Problem 25 - Advanced Engineering Mathematics

Question

Evaluate the given iterated integral by changing to polar coordinates. $$ \int_{-3}^{3} \int_{0}^{\sqrt{9-x^{2}}} \sqrt{x^{2}+y^{2}} d y d x $$

Step-by-Step Solution

Answer

The value of the integral is \(9\pi\).

Step 1: Analyze the Region of Integration

The given integral is defined with limits for \(x\) from \(-3\) to \(3\) and for \(y\) from \(0\) to \(\sqrt{9-x^2}\). This represents the upper half of a circle centered at the origin with radius 3 in the Cartesian plane.

Step 2: Set Up Polar Coordinates

To convert to polar coordinates, use the transformations: \(x = r \cos \theta\) and \(y = r \sin \theta\). The integration region in polar coordinates is for \(r\) from 0 to 3 and \(\theta\) from 0 to \(\pi\) (covering the upper half circle).

Step 3: Rewrite the Integral in Polar Coordinates

The given function \(\sqrt{x^2 + y^2}\) simplifies to \(r\) in polar coordinates. The differential area element \(dydx\) becomes \(rdrd\theta\). So, the integral becomes: \[\int_{0}^{\pi} \int_{0}^{3} r \cdot r \, dr \, d\theta.\]

Step 4: Evaluate Inner Integral Over r

Calculate the inner integral with respect to \(r\):\[\int_{0}^{3} r^2 \, dr = \left[ \frac{r^3}{3} \right]_{0}^{3} = \frac{27}{3} = 9.\]

Step 5: Evaluate Outer Integral Over θ

With the result from the inner integral now being 9, we compute the outer integral:\[\int_{0}^{\pi} 9 \, d\theta = 9\theta \bigg|_0^{\pi} = 9\pi.\]

Step 6: Conclusion

The value of the given iterated integral, after converting to polar coordinates and evaluating, is found to be \(9\pi\).

Key Concepts

Polar Coordinates

Polar coordinates are a way of representing points in two-dimensional space. Instead of using horizontal and vertical distances (like Cartesian coordinates), they use a distance from the origin and an angle from a reference direction. This is particularly useful for circular or symmetric shapes. In the given problem, we changed the rectangular coordinates to polar ones to simplify the area of integration, which was a circle in this case.
Polar coordinates are represented as

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)
• \( r \) is the radius, or distance from the origin.
• \( \theta \) is the angle from the positive x-axis.

This method allows us to express any point in terms of \( r \) and \( \theta \), which is quite handy for integration over circles.

Changing Variables

Changing variables, or substitution, in integration simplifies solving integrals, especially when the initial bounds and functions appear complex. In this problem, it made the integral simpler by converting coordinates.
When switching from Cartesian to polar coordinates:

• The integration limits were changed from lines and parabolic bounds to circular bounds, better suited for solving with \( r \) and \( \theta \).
• The differential elements change. The original \( dxdy \) is replaced by \( rdrd\theta \), reflecting the area in polar coordinates.

This transformation is often a critical step in simplifying complex integrals.

Integration Techniques

Integration techniques involve strategies that simplify or allow solving integration problems. In this exercise, the task was to evaluate an iterated integral. Specific techniques were used, like changing to polar coordinates, to ease computation.
Steps involved:

• Identifying the integral's region and converting it into a suitable form for calculation.
• Using polar coordinates made the symmetric region of integration much clearer and viable.
• After conversion, solving the inner integral initially, then the outer integral.

These techniques help handle integrals over complex shapes and limits, making solving them much more approachable.

Mathematics Education

Mathematics education aims to build a deep understanding of concepts and procedures. In this task, it focused on how converting coordinates and simplifying integrals helps solve real-world problems.
Essential learning points include:

• Understanding the utility and theory behind changing to polar coordinates for symmetric areas.
• Recognizing when substitution and changing variables aid in simplifying integrals enough to solve them easily.
• Practicing these concepts through examples ensures that learners grasp how and when to apply these techniques.

Focusing on these strategies helps students develop the problem-solving skills necessary in advanced mathematics.