Question

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

Short answer

\(I = \frac{\pi}{4} (1 - e^{-1})\)

Step-by-step solution

Understand the region of integration

The double integral is over the region where \(0 \leq x \leq 1\) and \(0 \leq y \leq \sqrt{1 - x^2}\). This describes a quarter circle in the first quadrant with radius 1.

Convert Cartesian coordinates to polar coordinates

In polar coordinates, the region described by \(0 \leq x \leq 1\) and \(0 \leq y \leq \sqrt{1 - x^2}\) is transformed to \(0 \leq r \leq 1\) and \(0 \leq \theta \leq \frac{\pi}{2}\). Use the transformations \(x = r \cos \theta\) and \(y = r \sin \theta\).

Write the integral in polar coordinates

The integral in Cartesian coordinates \(I = \int_{0}^{1} dx \int_{0}^{\sqrt{1-x^2}} e^{-x^2-y^2} dy\) becomes \(I = \int_{0}^{\frac{\pi}{2}} d\theta \int_{0}^{1} e^{-r^2} r \, dr\) in polar coordinates. The extra \(r\) term comes from the Jacobian determinant when changing to polar coordinates.

Evaluate the inner integral

First evaluate the inner integral \(\int_{0}^{1} e^{-r^2} r \, dr\). Use the substitution \(u = r^2\), hence \(du = 2r \, dr\). This simplifies to \(\frac{1}{2} \int_{0}^{1} e^{-u} du\). Evaluating this gives \(\frac{1}{2} [ -e^{-u} ]_0^1 = \frac{1}{2} (1 - e^{-1})\).

Evaluate the outer integral

Evaluate the outer integral \(\int_{0}^{\frac{\pi}{2}} d\theta\) which is simply \(\theta\) evaluated from 0 to \(\frac{\pi}{2}\), giving \(\frac{\pi}{2}\).

Combine the results

Combine the results of the integrals: \(I = \frac{\pi}{2} \times \frac{1}{2} (1 - e^{-1}) = \frac{\pi}{4} (1 - e^{-1})\).

Key concepts

Polar Coordinates Transformation

Transforming integrals from Cartesian coordinates to polar coordinates can make complex integrations simpler. In our given exercise, the original integral is expressed in Cartesian coordinates as \textcolor{blue}{\text{\(I = \text{\textarrow{\text{\textofourtext{\text{\textofourcomponent{\text{\textda{x I = e^{-x^{2}-y^{2}} hyd \theta integral.$$y^-degrees}}}}}}}}}}}} }}}integral}}). To convert this into polar coordinates, we need to make use of the following transformations:

• \)x = r \text{cos} \theta\(
• \)y = r \text{sin} \theta\(

Here, \)r\( represents the radial distance from the origin, while \)\theta\( is the angle measured counterclockwise from the positive x-axis. The region described by \text{0 to \) \text{integral \theta}dx}dx)(0 to is transformed to be positive x-{0 to first \(1}\theta\)}\(}}}}}}}}}})}}})}}))))}}}))}})0}result)} is transformed such that \)0 <= r <= 1\( and \)0 <= \theta <={ \frac{{$\textdict \frac{:]{\theta}$$, from