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Chapter 7: Problem 10

Set up the integral for both orders of integration and use the more convenient order to evaluate the integral over the region \(R .\) \(\int_{R} \int x d A\) \(R:\) semicircle bounded by \(y=\sqrt{25-x^{2}}\) and \(y=0\)

Short Answer

Expert verified
The value of the integral over the semicircular region is 0.

Step by step solution

01

Formulate the integration in rectangular coordinates

Given the limits, set the integration in rectangular coordinates for the semicircular region from the positive x-axis to the represents the semicircular region from the positive x-axis to the negative x-axis: \(\int^{5}_{-5} \int^{\sqrt{25-x^{2}}}_{0} x dy dx\).
02

Convert the rectangular coordinates to polar coordinates

Converting to polar coordinates simplifies the integration. In polar coordinates, \(x= rcos(\theta)\), \(y= rsin(\theta)\), \(dx dy = rdr d\theta\), and the semi-circle is bounded by \(0 \le \theta \le \pi\) and \(0 \le r \le 5\). Replace \(x\) and \(dx dy\) with their polar equivalents: \(\int^{\pi}_{0} \int^{5}_{0} rcos(\theta) r dr d\theta\).
03

Evaluate the integral

Evaluate the integral: First handle the inner integral: \(\int r^{2}cos(\theta) dr = \frac{1}{3}r^{3} cos(\theta) \Big |^{5}_{0} = \frac{125}{3} cos(\theta)\). Second, handle the outer integral: \(\int cos(\theta) d\theta = sin(\theta) \Big|^{\pi}_{0} = 0 - 0 = 0\).
04

Conclusion

The final answer is zero. The line integral over the shape is zero because the positive and negative contributions symmetrically cancel each other out.

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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 an elegant way to represent points in a plane. Instead of using traditional x and y axes, polar coordinates use a radius and an angle. In this system, any point is described by the distance from the origin
  • \( r \) represents the radius, or the distance from the origin.
  • \( \theta \) is the angle measured from the positive x-axis in a counterclockwise direction.
This method can simplify integrals over circular regions since it aligns with the natural circular shape. Rather than working with complex square roots and trigonometric functions, the transformation to polar coordinates replaces these with more straightforward multiplicative terms.
This often reduces the complexity of calculating double integrals, especially in problems involving circular symmetry.
Rectangular Coordinates
In rectangular coordinates, also known as cartesian coordinates, every point is described by an x and y value. This makes plotting and visualizing many standard shapes straightforward. For the semicircular region discussed, the bounds in rectangular coordinates are defined by
  • \( y = \sqrt{25-x^2} \)
  • \( y = 0 \)
Here, the circle equation \( x^2 + y^2 = 25 \) translates to the semicircular region in the plane.
While rectangular coordinates suffice for many calculations, they can become cumbersome when dealing with rotations or circular shapes. That's why they often get converted to polar coordinates for such integrals, which naturally fit the symmetry of circular shapes.
Orders of Integration
The order of integration refers to the sequence of performing the integrals when evaluating a double integral. In mathematical terms, given a function \( f(x,y) \), you may choose to integrate with respect to y first and x second, or vice versa.
This choice can significantly impact the ease of the calculation.
  • The inner integral's bounds depend on the variable of integration, so carefully setting these bounds is crucial.
  • Switching the order can sometimes simplify the integral significantly or even make it possible to solve analytically, rather than numerically.
In our semicircular region example, converting to polar coordinates was the effective approach, simplifying the computation and making one of the orders much easier to evaluate.
Semicircular Region
A semicircular region is a half-disc, typically defined by slicing a full circle along a line, such as the y-axis or the x-axis. In this specific exercise, the semicircle lies above the x-axis, encapsulated by
  • \( y = \sqrt{25 - x^2} \) which describes the upper semi-circle of a circle with radius 5
  • \( y = 0 \).
With such a symmetric region, calculations can become straightforward when using polar coordinates.
By virtue of symmetry, when integrating over such regions, contributions from one half of the region might cancel out the contributions from the other half, as seen in our solution, leading to an integral result like zero.

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