Thus, the integral become:

$$\begin{gathered} \mathrm{I}={\int }_0^1\mathrm{dx}{\int }_0^{\sqrt{1-{\mathrm{x}}^2}}{\mathrm{e}}^{-{\mathrm{x}}^2-{\mathrm{y}}^2}\mathrm{dy} \\ \mathrm{I}={\int }_{r=0}^1{\int }_{\theta =0}^{\tau \tau /2}{\mathrm{e}}^{-\mathrm{r}}\mathrm{rdrd\theta } \\ \mathrm{I}=\frac{\mathrm{\tau \tau }}{2}{\int }_0^1{\mathrm{e}}^{-{\mathrm{r}}^2}\mathrm{d}\left(-{\mathrm{r}}^2\right)\left(\frac{1}{2}\right) \\ \mathrm{I}=\frac{\mathrm{\tau \tau }}{4}\left(1-{\mathrm{e}}^{-1}\right) \end{gathered}$$

The value of Integral is $\mathrm{I}=\frac{\mathrm{\tau \tau }}{4}\left(1-{\mathrm{e}}^{-1}\right)$.