Problem 27 - Advanced Engineering Mathematics

Question

In Problems, evaluate the given iterated integral by changing to polar coordinates. $$ \int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} e^{x^{2}+y^{2}} d x d y $$

Step-by-Step Solution

Answer

\(\frac{\pi}{4} (e - 1)\)

Step 1: Understand the Region of Integration

The region of integration is given by the bounds: \(0 \leq y \leq 1\) and \(0 \leq x \leq \sqrt{1 - y^2}\). This describes a quarter-circle (in the first quadrant) with radius 1.

Step 2: Change to Polar Coordinates

In polar coordinates, \(x = r \cos \theta\) and \(y = r \sin \theta\), with the area element \(dA = r \, dr \, d\theta\). The circular region translates to \(0 \leq r \leq 1\) and \(0 \leq \theta \leq \frac{\pi}{2}\).

Step 3: Transform the Integral

Substitute the polar coordinates into the integral. Here, \(e^{x^2 + y^2} = e^{r^2}\), so the integral becomes: \[ \int_{0}^{\pi/2} \int_{0}^{1} e^{r^2} r \, dr \, d\theta \].

Step 4: Evaluate the Inner Integral with Respect to r

Evaluate \(\int_{0}^{1} e^{r^2} r \, dr\). Let \(u = r^2\) so \(du = 2r \, dr\). The integral becomes: \[ \frac{1}{2} \int_{0}^{1} e^{u} \, du = \frac{1}{2} \left[ e^{u} \right]_{0}^{1} = \frac{1}{2} (e - 1) \].

Step 5: Evaluate the Outer Integral with Respect to θ

Substitute the result from Step 4 into the outer integral: \[ \int_{0}^{\pi/2} \frac{1}{2} (e - 1) \, d\theta \]. This simplifies to \(\frac{1}{2} (e - 1) \theta \) evaluated from \(0\) to \(\frac{\pi}{2}\), which gives \(\frac{1}{2} (e - 1) \frac{\pi}{2}\) = \(\frac{\pi}{4} (e - 1)\).

Key Concepts

Iterated Integral

The term "iterated integral" refers to performing multiple integrals in sequence over a multi-dimensional region. In the context of the exercise, the iterated integral is expressed as a double integral, which means integrating with respect to two variables: \( x \) and \( y \). A nested approach is applied, where you first integrate with respect to one variable and, subsequently, proceed with the other variable.

In simpler terms, for our given problem, we initially integrate \( e^{x^2+y^2} \) with respect to \( x \), and then integrate the resulting function with respect to \( y \). This technique is essential when dealing with functions over areas in two-dimensional space, and it helps in calculating the accumulated value of a function across a specified region.

Region of Integration

The region of integration defines where your function is integrated over in the coordinate plane. It consists of all the points \( (x, y) \) where the function is defined and the integral needs to be evaluated. For our problem, this region is determined by the bounds: \( 0 \leq y \leq 1 \) and \( 0 \leq x \leq \sqrt{1 - y^2} \).

This describes part of a quarter-circle in the first quadrant of the Cartesian plane with a radius of 1. The bounds for \( x \) and \( y \) impose specific constraints that outline this particular section of the circle. Visualizing this can significantly assist in understanding where the integration takes place, which eventually aids in the application of the appropriate method to solve the integral.

Change of Variables

Changing variables is a technique used to simplify the process of integration by converting the given variables into a different set that better describes the region. In this case, Cartesian coordinates are transformed into polar coordinates.

The conversion formulas are: \( x = r \cos \theta \) and \( y = r \sin \theta \), where \( r \) is the radius from the origin, and \( \theta \) is the angle from the positive \( x \)-axis. The integration element \( dx \, dy \) is converted to \( r \, dr \, d\theta \). These changes accommodate the nature of the circular integration region, making the calculation easier and more intuitive. Through this transformation, the bounds in Cartesian coordinates translate to \( 0 \leq r \leq 1 \) and \( 0 \leq \theta \leq \frac{\pi}{2} \), which corresponds more naturally to the quarter-circle shape we are integrating over.

Evaluating Integrals

Evaluating integrals involves finding the value of the integral over its specified range. In the final steps of the exercise, we evaluate an inner and an outer integral.

First, consider the inner integral: \( \int_{0}^{1} e^{r^2} r \, dr \). To evaluate this, use substitution: let \( u = r^2 \), which simplifies the integral to \( \frac{1}{2} \int_{0}^{1} e^{u} \, du \). Calculate this integral to obtain \( \frac{1}{2} (e - 1) \).

Next, evaluate the outer integral: \( \int_{0}^{\pi/2} \frac{1}{2} (e - 1) \, d\theta \). This is relatively straightforward, yielding \( \frac{1}{2} (e - 1) \cdot \frac{\pi}{2} \). The final result is \( \frac{\pi}{4} (e - 1) \), representing the accumulated value over the defined region.