Chapter 5: Problem 14

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

Expert verified

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

Step by step solution

- Convert the Cartesian limits to polar coordinates

The given integral is oindent $ I = \int_{0}^{1} dx \int_{0}^{\sqrt{1-x^{2}}} e^{-x^{2}-y^{2}} dy $oindent The region of integration is inside the part of the unit circle in the first quadrant. In polar coordinates, this region is described by $0 \leq r \leq 1$ and $0 \leq \theta \leq \frac{\pi}{2}$.

- Express the integrand in polar coordinates

In polar coordinates: oindent$x = r\cos\theta$$y = r\sin\theta$$dx\,dy = r\,dr\,d\theta$ and the exponential function oindent$e^{-x^{2}-y^{2}} = e^{-r^{2}} $.

- Rewrite the integral in polar coordinates

Substitute the polar coordinates into the integral:oindent$I = \int_{0}^{\pi/2} d\theta \int_{0}^{1} e^{-r^2} r dr$.

- Evaluate the inner integral

Evaluate the integral with respect to $r$:oindent$\int_{0}^{1} r e^{-r^2} dr = -\frac{1}{2} e^{-r^2} \bigg|_{0}^{1} = -\frac{1}{2} \,\left(e^{-1} - e^{0}\right) = \frac{1}{2}\left( 1 - e^{-1} \right) $.

- Evaluate the outer integral

Evaluate the integral with respect to $\theta$:oindent$ I = \int_{0}^{\pi/2} \frac{1}{2} (1 - e^{-1}) d\theta = \frac{1}{2} (1 - e^{-1}) \left[ \theta \right]_{0}^{\pi/2} = \frac{1}{2}(1 - e^{-1}) \cdot \frac{\pi}{2} = \frac{\pi}{4} (1 - e^{-1}) $.oindent This gives the final result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polar Coordinates

Polar coordinates transform Cartesian coordinates $x $ and $y$ into $r$ and $\theta$. Instead of dealing with horizontal and vertical distances, polar coordinates use the distance from the origin ($r$) and the angle from the positive x-axis ($\theta$). This can simplify integrals, especially when dealing with circular or radial symmetry.

To convert, use:

$x = r \cos\theta$
$y = r \sin\theta$

The area element $dx dy$ in Cartesian coordinates becomes $r \, dr \, d\theta$ in polar coordinates. This conversion accounts for the change in scale as we move away from the origin.

Double Integrals

Double integrals extend the concept of single integrals to functions of two variables. They compute the volume under a surface. The integral $\int \int f(x, y) \, dx \, dy$ evaluates the function $f(x, y)$ across a region in the $xy$-plane.

When switching to polar coordinates, the double integral changes to $\int \int f(r, \theta) \, r \, dr \, d\theta$. The additional $r$ term accounts for the 'stretching' of areas as $r$ increases.

This simplifies problems involving circular regions, as the limits of $r$ and $\theta$ directly describe circles and arcs.

Exponential Functions

Exponential functions have the form $e^x$, where 'e' is a mathematical constant approximately equal to 2.71828. They grow quickly and appear often in mathematical modeling.

In this problem, the integrand is $e^{-x^2 - y^2}$, which becomes $e^{-r^2}$ in polar coordinates. The symmetry of the exponential function around the origin makes it particularly suited for polar coordinates.

Note: For the integral of given exponential functions involving powers of $r$, you can use substitution or recognize patterns to simplify the evaluation.

Coordinate Transformation

Coordinate transformation is the process of converting coordinates from one system to another. Here, switching from Cartesian to polar coordinates transforms the problem into a simpler one by taking advantage of the circular symmetry.

Steps:

Identify the region of integration in Cartesian coordinates.
Convert the integral limits to polar coordinates: $0 \leq r \leq 1$ and $0 \leq \theta \leq \frac{\pi}{2}$.
Substitute $x = r \cos\theta$ and $y = r \sin\theta$ into the integrand.
Adjust for the area element by multiplying by $r$.

These transformations streamline the integration process, making the integral more manageable.

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.