**Evaluate** \[\int_{-2}^{2} \int_{0}^{\sqrt{4-x^2}} e^{-x^2-y^2} \, dy \, dx\]**by converting to polar coordinates.**This problem involves evaluating a double integral of the function $ e^{-x^2-y^2} $ over a specified region in the Cartesian coordinate plane. The region of integration is bounded by $ x = -2 $ to $ x = 2 $ in the $ x $-direction, and from $ y = 0 $ to $ y = \sqrt{4-x^2} $ for each $ x $.By converting to polar coordinates, the variables $ x $ and $ y $ are described in terms of $ r $ (the radial distance from the origin) and $ \theta $ (the angle from the positive $ x $-axis). The transformations are $ x = r \cos \theta $ and $ y = r \sin \theta $. The bounds of integration should be adjusted according to the shape and size of the region described in Cartesian coordinates.Additionally, when converting to polar coordinates, the integrand $ e^{-x^2-y^2} $ should be rewritten as $ e^{-r^2} $, and the differential area element $ dy \, dx $ becomes $ r \, dr \, d\theta $.

Answered: Evaluate e-y dy da by converting to…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

**Evaluate**
\[
\int_{-2}^{2} \int_{0}^{\sqrt{4-x^2}} e^{-x^2-y^2} \, dy \, dx
\]
**by converting to polar coordinates.**

This problem involves evaluating a double integral of the function $ e^{-x^2-y^2} $ over a specified region in the Cartesian coordinate plane. The region of integration is bounded by $ x = -2 $ to $ x = 2 $ in the $ x $-direction, and from $ y = 0 $ to $ y = \sqrt{4-x^2} $ for each $ x $.

By converting to polar coordinates, the variables $ x $ and $ y $ are described in terms of $ r $ (the radial distance from the origin) and $ \theta $ (the angle from the positive $ x $-axis). The transformations are $ x = r \cos \theta $ and $ y = r \sin \theta $. The bounds of integration should be adjusted according to the shape and size of the region described in Cartesian coordinates.

Additionally, when converting to polar coordinates, the integrand $ e^{-x^2-y^2} $ should be rewritten as $ e^{-r^2} $, and the differential area element $ dy \, dx $ becomes $ r \, dr \, d\theta $.

Expert Solution

Step 1

Given:

$\int_{- 2}^{2} \int_{0}^{\sqrt{4 - x^{2}}} e^{- x^{2} - y^{2}} d y d x$

We want to evaluate above integral by using polar coordinates

Step 2

Calculation:

We want to evaluate $\int_{- 2}^{2} \int_{0}^{\sqrt{4 - x^{2}}} e^{- x^{2} - y^{2}} d y d x$ that is $\int_{- 2}^{2} \int_{0}^{\sqrt{4 - x^{2}}} \begin{array}{rcl} e \end{array} \begin{array}{rcl} ^{- (x^{2} + y^{2})} \end{array} \begin{array}{rcl} d \end{array} \begin{array}{rcl} y \end{array} \begin{array}{rcl} d \end{array} \begin{array}{rcl} x \end{array}$

We can express the region as

$D = \{(x, y) / - 2 \leq x \leq 2, 0 \leq y \leq \sqrt{4 - x^{2}}\}$

Now

$y = \sqrt{4 - x^{2}} \Rightarrow y^{2} = 4 - x^{2} \Rightarrow x^{2} + y^{2} = 4$

Therefore By substituting polar coordinates $x = r \cos θ, y = r \sin θ$

$\begin{array}{rcl} x^{2} + y^{2} & = & {(r \cos θ)}^{2} + {(r \sin θ)}^{2} \\ = & r^{2} (\cos^{2} θ + \sin^{2} θ) \\ x^{2} + y^{2} & = & r^{2} \end{array}$

Now we find Jacobean

$\begin{array}{rcl} J & = & \frac{\partial (x, y)}{\partial (r, θ)} \\ = & |\begin{matrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial θ} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial θ} \end{matrix}| \\ = & |\begin{matrix} \cos θ & - r \sin θ \\ \sin θ & r \cos θ \end{matrix}| \\ J & = & r \end{array}$

The graph is in first and second quadrant hence $0 \leq θ \leq π$

And $0 \leq r \leq 2$

Therefore

$\begin{array}{rcl} \int_{- 2}^{2} \int_{0}^{\sqrt{4 - x^{2}}} e^{- x^{2} - y^{2}} d y d x & = & \int_{- 2}^{2} \int_{0}^{\sqrt{4 - x^{2}}} e^{- (x^{2} + y^{2})} d y d x \\ = & \int_{0}^{π} \int_{0}^{2} e^{- (r^{2})} |J| d r d θ \\ = & \int_{0}^{π} \int_{0}^{2} e^{- (r^{2})} r d r d θ \\ = & [\int_{0}^{π} d θ] [\int_{0}^{2} e^{- (r^{2})} r d r] \\ = & {[θ]}_{0}^{π} [\int_{0}^{2} e^{- (r^{2})} r d r] \\ \int_{- 2}^{2} \int_{0}^{\sqrt{4 - x^{2}}} e^{- x^{2} - y^{2}} d y d x & = & π [\int_{0}^{2} e^{- (r^{2})} rdr] \end{array}$

Now substituting

$r^{2} = u \Rightarrow 2 r d r = d u \Rightarrow r d r = \frac{d u}{2} W h e n r = 0 \Rightarrow u = 0 W h e n r = 2 \Rightarrow u = 4$

$\begin{array}{rcl} \int_{- 2}^{2} \int_{0}^{\sqrt{4 - x^{2}}} e^{- x^{2} - y^{2}} d y d x & = & π [\int_{0}^{4} e^{- u} \frac{du}{2}] \\ = & \frac{π}{2} [\int_{0}^{4} e^{- u} du] \\ = & \frac{π}{2} {[\frac{e^{- u}}{- 1}]}_{0}^{4} \\ = & \frac{π}{2} [\frac{e^{- 4}}{- 1} - \frac{e^{0}}{- 1}] \\ = & \frac{π}{2} [- \frac{1}{e^{4}} + 1] \\ \int_{- 2}^{2} \int_{0}^{\sqrt{4 - x^{2}}} e^{- x^{2} - y^{2}} dydx & = & \frac{(e^{4} - 1) π}{2 e^{4}} \end{array}$

Therefore,

$\begin{array}{rcl} \int \end{array} \begin{array}{rcl} - 2 2 \end{array} \begin{array}{rcl} \int \end{array} \begin{array}{rcl} 0 \sqrt{4 - x^{2}} \end{array} \begin{array}{rcl} e \end{array} \begin{array}{rcl} \end{array}$ ^-x2-y2dydx=e4-1π2e4

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