Chapter 14: Problem 27
Evaluate the iterated integral by converting to polar coordinates. \(\mathrm{M}\) $$ \int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}}\left(x^{2}+y^{2}\right) d y d x $$
Short Answer
Expert verified
The iterated integral evaluates to \( \frac{\pi}{8} \).
Step by step solution
01
Identify the Region of Integration
The given iterated integral is \( \int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} (x^2 + y^2) \, dy \, dx \). This represents a region in the first quadrant under the curve \( y = \sqrt{1-x^2} \), which is part of a circle with radius 1 centered at the origin. The limits for \( x \) are from 0 to 1, and for \( y \) they go from 0 to \( \sqrt{1-x^2} \).
02
Convert to Polar Coordinates
In polar coordinates, \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \). The expression \( x^2 + y^2 \) becomes \( r^2 \). The Jacobian of the transformation is \( r \), so the area element is \( r \, dr \, d\theta \). The region's limits are \( r = 0 \) to 1, and \( \theta = 0 \) to \( \frac{\pi}{2} \).
03
Set Up the Polar Integral
Substitute into the integral using polar coordinates: \( \int_{0}^{\pi/2} \int_{0}^{1} r^2 \cdot r \, dr \, d\theta \). Simplifies to \( \int_{0}^{\pi/2} \int_{0}^{1} r^3 \, dr \, d\theta \).
04
Solve the Inner Integral
Evaluate the integral \( \int_{0}^{1} r^3 \, dr \). This results in \( \left[ \frac{r^4}{4} \right]_{0}^{1} = \frac{1}{4} \).
05
Solve the Outer Integral
Integrate the constant \( \frac{1}{4} \) with respect to \( \theta \) from 0 to \( \frac{\pi}{2} \): \( \int_{0}^{\pi/2} \frac{1}{4} \, d\theta = \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8} \).
06
Conclusion and Final Answer
Thus, the value of the iterated integral by converting to polar coordinates is \( \frac{\pi}{8} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar Coordinates
Polar coordinates provide a useful way to describe the location of a point in a plane. Unlike Cartesian coordinates that use
- a pair of perpendicular axes (x and y)
- to specify a point's location,
- the distance from a fixed point (the origin), called the radius \( r \)
- and the angle \( \theta \), measured from a fixed direction.
- \( x = r \cos(\theta) \)
- \( y = r \sin(\theta) \)
Region of Integration
The region of integration refers to the area over which we evaluate a given integral. In the context of an iterated integral, it defines the domain for
- the variables we integrate over.
- circular or sector-shaped regions are often the result.
Jacobian Transformation
The Jacobian transformation is a crucial step when changing variables in multiple integrals. It relates to the determinant of the matrix of all first-order partial derivatives of a transformation. This determinant gives us a scaling factor
- to adjust the differential area or volume elements
- when transforming from one coordinate system to another.
- the Jacobian becomes \( r \).
Integral Calculus
Integral calculus is a fundamental branch of calculus that deals with
- the concept of integration
- and its applications.
- applying it to determine areas, volumes, central points, and other useful quantities.
- evaluating multiple integrals in a specific order to find the volume under a surface
- or within a region.