Chapter 9: Problem 26
In Problems, evaluate the given iterated integral by changing to polar coordinates. $$ \int_{0}^{\sqrt{2} / 2} \int_{0}^{\sqrt{1-y^{2}}} \frac{y^{2}}{\sqrt{x^{2}+y^{2}}} d x d y $$
Short Answer
Expert verified
The integral evaluates to \( \frac{\pi}{24} \).
Step by step solution
01
Identify the Region of Integration
The integral is given in Cartesian coordinates, with the inner integral for \( x \) ranging from \( 0 \) to \( \sqrt{1-y^2} \) and the outer integral for \( y \) ranging from \( 0 \) to \( \frac{\sqrt{2}}{2} \). This describes the first quadrant of the unit circle from \( y=0 \) to \( y=\frac{\sqrt{2}}{2} \).
02
Convert to Polar Coordinates
In polar coordinates, \( x = r \cos \theta \) and \( y = r \sin \theta \) and \( dx \ dy = r \ dr \ d\theta \). The range for \( r \) is from 0 to 1, and \( \theta \) is from 0 to \( \frac{\pi}{4} \), since this corresponds to \( y = x \).
03
Substitute and Simplify the Integrand
The integrand \( \frac{y^2}{\sqrt{x^2 + y^2}} \) becomes \( \frac{(r \sin \theta)^2}{r} \) when converted to polar coordinates; this simplifies to \( r \sin^2 \theta \). The integral becomes \( \int_0^{\frac{\pi}{4}} \int_0^1 r^2 \sin^2 \theta \, dr \, d\theta \).
04
Integrate with Respect to \( r \)
Integrate the expression \( r^2 \sin^2 \theta \) with respect to \( r \):\[\int_0^1 r^2 \sin^2 \theta \, dr = \sin^2 \theta \int_0^1 r^2 \, dr = \sin^2 \theta \left[ \frac{r^3}{3} \right]_0^1 = \frac{1}{3} \sin^2 \theta\]
05
Integrate with Respect to \( \theta \)
Now, integrate the result \( \frac{1}{3} \sin^2 \theta \) with respect to \( \theta \):\[\int_0^{\frac{\pi}{4}} \frac{1}{3} \sin^2 \theta \, d\theta = \frac{1}{3} \int_0^{\frac{\pi}{4}} \sin^2 \theta \, d\theta\]Using the identity \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \), the integral becomes:\[\frac{1}{6} \int_0^{\frac{\pi}{4}} (1 - \cos(2\theta)) \, d\theta = \frac{1}{6} \left[\theta - \frac{1}{2} \sin(2\theta) \right]_0^{\frac{\pi}{4}}\]Evaluate the expression:\[= \frac{1}{6} \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{24}\]
06
Conclusion
The value of the given iterated integral is \( \frac{\pi}{24} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
polar coordinates
Polar coordinates offer a way to represent a point in a plane using two values: the radial distance from a reference point and the angle from a reference direction. This is particularly helpful when dealing with circular or radial symmetry. The radial distance, \( r \), indicates how far away the point is from the origin, while the angle, \( \theta \), tells us the direction. In conversion from Cartesian coordinates, where a point is represented by \( (x, y) \), to polar coordinates, we use:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
iterated integrals
Iterated integrals are a way to compute the volume under a surface or to evaluate the integral of a function in two dimensions, over a region defined in the plane. These involve integrating a function of two variables by performing integration successively – one variable at a time. Starting from the innermost integral and working outward, each variable is treated independently.
- First, identify the inner and outer limits of integration to specify the region of interest.
- Integrate the function with respect to the innermost variable while treating other variables as constants.
- Next, proceed to integrate the resulting expression with respect to the next variable.
unit circle integration
Unit circle integration involves calculating an integral over a circular region with a radius of 1, centered at the origin. This is often expressed using polar coordinates because of the natural match with circular symmetry.For a unit circle, the range for \( r \) – the distance from the origin – is from 0 to 1. The angle \( \theta \) varies from 0 to \( 2\pi \) to cover the whole circle, or a fraction of \( 2\pi \) for sectors or quadrants. When integrating over a sector, such as a quadrant, \( \theta \) spans an angle less than \( 2\pi \), such as from 0 to \( \frac{\pi}{4} \).In this exercise, the setup involves the first quadrant of the unit circle, which simplifies the bounds: \( r \) from 0 to 1, and \( \theta \) from 0 to \( \frac{\pi}{4} \). This way, the challenging bounds and complicated path definition become simple single-valued ranges, easing integration. By choosing polar coordinates for unit circle integration, calculations align perfectly with the natural geometry, making the integral simpler to solve.