Problem 30 - Advanced Engineering Mathematics

Question

Evaluate the given iterated integral by changing to polar coordinates. $$ \int_{0}^{1} \int_{0}^{\sqrt{2 y-y^{2}}}\left(1-x^{2}-y^{2}\right) d x d y $$

Step-by-Step Solution

Answer

The given integral evaluates to \( \frac{\pi}{4} \).

Step 1: Identify the Region

First, identify the region of integration. The limits of integration indicate that for each fixed \( y \), \( x \) goes from 0 to \( \sqrt{2y-y^2} \). This describes the region under a semi-circle of radius 1, with the circle centered at \((1,0)\). The given region is in the upper half of the circle centered at \((1,0)\) with radius 1.

Step 2: Transform to Polar Coordinates

Introduce polar coordinates by letting \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \). The expression \( 2y - y^2 = 1 - (x-1)^2 - y^2 \) suggests using a coordinate shift. The center of the circle translates to \((1,0)\), hence, \( x = 1 + r\cos(\theta) \) and \( y = r\sin(\theta) \).

Step 3: Set Up Polar Integral

The limits for \( r \) are from 0 to 1. The angle \( \theta \) varies from 0 to \( \pi \) owing to the semicircle lying on the upper half of the plane. The expression within the integral becomes \( 1 - (x - 1)^2 - y^2 = 1 - ((1 + r \cos(\theta)) - 1)^2 - (r \sin(\theta))^2 = 1 - r^2 \). Thus, the double integral in polar coordinates is:\[\int_{0}^{\pi} \int_{0}^{1} (1 - r^2)r \, dr \, d\theta\]

Step 4: Integrate with Respect to \( r \)

Compute the integral with respect to \( r \):\[\int_{0}^{1} (1 - r^2)r \, dr = \int_{0}^{1} (r - r^3) \, dr = \left[ \frac{r^2}{2} - \frac{r^4}{4} \right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}\]

Step 5: Integrate with Respect to \( \theta \)

Now integrate with respect to \( \theta \):\[\int_{0}^{\pi} \frac{1}{4} \, d\theta = \frac{1}{4} \left[\theta\right]_0^{\pi} = \frac{1}{4} \cdot \pi = \frac{\pi}{4}\]

Step 6: Conclusion

The value of the iterated integral is \( \frac{\pi}{4} \). Therefore, the integral evaluates to \( \frac{\pi}{4} \).

Key Concepts

Double Integrals

A double integral is a way to integrate over two variables, typically over a defined region in the plane. In our exercise, the region in question is defined by the boundaries for variables \( x \) and \( y \). Utilizing double integrals allows us to calculate the cumulative effect of a function over a particular area, whether that might be volume, area, or other physical quantities.

In the given problem, we start by evaluating an iterated integral, which means integrating one variable at a time: first \( x \), then \( y \). This layered approach helps simplify the calculation by breaking it down into manageable steps.

When dealing with double integrals, finding the region of integration is crucial. It helps you visualize where the function is being integrated, which in our problem is beneath a semi-circle. This visualization becomes essential when transforming coordinates.

Coordinate Transformation

A coordinate transformation is the process of changing from one coordinate system to another. This is particularly useful in calculus since some regions are easier to describe in one system than another. For example, regions that are circular, or parts of circles, can often be more easily managed in polar coordinates.

In the given problem, the rectangular coordinates \( x \) and \( y \) are transformed into polar coordinates with \( r \) and \( \theta \). Specifically, the coordinates are shifted so that they are centered around the new origin, due to the given region being part of a circle centered at \((1,0)\).

• Using polar coordinates, we set \( x = 1 + r\cos(\theta) \)
• And \( y = r\sin(\theta) \)

This adjustment simplifies the integral by aligning it more closely with the shape of the region.

Circular Region

The exercise mentions a circular region, which is key to understanding why we switch to polar coordinates. Circular regions, particularly those not centered at the origin, can be complex to describe in rectangular coordinates.

The original integration limits reveal a semi-circle region sitting at the top half of a circle centered at \((1,0)\). By transforming the coordinates, we account for this circular shift and simplify the integration using the familiar form of polar coordinates.

This semi-circular transformation is pivotal:

• Represent the circle accurately by switching from straight-line limits to angular limits \( \theta \).
• The radius \( r \) changes from 0 to the edge of the circle segment, simplifies calculations.

Recognizing and leveraging these circular symmetries helps solve problems more efficiently.

Iterated Integrals

Iterated integrals simplify the process of evaluating multi-variable integrals by working one dimension at a time. The operation involves integrating with respect to one variable while treating others as constants, and then proceeding to the next.

In this problem, the transition to polar coordinates leads to an iterated integral over \( r \) first, followed by \( \theta \). Calculating in this order is advantageous because the limit for \( \theta \) does not depend on \( r \), making it a straightforward computation.

This step-by-step integration appears as follows:

• First, integrate \( (1 - r^2)r \) with respect to \( r \).
• Followed by integrating the result with respect to \( \theta \).

Each separate step not only makes the process easier but also clearer, providing a clearer path to the solution, which in our case is \( \frac{\pi}{4} \). This process underlies how double integrals unfold into more approachable single-variable integrations.